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http://oeis.org/A001158 | 1,542,378,517,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743046.43/warc/CC-MAIN-20181116132301-20181116154301-00220.warc.gz | 233,060,138 | 6,352 | This site is supported by donations to The OEIS Foundation.
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A001158 sigma_3(n): sum of cubes of divisors of n. (Formerly M4605 N1964) 109
1, 9, 28, 73, 126, 252, 344, 585, 757, 1134, 1332, 2044, 2198, 3096, 3528, 4681, 4914, 6813, 6860, 9198, 9632, 11988, 12168, 16380, 15751, 19782, 20440, 25112, 24390, 31752, 29792, 37449, 37296, 44226, 43344, 55261, 50654, 61740, 61544, 73710, 68922, 86688 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS If the canonical factorization of n into prime powers is the product of p^e(p) then sigma_k(n) = Product_p ((p^((e(p)+1)*k))-1)/(p^k-1). Sum_{d|n} 1/d^k is equal to sigma_k(n)/n^k. So sequences A017665-A017712 also give the numerators and denominators of sigma_k(n)/n^k for k = 1..24. The power sums sigma_k(n) are in sequences A000203 (k=1), A001157-A001160 (k=2,3,4,5), A013954-A013972 for k = 6..24. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 05 2001 Also the eigenvalues of the Hecke operator T_n for the entire modular normalized Eisenstein form E_4(z) (see A004009): T_n E_4 = a(n) E_4, n >= 1. For the Hecke operator T_n and eigenforms see, e.g., the Koecher-Krieg reference, p. 207, eq. (5) and p. 211, section 4, or the Apostol reference p. 120, eq. (13) and pp. 129 - 133. - Wolfdieter Lang, Jan 28 2016 REFERENCES M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math.Series 55, Tenth Printing, 1972, p. 827. T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 38. T. M. Apostol, Modular Functions and Dirichlet Series in Number Theory, Second edition, Springer, 1990, pp. 120, 129 - 133. G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, AMS Chelsea Publishing, Providence, Rhode Island, 2002, p. 166. Max Koecher and Aloys Krieg, Elliptische Funktionen und Modulformen, 2. Auflage, Springer, 2007, pp. 207, 211. N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence). N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). Zagier, Don. "Elliptic modular forms and their applications." The 1-2-3 of modular forms. Springer Berlin Heidelberg, 2008. 1-103. See p. 17, G_4(z). LINKS T. D. Noe, Table of n, a(n) for n = 1..10000 M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy]. M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math.Series 55, Tenth Printing, 1972, p. 827. Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1. Eric Weisstein's World of Mathematics, Divisor Function. FORMULA Multiplicative with a(p^e) = (p^(3e+3)-1)/(p^3-1). - David W. Wilson, Aug 01 2001 Dirichlet g.f. zeta(s)*zeta(s-3). - R. J. Mathar, Mar 04 2011 G.f.: sum(k>=1, k^3*x^k/(1-x^k)). - Benoit Cloitre, Apr 21 2003 Equals A051731 * [1, 8, 27, 64, 125, ...] = A127093 * [1, 4, 9, 16, 25, ...]. - Gary W. Adamson, Nov 02 2007 L.g.f.: -log(Product_{j>=1} (1-x^j)^(j^2)) = (1/1)*z^1 + (9/2)*z^2 + (28/3)*z^3 + (73/4)*z^4 + ... + (a(n)/n)*z^n + ... - Joerg Arndt, Feb 04 2011 a(n) = Sum{d|n} tau_{-2}^d*J_3(n/d), where tau_{-2} is A007427 and J_3 is A059376. - Enrique Pérez Herrero, Jan 19 2013 a(n) = A004009(n)/240. - Artur Jasinski, Sep 06 2016. See, e.g., Hardy, p. 166, (10.5.6), with Q = E_4, and with present offset 0. - Wolfdieter Lang, Jan 31 2017 8*a(n) = sum of cubes of even divisors of 2*n. - Wolfdieter Lang, Jan 07 2017 EXAMPLE G.f. = x + 9*x^2 + 28*x^3 + 73*x^4 + 126*x^5 + 252*x^6 + 344*x^7 + ... MAPLE seq(numtheory:-sigma[3](n), n=1..100); # Robert Israel, Feb 05 2016 MATHEMATICA Table[DivisorSigma[3, n], {n, 100}] (* corrected by T. D. Noe, Mar 22 2009 *) PROG (PARI) N=99; q='q+O('q^N); Vec(sum(n=1, N, n^3*q^n/(1-q^n))) /* Joerg Arndt, Feb 04 2011 */ (Sage) [sigma(n, 3) for n in xrange(1, 40)] # Zerinvary Lajos, Jun 04 2009 (Maxima) makelist(divsum(n, 3), n, 1, 100); /* Emanuele Munarini, Mar 26 2011 */ (MAGMA) [DivisorSigma(3, n): n in [1..40]]; // Bruno Berselli, Apr 10 2013 (Haskell) a001158 n = product \$ zipWith (\p e -> (p^(3*e + 3) - 1) `div` (p^3 - 1)) (a027748_row n) (a124010_row n) -- Reinhard Zumkeller, Jun 30 2013 (PARI) {a(n) = if( n<1, 0, sumdiv(n, d, d^3))}; /* Michael Somos, Jan 07 2017 */ CROSSREFS Cf. A000005, A000203, A001157. Cf. A051731, A127093. Cf. A027748, A124010. Cf. A004009. Sequence in context: A226333 A017669 A277065 * A171215 A296601 A294567 Adjacent sequences: A001155 A001156 A001157 * A001159 A001160 A001161 KEYWORD nonn,easy,nice,mult AUTHOR STATUS approved
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Last modified November 16 09:21 EST 2018. Contains 317268 sequences. (Running on oeis4.) | 1,967 | 5,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2018-47 | latest | en | 0.53965 |
https://scoins.net/cie-maths/sixth-form-maths/sixth-form-pure/x-graph-sketching-sixth/harder-sketching.html | 1,723,727,050,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641291968.96/warc/CC-MAIN-20240815110654-20240815140654-00388.warc.gz | 396,084,580 | 29,734 | Harder Sketching | Scoins.net | DJS
## Harder Sketching
Basic questions to consider when looking at sketching a function:
Do I recognise this type / class of function?
Where is x zero?
Where is y zero?
Are there any (other) interesting places?
Where is the gradient zero? [called a turning point]
Does something interesting happen as x or y go to infinity? (called an asymptote)
Does the curvature change from ) positve to ( negative ? [called an inflexion]
Let’s look at those slowly:
An explicit function means y is some expression in x, such as y = x/(2-x), y = sin(x+2). An implicit function has x and y mixed up, such as xy = 6, sin(xy) - x = 1. Generally we work with explicit functions or ones we can easily rewrite as explicit.
The type or class of function: you might well recognise some of these
Straight lines ax + by = c, y = mx + c
Parabolas y = x² , y = ax² + bx + c Parabola means ‘like a ball’
Hyperbolas y = a/x, xy = c,
Circles x² + y² = r² and Ellipses ax² + by² = c²
Cubics y = x³ y = ax³ + bx² + cx + d
Where is x or y zero? With harder expressions this requires some algebra, as in y = (x-2)(x+3); sometimes it answers one of the later questions, as in y = 2 + 1/x or xy = 6
Are there any (other) interesting places? Some of the answers to this have probably been found by now, but, for example, x = 2 is interesting in any expression which includes (x-2) in it - because it gives a zero value.
Where is the gradient zero? Heading towards seriously hard maths; you can find a (usually slightly less difficult) formula for the gradient of a curve: the simple version says that every axn term in your base function becomes an anxn-1 . When this gradient function is zero, [the first differential]. you have a turning point.
Does something interesting happen as x or y go to infinity? Easier to spot and valid maths for Middle school, you’re looking for values of x that make y infinite, such as x=2 in y = 1/(2-x) or for trends such as y never getting over 6 in y = 6 - 1/(2-x) while the size of x is bigger than two.
Does the curvature change sign? Hard maths: the simplest version of this says that the tangent crosses the line; the slightly harder way (that finds the place of change) says that every ax term in your base function becomes of the form an(n-1)xⁿ⁻² . When this function is zero, [the second differential]. you have an inflexion. The inflexion just might also be a turning point; lots of the Sixth form are confused by this idea.
2013 additions, having returned to teaching middle and upper school.
Higher level GCSE students will recognise y=1/x and be able to sketch it in the air (Air Maths, like Air Guitar-playing); it is a curve that, in the positive quadrant, comes down the y-axis and creeps off along the x-axis, also symmetrical around x+y=0. It has asymptotes on both axes. The same students will not so easily recognise y-3 = 1/(x-2), which is the same curve, but with the asymptotes moved to centre on (2,3). The (usually good Y12) students who do ‘see’ this often show incomprehension (perhaps feigned) at the lack of vision of their colleagues. I suggest that there is a moment within the reading where some equivalent to ‘standing back from a painting to try to see the whole’ applies to the reading of the equation.
Also, a recent example from an adjacent paper in the same syllabus was something like y=4x+8 + 1/(x-2). While x is large, this curve is trying to be y=4x+8, a nice (easy) straight line; the reciprocal 1/(x-2) is irrelevant. When x gets close to 2, though, the 4x+8 becomes irrelevant (around 16) and the curve is trying to be the y=1/x curve, centred on the asymptote at x=2. A little checking to see what happens each side of x=2 (x<2 is negative, x>2 is positive) and the curve is ready to sketch. Yes; it is that easy – once you ‘see’ (or are persuaded) how the elements go together.
The similar curve y=4x-8 + 1/(x-2) presents, potentially, more confusion near x=2, but becomes clear if you make translations of two units on both axes. Plotting points as you might have done at GCSE is too slow and shows no understanding of the subject. At A-level, it is understanding that is expected to be evident. if you're having trouble here, you need to recognise the clear need to raise your game, to jump to a higher level of understanding.
Covid Email: David@Scoins.net © David Scoins 2021 | 1,181 | 4,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-33 | latest | en | 0.828976 |
https://www.techtud.com/doubt/subset-language-regular-expression | 1,607,220,202,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141753148.92/warc/CC-MAIN-20201206002041-20201206032041-00156.warc.gz | 867,126,390 | 17,999 | ##### Subset of a language in Regular Expression
(1) Subset of a language in RE is always RE.
(2) A language is not RE is uncountably infinite.
statements are true/false and why?
Ashish Kumar Goyal
30 Jan 2017 06:20 pm
1st->false and 2nd->true....
A regular language is not closed under subset operation.
and any finite language is regular. regular contains all finite languages and some infinite. so, a lang not RE is obviously infinite
Shobhit
30 Jan 2017 08:38 pm
^ how second one is true?
Ashish Kumar Goyal
30 Jan 2017 08:54 pm
@sudsho
any finite language is regular and a FA can be designed for it. So a lang not regular is definately infinite.
Shobhit
30 Jan 2017 09:16 pm
language not regular is definitly infinite...fine
(2) A language is not RE is uncountably infinite.
there is no difference between countable infinite and uncountable infinite?
Shobhit
30 Jan 2017 09:29 pm
(1) Subset of a language in RE is always RE.
false..RE are not closed under subset operation
(2) A language is not RE is uncountably infinite
false...we know that sigma star is countable...a language is nothing but a subset of sigma star..every subset of a countable set is countable..hence all languages are countable...though set of languages which are not RE are uncountable....mind my word here.."set" | 342 | 1,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.92274 |
https://cs.stackexchange.com/tags/curry-howard/hot | 1,716,174,724,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058147.77/warc/CC-MAIN-20240520015105-20240520045105-00526.warc.gz | 153,253,742 | 24,976 | # Tag Info
Accepted
### Is there an isomorphism between (subset of) category theory and relational algebra?
Let me articulate the Curry-Howard-Lambek correspondence with a bit of jargon which I'll explain. Lambek showed that the simply typed lambda calculus with products was the internal language of a ...
• 12.1k
### Testing whether an arbitrary proof is circular?
The vast majority of proof systems don't allow for infinite, circular proofs, but they do so by making their langauges non-Turing complete. In a normal functional language, the only way to make a ...
• 29.8k
Accepted
### Can a type system serve as a proof assistant for foreign functions?
Long story short: no you can't. A foreign function is like a black box and the type you ascribe to it is a promise you make: in the Curry-Howard correspondence that would correspond to adding an axiom ...
• 573
### Curry-Howard isomorphism and non-constructive logic
I think people sometimes disagree on what exactly Curry-Howard is. But, one way to look at it is an exact correspondence between the syntactic rules for logic and for type theory. For the ...
• 2,697
Accepted
### Curry-Howard, void, and type checking in Haskell
I would find a different tutorial because the author of that one is fundamentally confused. They wrongly claim that $\neg a$ and $\bot\to a$ are equivalent ($a\to\bot$ would be correct), and also ...
• 2,147
Accepted
### Implementing mathematical theory of arithmetic in Haskell via Curry-Howard correspondence
Proofs in Haskell? Okay, first let's talk about the Curry-Howard correspondence. This says that one can view theorems as types and proofs as programs. However, it says nothing about which specific ...
• 29.8k
Accepted
### Why algorithms calculating non-tirivial zeros can't be used as proofs of Riemann Hypothesis?
The fact of the matter is, if a proof exists, then a Curry-Howard version of the program exists too. That doesn't mean that it's easy to find, though. Undecidability still holds for Curry-Howard: if ...
• 29.8k
Accepted
### How logic programming (especially ASP) is related to the reasoning in (first-order) logic?
Logic programming is proof search for some logic. Traditionally, this is the Horn clause fragment of first-order logic. Languages like lambdaProlog extend this to (intuitionistic) hereditary Harrop ...
• 12.1k
Accepted
### forall a b, a -> b
Think of this in terms of the Curry-Howard isomorphism. What would this type look like as a theorem? For any propositions $A$ and $B$, $A \implies B$. Clearly this is not true, if it were, then we ...
• 29.8k
### forall a b, a -> b
I know for pretty sure that there is a function with the type $f: \forall \alpha, \beta . \alpha \rightarrow \beta$, but I can't wrap my head over it. No, that type is not inhabited. There are no ...
• 14.6k
Accepted
### Why Church-encoded types aren't sufficient to express inductive proofs?
How would you prove inside the pure CoC that the induction principle holds of the Church numerals? See Thomas Streicher's, Independence of the induction principle and the axiom of choice in the pure ...
• 8,328
Accepted
### Uses of the type Unit
Some other usages of the type Unit (I'm sure the list is not exhaustive): (1) The value of type Unit is used to simulate ...
• 3,479
Accepted
### Can lambda-calculus be used for knowledge representation?
The λ-calculus was invented to be a logic and foundation of mathematics (1-4). The most well-known logic to use λ-calculus for formulae (as opposed to proofs in the Curry-Howard ...
• 8,328
Accepted
### Is implication(function) more fundamental than lets say conjunction(product) in type theory?
For Agda, I think, as stated in the other question, that the fact that function types (or rather Π-types) are built-in while pairs (or even Σ-types) aren't so much is a reasonable argument that they ...
• 12.1k
### Lambda calculus as the language of universal logic - connectives vs functions in lambda calculus?
You are on your way to discovering the Curry-Howard correspondence.
• 30.5k
### Curry–Howard correspondence and functional programming "reliability"
The dependent types allow you to specify what properties your function should have, not just what its domain and codomain are. This way it becomes impossible to accidentally use the wrong function. ...
• 30.5k
### Does the underlying computational calculus in type theories affect decidability?
I don't think that the term "modern" helps distinguish anything. One way to explain is to draw a distinction between "behavioral", or "semantic", type theories and "formal", or "syntactic", type ...
### Does the underlying computational calculus in type theories affect decidability?
You are probably looking for computational type theory and you should probably look into realizability theory as well, which explains how to interpret type theory (and higher-order logic) on top of ...
• 30.5k
Accepted
### What are some examples of proofs that are also themselves "useful" programs?
Many constructive proofs could be useful. For example, to prove that every set of points in the plane (in general position) have a Delaunay triangulation, you would give the algorithm for producing ...
• 156
### Relation between Curry-Howard isomorphism and Kripke semantics for intuitionistic logic
Both Kripke semantics and propositions-as-types interpretation are sound and complete for the intuitionistic propositional calculus. In this sense they are equivalent. However, there are formulas in ...
• 30.5k
### Curry-Howard isomorphism and non-constructive logic
While proof assistants typically use constructive mathematics, the Curry-Howard-Correspondence does not necessarily require constructivism. The important aspect of the correspondence is that the ...
• 131
### Curry Howard correspondence to Predicate Logic?
To explain why I'm uncomfortable with Newsham's and (especially) Piponi's data wrappers ... (This is going to be more question than answer, but perhaps it'll work ...
• 487
### Curry-Howard isomorphism and non-constructive logic
The Curry-Howard correspondence is not crucial for functioning of a proof assistant. Unforunately, the term "Curry-Howard correspondence" seems to be misused nowadays for all sorts of things....
• 30.5k
### Why algorithms calculating non-tirivial zeros can't be used as proofs of Riemann Hypothesis?
The programs that you describe are very good at searching for zeroes in an interval; they can find all of the zeroes of the form $s+it$ between $t=0$ and $t=10^9$, say, and show that all these zeroes ...
• 2,519
### Uses of the type Unit
I can think of a couple: If a language makes a distinction between functions that return a value, and those that don't, it becomes difficult to stitch functions together. You have one set of ...
• 21
### Relation between Curry-Howard isomorphism and Kripke semantics for intuitionistic logic
For propositional logic, you're probably right. As far as quantifier logic goes: the Curry-Howard correspondence never included quantifiers and there isn't really any consensus or standard treatment ...
• 349
### On the logical and categorical interpretation of lambda calculi and type systems
(My answer is only about how non-termination relates to inconsistency.) The general idea of the Curry-Howard correspondence is that a term having a type, corresponds to a proof for a theorem. A term ...
### What is "the ability of classical control operators to return multiple times from a single term"?
The control operators in question allow you to capture continuations such that calling the continuation is in some sense equivalent to "returning from" the expression. so, for instance, in (...
• 2,697
1 vote
### What are the differences between LCF's Theorem and Automath's Prop?
The difference is in what kind of type theorem/Prop is. In Isabelle, theorem is a type in ...
1 vote
### Curry–Howard correspondence and functional programming "reliability"
The type system in FP languages helps you avoid a certain number of errors. The more you use various features of the type system, the more errors you can exclude. For example, if you need the list of ...
• 281
Only top scored, non community-wiki answers of a minimum length are eligible | 1,864 | 8,265 | {"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} | 2.625 | 3 | CC-MAIN-2024-22 | latest | en | 0.917404 |
http://kidscancode.org/godot_recipes/math/interpolation/ | 1,585,504,306,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370495413.19/warc/CC-MAIN-20200329171027-20200329201027-00064.warc.gz | 92,955,252 | 8,146 | # Interpolation
Interpolation, or its commonly-used abbreviation lerp, is a term that comes up often in game development. If you’ve never come across it before it can seem mysterious and highly-technical, but as you’ll see in this tutorial, it’s actually a straightforward concept with a wide variety of applications in game programming.
## Numeric Interpolation
The core formula for linear interpolation is this:
``````func lerp(a, b, t):
return (1 - t) * a + t * b``````
In this formula, `a` and `b` represent the two values and `t` is the amount of interpolation, typically expressed as a value between `0` (which returns `a`), and `1` (which returns `b`). The function finds a value the given amountFor example:
``````x = lerp(0, 1, 0.75) # x is 0.75
x = lerp(0, 100, 0.5) # x is 50
x = lerp(10, 75, 0.3) # x is 29.5
x = lerp(30, 2, 0.75) # x is 9``````
It’s called linear interpolation because the path between the two points is a straight line.
You can animate a node’s properties with `lerp()`. For example, if you divide the elapsed time by the desired duration, you’ll get a value between zero and one you can use to alter a property smoothly over time. This script scales a sprite up to five times its starting size while fading it out (using `modulate.a`) over two seconds:
``````extends Sprite
var time = 0
var duration = 2 # length of the effect
func _process(delta):
if time < duration:
time += delta
modulate.a = lerp(1, 0, time / duration)
scale = Vector2.ONE * lerp(1, 5, time / duration)``````
## Vector interpolation
You can also interpolate between vectors. Both `Vector2` and `Vector3` provide `linear_interpolate()` methods for this.
For example, to find a vector that’s halfway between a `Spatial` node’s forward and left direction vectors:
``````var forward = -transform.basis.z
var left = transform.basis.x
var forward_left = forward.linear_interpolate(left, 0.5)``````
The following example moves a Sprite node towards the mouse click position. Each frame the node moves 10% of the way to the target. This results in an “approach” effect, where the object’s speed becomes slower the closer it gets to the target.
``````extends Sprite
var target
func _input(event):
if event is InputEventMouseButton and event.pressed:
target = event.position
func _process(delta):
if target:
position = position.linear_interpolate(target, 0.1)``````
For more advanced applications of interpolation, see `Tween`. | 650 | 2,446 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-16 | longest | en | 0.828618 |
https://encyclopediaofmath.org/index.php?title=Statistical_mechanics,_mathematical_problems_in&oldid=18154 | 1,713,987,233,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819847.83/warc/CC-MAIN-20240424174709-20240424204709-00001.warc.gz | 207,133,927 | 13,504 | # Statistical mechanics, mathematical problems in
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The totality of problems of mathematical physics that arise from trying to interpret accurately the basic concepts and facts of statistical mechanics. These problems can be conditionally divided into the following groups:
1) the foundations of the basic principles of statistical mechanics;
2) equilibrium ensembles at the thermodynamic limit, inference of thermodynamic relations;
3) phase transitions;
4) the evolution of ensembles, the relaxation problem, studies on kinetic and hydrodynamic equations;
5) basic states, elementary excitations (in the case of a quantum system).
In statistical mechanics, systems consisting of a large number of (microscopic) particles contained in a large (relative to the dimensions of the particles) domain of the space are studied. Statistical mechanics, depending on the method used to describe the system, is divided into classical and quantum mechanics.
The description of the classical system contained in involves giving the space of possible states of each separate particle (single-particle space), as well as the totality of admissible configurations (, , ) of a finite number of particles within , and defining the energy for every configuration and the law of the evolution of the system in time (also called the dynamics), i.e. the semi-group (more often an entire group) of transformations , , of onto itself which preserve the energy :
for any and any . In many cases, naturally has a symplectic structure, and the transformations are produced by using solutions of the so-called Hamiltonian system of equations of motion, generated by the Hamilton function (see [1]). Moreover, a natural measure usually exists in the space such that the measure in ( is the product measure on ) is invariant relative to the evolutions . However, for macroscopic systems consisting of a large number of particles, such a detailed description of their states and of the dynamics of these states (i.e. a description of the trajectories of each individual particle) proves to be almost impossible, and of no use anyway from the point of view of studying the macroscopic properties of the whole system. These properties are determined only by certain mean characteristics of the configuration , and of its evolution , , in time: for example, the fractions , , of particles in whose states belong to a given set of a single-particle space , or the fractions of particles whose states at the moment of time belong to the set , and at the moment of time belong to the set , etc.
These concepts have given rise to the following radical idea: The state of a macroscopic system must be defined by a probability distribution on the phase space , whereby the evolution , , of this distribution in time is generated by the initial development of the system itself:
(1)
here is the complete inverse image of the set under the mapping . This stipulation is complemented by the following postulate: For every "good" probability distribution on the phase space and an appropriate physical quantity (i.e. a real-valued function on ), the values this quantity takes are, with probability close to one (calculated with respect to ), close to its mean value . One of the problems related to statistical mechanics is to put this assertion into an exact form. One of the possible results is of the following kind: Let the distribution on possess the property of rapidly-decreasing dependence (i.e. the probability distributions it generates of configurations for two subsystems, situated far apart, are almost independent), while the physical quantity is summable, i.e.
(2)
where is arbitrary, is a certain symmetric "local" function on (i.e. tends rapidly to zero as the points move further away from each other), and if . In this case , and the fluctuation (with probability close to one for large ), while the distribution of is close to normal (as before, when , see [2]).
A probability distribution on a phase space is said to be an equilibrium distribution if it is invariant relative to the dynamics . Apart from the energy , let there be several more so-called integrals of motion , i.e. functions on that are invariant relative to (for example, the number of particles in the system, the total momentum of the particles, the total spin, etc.). Every distribution on of the form
where is the invariant measure on and is a function (possibly generalized), is an equilibrium distribution. An equilibrium distribution defined by a density of the form
(3)
( is a normalizing factor), is called a microcanonical distribution (or microcanonical ensemble), concentrated on the surface
(4)
of constancy of first integrals.
In statistical mechanics it is postulated that the microcanonical distribution (3) is an equilibrium distribution (i.e. the mean values of the physical variables calculated by means of (3) coincide with great accuracy with the experimentally-measured values). For a long time it was believed that in order to prove this postulate, an unknown ergodic hypothesis had to be demonstrated: If is a complete set of (smooth) integrals of motion, the microcanonical distribution is the unique (smooth) equilibrium distribution of any surface . Attempts to prove this hypothesis gave rise to modern ergodic theory (see [3], [4]). Now, however, it has become clear that the ergodicity of finite systems is an unnecessarily-strict assumption: In order to prove the postulate on microcanonical distributions it is sufficient to establish the ergodicity of the system at the thermodynamic limit . Apart from the microcanonical distribution, the Gibbs equilibrium distribution (sometimes called a grand canonical ensemble) is often studied. It is defined by the density
(5)
where is a normalizing factor, and are arbitrary real parameters (the parameter , where is the absolute temperature and is an absolute constant, the so-called Boltzmann constant). Intermediate distributions are also sometimes studied (small canonical ensembles); these have a density of the form
(6)
where and are two complementary subsets of indices in . The Gibbs distribution (5), as well as a distribution (6), is in many ways more convenient than the microcanonical distribution (3), and the use of both is justified by the following hypothesis — the so-called equivalence principle of ensembles: For "suitable" physical quantities on (for example, for a summable quantity of the form (2)), given values of the parameters for which only one equilibrium phase exists, the mean , calculated using the Gibbs distribution (5) for large , is close to the mean , calculated using a microcanonical ensemble on the surface , where . The proof of this equivalence also constitutes one of the general mathematical problems of statistical mechanics and thermodynamics (see [5], [6], [7]).
The accepted method of describing systems in statistical mechanics is justified when the domain has sufficiently-large volume, in other words, statistical mechanics studies the asymptotic properties of systems in the limit (i.e. a sequence of systems of one and the same particles, contained respectively in the volumes , where ). This passage to the limit is usually called a thermodynamic passage to the limit. One of the primary problems related to the thermodynamic limit is, beginning with equilibrium ensembles, to define the so-called thermodynamic potentials and relations. It proves possible to find all thermodynamic potentials, when , from the asymptotics of the normalizing factors , , , etc., in the ensembles (3), (5), (6); for example, the Gibbs thermodynamic potential is equal to
(7)
where is the normalizing factor in the Gibbs ensemble (5). Other thermodynamic functions are introduced in the same way, and relations linking them are established. The majority of mathematical problems which arise here (existence of a limit, properties of thermodynamic potentials, etc.) have been studied fairly comprehensively, although a number of unsolved problems do remain (see, for example, [7]).
Since the end of the 1960's, the following general approach has found favour in mathematical statistical mechanics: instead of studying the asymptotic properties of finite systems in a thermodynamic passage to the limit, one should consider idealized infinite systems constructed in a particular way, whose characteristics coincide with the asymptotics being studied (this point of view was encountered in earlier works in less obvious form). Examining infinite systems gives a visible meaning to the somewhat formal procedure of a thermodynamic passage to the limit, and makes it possible to do without it altogether. The phase space of an infinite system consists of infinite configurations of particles , , located throughout , while their dynamics , , is constructed as the limit of the dynamics of finite systems as . The macroscopic states of an infinite system are defined by probability distributions on the space , which develop in line with the dynamics in (see [1]). Limit Gibbs distributions are introduced on . They are constructed in a particular way by means of the Gibbs distributions (5) on finite systems (see [5], [9]). Although the introduction of infinite systems is a generally-accepted and fruitful method, it leads to complex intrinsic mathematical problems, which remain for the most part unsolved. These problems include, for example, the construction of a dynamics , the construction of limit Gibbs distributions, studies on their properties, etc.
One of the main problems of statistical mechanics is the study of so-called phase transition, i.e. a sharp change of properties of a macroscopic system in a state of equilibrium, given a small change of the parameters describing this equilibrium — temperature, density of the particles, pressure, etc. In the modern mathematical approach, the problem of phase transition can be described in terms of limit Gibbs distributions in the following way: Given certain values of the parameters , it is possible, in general, to create a number of Gibbs distributions on that are invariant relative to the action of the group of shifts in (or of a subgroup such that the quotient group is compact), and are ergodic relative to this group (so-called pure phases). A point of the parameter space is said to be regular if it has a sufficiently small neighbourhood in which the structure of the set of pure phases, and their basic qualitative properties (for example, the character of decrease of the correlations), remains unaltered. It is assumed that all numerical characteristics of these distributions (correlation functions, semi-invariants, etc.) in a neighbourhood of regular points depend analytically on the parameters . All other (not regular) points in the parameter space are also points of phase transition. Thus, either a sharp change in the structure of the Gibbs distributions (say a new phase disappears or arises) or in their properties (for example, decrease of the correlations changes from exponential to gradual) takes place at such points of the phase space. Each of the characteristics of the distribution, as a function of , is considered to have a singularity at a point of phase transition. Describing the structure of phases and their properties for every concrete system, defining points of phase transition and the character of the singularities at these points, as well as other tasks, also constitute the problem of phase transition. Although a large class of model systems exists for which (at low temperatures) a number of general methods of solving this problem have been worked out (see [9]), the theory of phase transition is still far from being perfect. The study of the so-called critical points is particularly complex (roughly speaking, the fusion of various phases takes place at these points; see [10]), in that, at these points, the decrease of the correlations of the Gibbs distribution is very slow.
Many problems of statistical mechanics are related to the study of the time-development of distributions on a phase space, and particularly to the problem of relaxation, i.e. of approach to equilibrium. It is thought that, over a long period of time, every distribution on a phase space is approximated by an equilibrium (Gibbs) distribution. Despite the fact that many general representations of the mechanism of this process have been worked out, and that a number of simplified models have been studied, no complete theory exists yet (1991). The basic representations of the relaxation process (for the most part, they are still hypothetical) reduce to the fact that this process has three stages. At the first stage (while collisions of several particles are taking place), the distribution leads to a development of the system which is completely defined by a change of the first correlation function (i.e. by a distribution in a single-particle space ). At the second stage — the kinetic stage, which lasts throughout the "free path" of the particle — the change of the first correlation function crosses over to a development in which everything depends only on the mean values of the particles, their speed, density, energy, etc. Lastly, there is the hydrodynamic stage, during which (as with macroscopic time) these mean values of density, speed, etc., are approximated by equilibrium values (see [11], [12]). Proving this scenario as a whole or in its separate parts is a complex mathematical problem, which is far from being completely solved (1991). Various systems of kinetic equations constitute a fundamental method of research. They are both exact, i.e. resulting directly from the definition of the Liouville equation (the hierarchical BBGKY chain — Bogolyubov, Born, Green, Kirkwood, Yvon) and approximate (the equations of Boltzmann, Chapman–Enskog, Vlasov–Landau, the hydrodynamic equations, etc.). These equations and their relation with the true picture of the evolution are also the subject of intensive mathematical study (cf. Bogolyubov chain of equations; Boltzmann equation; Chapman–Enskog method; Vlasov kinetic equation).
Quantum statistical mechanics is based on the same principles as classical mechanics. The quantum description of a system of particles located within a domain requires that a Hilbert space be given (the state space of the system), and that a self-adjoint operator , operating on , be given (the energy operator of the system). The dynamics of the system is defined by the group , , of unitary operators operating on , whereby the dynamics generates a group of automorphisms of the algebra of bounded operators (observables) on :
The change to the statistical description in the quantum case plays a role in defining a certain "mean" on the algebra , i.e. of a positive linear functional on this algebra, usually called a state. Every state on can be written in the form
where is a positive nuclear operator from and . The operator is usually called the density matrix of the state . The evolution in time of the state is defined by the evolution of the algebra itself: . States that are invariant relative to this evolution are called equilibrium states, as before. For a system in which, apart from the energy , there are several pairwise-commuting integrals of motion , the equilibrium state with density matrix
is called a Gibbs state ( and are parameters, is a normalizing factor). An infinite quantum system is introduced in the same way as for a classical system by passage to the thermodynamic limit (see [5]). In order to describe this system, the -algebra is examined (the bar signifies closure in the uniform topology), which is called the algebra of quasi-local observables, while the evolution in is defined as the limit of the evolutions on the finite algebras . It is possible to introduce limit Gibbs states on the algebra in the same way as for classical systems (see [5]). The problem of phase transitions in quantum systems is formulated in the same way as for classical systems in terms of limit Gibbs states.
Finally, in quantum statistical mechanics there also exists a whole circle of kinetic problems; however, the mechanism of relaxation processes in quantum mechanics is more complicated than in classical mechanics, and is less studied.
There exist specific results on the so-called groundstate of a system in the quantum case (corresponding to zero temperature) and on perturbations of this state of finite energy. A number of interesting problems is related to this problem (superconductivity, superfluidity), emerging at low temperatures [13]. Problems on the construction and study of quantum fields may be investigated by means of the theory of Gibbs fields, as developed in statistical mechanics (see [14], [15]).
#### References
[1] V.I. Arnol'd, "Mathematical methods of classical mechanics" , Springer (1978) (Translated from Russian) [2] B.S. Nakhapetyan, "The central limit theorem for random fields with mixing property" R.L. Dobrushin (ed.) Ya.G. Sinai (ed.) , Multi-component random systems , M. Dekker (1980) pp. 531–547 (Translated from Russian) [3] I.P. [I.P. Kornfel'd] Cornfel'd, S.V. Fomin, Ya.G. Sinai, "Ergodic theory" , Springer (1982) (Translated from Russian) [4] N.S. Krylov, "Works on the foundation of statistical physics" , Princeton Univ. Press (1979) (Translated from Russian) [5] D. Ruelle, "Statistical mechanics: rigorous results" , Benjamin (1966) [6] A.M. [A.M. Khalfina] Halfina, "The limiting equivalence of the canonical and grand canonical ensembles (low density case)" Math. USSR Sb. , 9 : 1 (1969) pp. 1–52 Mat. Sb. , 80 (1969) pp. 3–51 [7] R.A. Minlos, A. Khaitov, "Limiting equivalence of thermodynamic ensembles in the case of one-dimensional systems" Trans. Moscow Math. Soc. , 32 (1975) pp. 143–180 Tr. Moskov. Mat. Obshch. , 32 (1975) pp. 147–186 [8] O.E. Lanford, , Gibbs states in statistical physics , Moscow (1978) pp. 159–218 (In Russian; translated from English) [9] Ya.G. Sinai, "Theory of phase transitions: rigorous results" , Pergamon (1982) (Translated from Russian) [10] H.E. Stanley, "Phase transitions and critical phenomena" , Clarendon Press (1971) [11] R. Libov, "Introduction to the theory of kinetic equations" , Wiley (1969) [12] R. Balescu, "Equilibrium and non-equilibrium statistical mechanics" , 1–2 , Wiley (1975) [13] L.D. Landau, E.M. Lifshitz, "Statistical physics" , Pergamon (1980) (Translated from Russian) [14] B. Simon, "The Euclidean (quantum) field theory" , Princeton Univ. Press (1974) [15] , Euclidean quantum field theory. The Markov approach , Moscow (1978) (In Russian; translated from English) [16] J. Glimm, A. Jaffe, "Quantum physics, a functional integral point of view" , Springer (1981) [17] V.A. Malyshev, R.A. Minlos, "Gibbs random fields" , Kluwer (1991) (Translated from Russian) | 3,952 | 18,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.924953 |
https://web2.0calc.com/questions/how-do-you-find-the-area-of-a-square-and-triangle | 1,561,409,817,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999709.4/warc/CC-MAIN-20190624191239-20190624213239-00486.warc.gz | 639,880,050 | 5,937 | +0
# how do you find the area of a square and triangle?
+1
312
2
+67
how do you find the area of a square?
How do you find the area of a triangle?
Jun 20, 2018
#1
+18329
+2
Area of square (or rectangle) = Length x Width units^2 (for a SQUARE L & W are equal)
Area of a triangle = 1/2 Base x height units^2
Jun 20, 2018
#2
+67
+1
Thank you.
AshleyGordo Jun 20, 2018 | 142 | 384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-26 | latest | en | 0.829942 |
http://datascience.sharerecipe.net/2019/02/19/cleaner-r-code-with-functional-programming/ | 1,579,779,338,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250610004.56/warc/CC-MAIN-20200123101110-20200123130110-00332.warc.gz | 45,617,241 | 8,788 | # Cleaner R Code with Functional Programming
What’s a data analyst to do?The solution is to use applies (also called maps).
Maps take a collection of things, and apply some function to each one of those things.
Here’s a diagram taken directly from RStudio’s purrr cheat sheet (Credit: Mara Averick):Sidenote: The dplyr package actually gets its name from applies.
dplyr = data + apply + R.
The purrr package contains a ridiculous number of maps from which to choose.
Seriously, check out that cheatsheet!Example, bringing it all together: Suppose I had a vector of strings, and I wanted to extract the longest word in each.
There is no vectorized function that will do this for me.
I will need to split the string by the space character and get the longest word.
For dramatic effect, I also upper-case the strings and paste them back together:library(tidyverse)library(purrr) sentences <- c( "My head is not functional", "Programming is hard", "Too many rules") getLongestWord <- function(words) { word_counts <- str_length(words) longest_word <- words[which.
max(word_counts)] return(longest_word)} sentences %>% toupper() %>% str_split(' ') %>% map_chr(getLongestWord) %>% str_c(collapse = ' ') # [1] "FUNCTIONAL PROGRAMMING RULES"Bonus Step: Know the LingoIn other languages, some of the lingo of FP is built-in.
Specifically, there are three higher-order functions that make their way into almost every language, functional or not: map (which we’ve already covered), reduce, and filter.
A higher-order function is a function that either takes a function as an argument, returns a function, or both.
Filtering in R is easy.
For data frames, we can use use tidyverse::filter.
For most other things, we can simply use R’s vectorization.
However, when all else fails, base R does indeed have a Filter() function.
Example:Filter(function(x) x %% 2 == 0, 1:10)# [1] 2 4 6 8 10Similarly, you probably won’t ever need Reduce() in R.
But just in case, here’s how it works: Reduce() will take a collection and a binary function (ie, takes two parameters), and successively applies that function two-at-a-time along that collection, cumulatively.
Example:wrap <- function(a, b) paste0("(", a, " ", b, ")")Reduce(wrap, c("A", "B", "C", "D", "E"))# [1] "((((A B) C) D) E)"Another beloved FP topic is that of currying.
Currying is the act of taking a function with many arguments, and breaking it out into functions that take a partial amount of those arguments.
These are sometimes called partial functions.
The following example uses a function factory to make partial functions:# Adder is a "function factory" – a function that makes new functions.
adder <- function(a) { return(function(b) a + b)} # Function factory pumping out new functions.
To make this slightly more readable, the functional library gives you an explicitly currying builder:library(functional)add <- function(a, b) a + badd3 <- Curry(add, a = 3)add5 <- Curry(add, a = 5) add3(add5(1))# 9Sidenote: The verb “currying” comes from Haskell Curry, famed Mathematician/Computer Scientist/Fellow Penn Stater.
SummaryDo you feel smarter?.More powerful?.Ready to torture your data with some of your new FP skills?.Here are some of the big takeaways:No more loops!.Ever!Anytime you want to use a loop, find the appropriate apply/map.
Integrate the Tidyverse into your workflow wherever possible.
Use the pipe (%>%) when applying several functions to one thing (eg, a data frame being manipulated in the Tidyverse).
Adhering to these mindsets while coding can greatly reduce ugly, difficult-to-maintain spaghetti code.
Bottling things in functions can leave you with clean, readable, modular ravioli code.
I’ll leave you with a famous quote from John Woods:Always code as if the [person] who ends up maintaining your code will be a violent psychopath who knows where you live.
. | 930 | 3,863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-05 | latest | en | 0.869711 |
https://philoid.com/question/78115-evaluate-each-of-the-following | 1,686,306,767,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656675.90/warc/CC-MAIN-20230609100535-20230609130535-00060.warc.gz | 474,346,954 | 10,988 | ## Book: RD Sharma - Mathematics (Volume 1)
### Chapter: 4. Inverse Trigonometric Functions
#### Subject: Maths - Class 12th
##### Q. No. 3 of Exercise 4.6
Listen NCERT Audio Books - Kitabein Ab Bolengi
3
##### Evaluate each of the following:
Now first of the principal value of
cosec–1
Let cosec–1 = y
cosec y =
= cosec
The range of principal value of cosec–1 is –{0}
and cosec
Therefore, the principal value of cosec–1 is …(1)
Now, the value of cot–1(–1)
Let cot–1(–1) = y
cot y = –1
= – cot = 1
= cot
= cot
The range of principal value of cot–1is (0, π)
and cot = –1
Therefore, the principal value of cot–1(–1) is …(2)
From (1) and (2) we can write the given equation as
=
=
=
1
1
1
1
2
3
3
3
3 | 255 | 724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2023-23 | longest | en | 0.830619 |
https://en.khanacademy.org/math/cc-fifth-grade-math/5th-multiply-fractions/area-of-rectangles-with-fraction-side-lengths/v/intuition-for-area-with-fractional-side-lengths | 1,708,945,368,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474659.73/warc/CC-MAIN-20240226094435-20240226124435-00250.warc.gz | 236,298,526 | 103,244 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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Lesson 5: Area of rectangles with fraction side lengths
# Finding area with fractional sides 2
Learn how to calculate the area of rectangles with fractional side lengths. Watch examples of this concept in action, and then see practice problems applying what was shown to solve similar problems.
## Want to join the conversation?
• What about if the problem had a mixed number in it and no other fraction? How would you solve it, then?
• its simple, all you have to do is multiply it normally
• I'm sorry that this is not a question, but the replay button does not work, and the video keeps on turning black when I attempted to watch the video.
• how does the area be similar to the fractions?
• What does it mean to square a meter.....?
that confuses me.
• When you "square" something by adding a ² at the end, it means you have multiplied the variable attached by itself.
So, say 2 is J. Then J² would be 2 x 2, or 4.
Hope this helps,
BIOHAZARD, The One and Only Radioactive Coder
• I don't understand why the denominator changes when you multiply two fractions and when you add two fractions, the denominator stays the same. For example, 4/5*410=16/50, but when you add 4/5+4/10, it equals 8/5 or 1, 3/5 . How is it different and can you also change the denominator in division and not subtraction?
(1 vote)
• 1m 1m 1m for every side sounds weird?
(1 vote)
• So it's basically like all the sides are 1m 1m 1m 1m?
(1 vote) | 418 | 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} | 3.9375 | 4 | CC-MAIN-2024-10 | latest | en | 0.932941 |
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# 9: Sets
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• 9.1: Basics
Object: any distinct entity
• 9.2: Defining sets
Remember that mathematical notation is about communicating mathematical information. Since a set is defined by its member objects, to communicate the details of a set of objects one needs to provide a means to decide whether any given object is or is not an element of the set.
• 9.3: Subsets and equality of sets
Often we want to distinguish a collection of certain “special” elements within a larger set of elements.
• 9.4: Complement, union, and intersection
First, it is often convenient to restrict the scope of the discussion.
• 9.5: Cartesian Product
the set of all possible ordered pairs of elements from two given sets A and B, where the first element in a pair is from A and the second is from B
• 9.6: Alphabets and words
any set can be considered an alphabet
• 9.7: Sets of sets
Sets can be made up of any kind of objects, even other sets! (But now we must be careful of the use of the phrase “contained in”.)
• 9.8: Activities
• 9.9: Exercises
This page titled 9: Sets is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Jeremy Sylvestre via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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Frequent Visitor
## Distance in a map to a non-fixed point.
Hi everyone,
I hope you could help me.
I have a dataset with a list of clients with their latitude and longitude.
I have created a map where I can show customers who are at a distance (in kilometers) from a specific point.
What I would like is, to be able to choose a client in a slicer. And then be able to show the clients who are at a chosen distance from him.
The formula for calculating the distance is as follows:
``````KLM: =
ACOS (
COS ( RADIANS ( 90 - 'clients_test'[lat_origine] ) )
* COS ( RADIANS ( 90 - 'clients_test'[latitude] ) )
+ SIN ( RADIANS ( 90 - 'clients_test'[lat_origine] ) )
* SIN ( RADIANS ( 90 - 'clients_test'[latitude] ) )
* COS ( RADIANS ( 'clients_test'[long_origine] - 'clients_test'[longitude] ))
) * 6371``````
Currently, I put the latitude and longitude in a column. What I want is that those two data come from a variable or a measure.
I leave you my power bi file HERE
Thank you very much in advance.
AlbLS
2 REPLIES 2
Community Support Team
## Re: Distance in a map to a non-fixed point.
You may try SELECTEDVALUE to add a measure.
https://community.powerbi.com/t5/Desktop/Mapping-question/m-p/365056#M165061
Community Support Team _ Sam Zha
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Frequent Visitor
## Re: Distance in a map to a non-fixed point.
Hello @v-chuncz-msft ,
Could you explain a little more in detail what you mean, please?
I need to select a client in a slicer. Then, choose a distance to that client and that only the clients that are at the chosen distance of the first client are shown on the map.
How do I have to use the SELECTVALUE formula?
Thank you
AlbLS
Announcements | 476 | 1,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-05 | latest | en | 0.894343 |
http://code.mathforge.org/MonteCarlo/ | 1,516,572,026,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890893.58/warc/CC-MAIN-20180121214857-20180121234857-00715.warc.gz | 68,626,878 | 3,576 | hands of cards from a deck of cards, (with bias ) -sided dice, coins,
updating every fliprolldeals,
until recording ,
repeat this experiment until recording .
### Help
This website allows you to run probability experiments. You can flip coins, roll dice, and deal cards. You can specify various parameters, explained below.
The conditions for stopping an experiment or run of experiments and the counts that can be recorded use a particular syntax. The basic element is `count(...)` which counts the number of occurences of an event within an experiment. For keeping track between experiments, there is `total(...)`. These can also be used in the "recording" fields. Logical and mathematical expressions using these can be used. Examples are given in the explanations below.
• Type of experiment: You can choose whether to flip coins, roll dice, or deal cards. Some of the other options depend on which you chose.
#### Coins
• Number of coins flipped in each individual experiment.
• (Un)Biased: It is possible to specify a bias for the coins. The bias is the probability of flipping a head.
• (Un)Ordered: When more than one coin is flipped, the results can be recorded either ordered or unordered. If unordered, the results are sorted so that heads are listed first.
#### Dice
• Number of dice rolled in each individual experiment.
• (Un)Biased: It is possible to specify a bias for the dice. The biases are given by listing the probabilities of getting each number in order separated by commas. Empty entries are allowed: the "left over" probability is equally distributed between any empty entries. Thus a bias of `,,,,,.5` for a 6-sided die assigns a probability of .5 to rolling a 6 with all other numbers equally likely.
• (Un)Ordered: When more than one dice is rolled, the results can be recorded either ordered or unordered. If unordered, the results are sorted so that smaller numbers are listed first.
• Sides: Various standard sized dice are allowed. Note that when, for example, a 12-sided die is used then if needed the values are reported with a leading zero to make them all the same length. So rolling an 11 then a 2 is distinguishable from rolling a 1 then a 12 (1102 versus 0112).
#### Cards
• Number of cards. With cards, the situation is that there is a pack of cards of a given size which is shuffled and then a number of hands of a fixed size are dealt from that pack (without replacement). So the details to specify are the number of hands, the size of each hand, and the size of the deck. At present, the largest deck size is 52. Hands are considered to be unordered, so are reported in the natural order with ace low. Cards are displayed as, for example, `4♥`, but when matching use `C`, `D`, `H`, `S` for the suits.
#### Other Details
• Updating every ...: The website lists all of the experiments, but with a long run it can be slow to update every flip/roll/deal.
• Until: This gives a condition for when a given run of experiments should be stopped. The examples are all in terms of coins.
• `count() = 10` Stop when the number of runs is 10.
• `count(h) = 10` Stop when the number of heads is 10.
• `count(h) = 10 or count(t) = 10` Stop when the number of heads is 10 or the number of tails is 10.
• `count(hht) = 3` Stop when the pattern "HHT" has been seen three times.
• `count(hh,tt) = 1` When flipping two coins, stop when two heads are followed by two tails.
• `count(hh) = 1 or count(th) = 1` Stop when either HH or TH is seen.
• Recording: This records various counts of things that happen during an experiment. Separate different counts with commas.
• `count()` Count the number of flips.
• `count(h), count(t)` Count the heads and the tails.
• `count(hh)/count(h)` Record the ratio of "HH" to "H".
• Repeat until: The basic experiment can be repeated, this gives a condition on when to stop.
• `total(count() > 10) = 3` Stop when there have been three experiments that have had at least 10 flips.
• `total(count(h) > count(t)) = 1` Stop when an experiment has more heads than tails.
• `total(1) = 10` Stop after ten experiments.
• Recording: This records various counts through the experiments.
• `total(count())` Record the total number of flips overall.
• `total(count(h))` Record the total number of heads overall.
• `total(count(hh) > count(th))` Record the total number of experiments where the number of HHs is bigger than the number of THs.
Some notes on how the counts and totals are implemented:
• The counts look for an exact match with the pattern at the end of the current list. (On the TODO is to make this more flexible.)
• The patterns are case insensitive.
• If the pattern is `HH` and the flips recorded are `HHH` then the pattern will match twice.
• In the totals, it is possible to put expressions that evaluate to true or false (such as ```count(h) > 3```). These are counted as 1 for true and 0 for false. So `total(count(h) > 3)` counts how many times the count of heads is bigger than 3.
### TODO
• Make the pattern matching more flexible
The code is released under the MIT Licence so if you want to improve it, feel free. It is available on github. | 1,225 | 5,131 | {"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} | 3.46875 | 3 | CC-MAIN-2018-05 | latest | en | 0.897702 |
http://www.numbersaplenty.com/555555555555555 | 1,556,058,406,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578613888.70/warc/CC-MAIN-20190423214818-20190424000818-00386.warc.gz | 278,592,122 | 4,619 | Search a number
555555555555555 = 353137412712906161
BaseRepresentation
bin111111001010001100101101…
…1100010101011100011100011
32200212001120222212122101100210
41332110121123202223203203
51040304210234210234210
65245322410543145203
7225006364365561342
oct17624313342534343
92625046885571323
10555555555555555
11151020873897a95
125238647b5a5203
131aacb883031389
149b29064b19b59
15443641cd4ce20
hex1f9465b8ab8e3
555555555555555 has 128 divisors (see below), whose sum is σ = 968908644974592. Its totient is φ = 271179601920000.
The previous prime is 555555555555529. The next prime is 555555555555557.
555555555555555 is nontrivially palindromic in base 10.
It is not a de Polignac number, because 555555555555555 - 29 = 555555555555043 is a prime.
It is a nude number because it is divisible by every one of its digits.
555555555555555 is a modest number, since divided by 55555555 gives 5555555 as remainder.
555555555555555 is a nontrivial repdigit in base 10.
It is a plaindrome in base 10.
It is a nialpdrome in base 10.
It is a zygodrome in base 10.
It is not an unprimeable number, because it can be changed into a prime (555555555555557) by changing a digit.
It is an upside-down number.
It is a polite number, since it can be written in 127 ways as a sum of consecutive naturals, for example, 189711675 + ... + 192617835.
It is an arithmetic number, because the mean of its divisors is an integer number (7569598788864).
Almost surely, 2555555555555555 is an apocalyptic number.
555555555555555 is a deficient number, since it is larger than the sum of its proper divisors (413353089419037).
555555555555555 is a wasteful number, since it uses less digits than its factorization.
555555555555555 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 2906549.
The product of its digits is 30517578125, while the sum is 75.
The spelling of 555555555555555 in words is "five hundred fifty-five trillion, five hundred fifty-five billion, five hundred fifty-five million, five hundred fifty-five thousand, five hundred fifty-five". | 627 | 2,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-18 | latest | en | 0.744318 |
https://studylib.net/doc/25626656/conic-sect-full | 1,642,513,101,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300849.28/warc/CC-MAIN-20220118122602-20220118152602-00708.warc.gz | 654,679,022 | 19,714 | conic sect full
```Chapter
11
CONIC SECTIONS
11.1 Overview
11.1.1 Sections of a cone Let l be a fixed vertical line and m be another line intersecting
it at a fixed point V and inclined to it at an angle α (Fig. 11.1).
Fig. 11.1
Suppose we rotate the line m around the line l in such a way that the angle α remains
constant. Then the surface generated is a double-napped right circular hollow cone
herein after referred as cone and extending indefinitely in both directions (Fig. 11.2).
Fig. 11.2
Fig. 11.3
18/04/18
CONIC SECTIONS
187
The point V is called the vertex; the line l is the axis of the cone. The rotating line m is
called a generator of the cone. The vertex separates the cone into two parts called
nappes.
If we take the intersection of a plane with a cone, the section so obtained is called a
conic section. Thus, conic sections are the curves obtained by intersecting a right
circular cone by a plane.
We obtain different kinds of conic sections depending on the position of the intersecting
plane with respect to the cone and the angle made by it with the vertical axis of the
cone. Let β be the angle made by the intersecting plane with the vertical axis of the
cone (Fig.11.3).
The intersection of the plane with the cone can take place either at the vertex of the
cone or at any other part of the nappe either below or above the vertex.
When the plane cuts the nappe (other than the vertex) of the cone, we have the
following situations:
(a) When β = 90o, the section is a circle.
(b) When α < β < 90o, the section is an ellipse.
(c) When β = α; the section is a parabola.
(In each of the above three situations, the plane cuts entirely across one nappe
of the cone).
(d) When 0 ≤ β < α; the plane cuts through both the nappes and the curves of
intersection is a hyperbola.
Indeed these curves are important tools for present day exploration of outer space and
also for research into the behaviour of atomic particles.
We take conic sections as plane curves. For this purpose, it is convenient to use equivalent
definition that refer only to the plane in which the curve lies, and refer to special points
and lines in this plane called foci and directrices. According to this approach, parabola,
ellipse and hyperbola are defined in terms of a fixed point (called focus) and fixed line
(called directrix) in the plane.
If S is the focus and l is the directrix, then the set of all points in the plane whose
distance from S bears a constant ratio e called eccentricity to their distance from l is a
conic section.
As special case of ellipse, we obtain circle for which e = 0 and hence we study it
differently.
11.1.2 Circle A circle is the set of all points in a plane which are at a fixed distance
from a fixed point in the plane. The fixed point is called the centre of the circle and the
distance from centre to any point on the circle is called the radius of the circle.
18/04/18
188
EXEMPLAR PROBLEMS – MATHEMATICS
The equation of a circle with radius r having
centre (h, k) is given by (x – h)2 + (y – k)2 = r2
The general equation of the circle is given by
x2 + y2 + 2gx + 2fy + c = 0, where g, f and c are
constants.
(a) The centre of this circle is (– g, – f )
(b) The radius of the circle is
g2 + f 2 − c
The general equation of the circle passing through
Fig. 11.4
the origin is given by x2 + y2 + 2gx + 2fy = 0.
2
2
General equation of second degree i.e., ax + 2hxy + by + 2gx + 2fy + c = 0 represent
a circle if (i) the coefficient of x2 equals the coefficient of y2, i.e., a = b ≠ 0 and (ii) the
coefficient of xy is zero, i.e., h = 0.
The parametric equations of the circle x2 + y2 = r2 are given by x = r cosθ, y = r sinθ
where θ is the parameter and the parametric equations of the circle (x – h)2 + (y – k)2 = r2
are given by
x – h = r cosθ, y – k = r sinθ
or
x = h + r cosθ, y = k + r sinθ.
Fig. 11.5
Note: The general equation of the circle involves three
constants which implies that at least three conditions are
required to determine a circle uniquely.
11.1.3 Parabola
A parabola is the set of points P whose distances from a
fixed point F in the plane are equal to their distances from
a fixed line l in the plane. The fixed point F is called focus
and the fixed line l the directrix of the parabola.
Fig. 11.6
18/04/18
CONIC SECTIONS
189
Standard equations of parabola
The four possible forms of parabola are shown below in Fig. 11.7 (a) to (d)
The latus rectum of a parabola is a line segment perpendicular to the axis of the
parabola, through the focus and whose end points lie on the parabola (Fig. 11.7).
Fig. 11.7
Forms of Parabolas
Axis
Directix
Vertex
Focus
Length of latus
rectum
Equations of latus
rectum
y2 = 4ax
y=0
x=–a
(0, 0)
(a, 0)
4a
y2 = – 4ax
y=0
x=a
(0, 0)
(– a, 0)
4a
x2 = 4ay
x=0
y=–a
(0, 0)
(0, a)
4a
x2 = – 4ay
x=0
y=a
(0, 0)
(0, – a)
4a
x=a
x=–a
y=a
y=–a
18/04/18
190
EXEMPLAR PROBLEMS – MATHEMATICS
Focal distance of a point
Let the equation of the parabola be y2 = 4ax and P(x, y) be a point on it. Then the
distance of P from the focus (a, 0) is called the focal distance of the point, i.e.,
FP =
=
( x − a )2 + y 2
( x − a) 2 + 4ax
= ( x + a) 2
= |x+a|
11.1.4 Ellipse An ellipse is the set of points in a plane, the sum of whose distances
from two fixed points is constant. Alternatively, an ellipse is the set of all points in the
plane whose distances from a fixed point in the plane bears a constant ratio, less than,
to their distance from a fixed line in the plane. The fixed point is called focus, the fixed
line a directrix and the constant ratio (e) the centricity of the ellipse.
We have two standard forms of the ellipse, i.e.,
(i)
x2 y 2
x2 y2
+
=
1
+
= 1,
and
(ii)
a 2 b2
b2 a 2
In both cases a > b and b2 = a2(1 – e2), e < 1.
In (i) major axis is along x-axis and minor along y-axis and in (ii) major axis is along yaxis and minor along x-axis as shown in Fig. 11.8 (a) and (b) respectively.
Fig. 11.8
18/04/18
CONIC SECTIONS
Forms of the ellipse
x2 y 2
+
=1
a 2 b2
x2 y2
+
=1
b2 a 2
a>b
y=0
2a
x=0
2b
a>b
x=0
2a
y=0
2b
Equation of major axis
Length of major axis
Equation of Minor axis
Length of Minor axis
a
e
a
e
Directrices
x=±
Equation of latus rectum
x = ± ae
y = ± ae
2b2
a
2b 2
a
(0, 0)
(0, 0)
Length of latus rectum
Centre
y=±
191
Focal Distance
x2 y 2
The focal distance of a point (x, y) on the ellipse 2 + 2 = 1 is
a
b
a – e | x | from the nearer focus
a + e | x | from the farther focus
Sum of the focal distances of any point on an ellipse is constant and equal to the length
of the major axis.
11.1.5 Hyperbola A hyperbola is the set of all points in a plane, the difference of
whose distances from two fixed points is constant. Alternatively, a hyperbola is the set
of all points in a plane whose distances from a fixed point in the plane bears a constant
ratio, greater than 1, to their distances from a fixed line in the plane. The fixed point is
called a focus, the fixed line a directrix and the constant ratio denoted by e, the ecentricity
of the hyperbola.
We have two standard forms of the hyperbola, i.e.,
(i)
x2 y 2
−
=1
a 2 b2
and
(ii)
y 2 x2
−
=1
a 2 b2
18/04/18
192
EXEMPLAR PROBLEMS – MATHEMATICS
Here b2 = a2 (e2 – 1), e > 1.
In (i) transverse axis is along x-axis and conjugate axis along y-axis where as in (ii)
transverse axis is along y-axis and conjugate axis along x-axis.
Fig. 11.9
x2 y 2
−
=1
a 2 b2
y 2 x2
−
=1
a 2 b2
Equation of transverse axis
y=0
x=0
Equation of conjugate axis
x=0
y=0
Length of transverse axis
2a
2a
Foci
(± ae, 0)
(0, ± ae)
Equation of latus rectum
x = ± ae
y = ± ae
2b2
a
2b 2
a
(0, 0)
(0, 0)
Forms of the hyperbola
Length of latus rectum
Centre
18/04/18
CONIC SECTIONS
193
Focal distance
The focal distance of any point (x, y) on the hyperbola
x2 y 2
−
= 1 is
a 2 b2
e | x | – a from the nearer focus
e | x | + a from the farther focus
Differences of the focal distances of any point on a hyperbola is constant and equal to
the length of the transverse axis.
Parametric equation of conics
Conics
(i)
Parabola : y2 = 4ax
(ii)
Ellipse :
(iii)
x2 y 2
Hyperbola : 2 − 2 = 1
a
b
x2 y 2
+
=1
a 2 b2
Parametric equations
x = at2, y = 2at; – ∞ < t < ∞
x = a cosθ, y = b sinθ; 0 ≤ θ ≤ 2π
x = a secθ, y = b tanθ, where
−
π
π π
3π
<θ< ;
<θ<
2
2 2
2
11.2 Solved Examples
Example 1 Find the centre and radius of the circle x2 + y2 – 2x + 4y = 8
Solution we write the given equation in the form (x2 – 2x) + ( y2 + 4y) = 8
Now, completing the squares, we get
(x2 – 2x + 1) + ( y2 + 4y + 4) = 8 + 1 + 4
(x – 1)2 + (y + 2)2 = 13
Comparing it with the standard form of the equation of the circle, we see that the
centre of the circle is (1, –2) and radius is 13 .
Example 2 If the equation of the parabola is x2 = – 8y, find coordinates of the focus,
the equation of the directrix and length of latus rectum.
Solution The given equation is of the form x2 = – 4ay where a is positive.
Therefore, the focus is on y-axis in the negative direction and parabola opens downwards.
18/04/18
194
EXEMPLAR PROBLEMS – MATHEMATICS
Comparing the given equation with standard form, we get a = 2.
Therefore, the coordinates of the focus are (0, –2) and the the equation of directrix is
y = 2 and the length of the latus rectum is 4a, i.e., 8.
Example 3 Given the ellipse with equation 9x2 + 25y2 = 225, find the major and minor
axes, eccentricity, foci and vertices.
Solution We put the equation in standard form by dividing by 225 and get
x2 y 2
+
=1
25 9
4
. Since the denominator
5
of x2 is larger, the major axis is along x-axis, minor axis along y-axis, foci are (4, 0) and
(– 4, 0) and vertices are (5, 0) and (–5, 0).
This shows that a = 5 and b = 3. Hence 9 = 25(1 – e2), so e =
Example 4 Find the equation of the ellipse with foci at (± 5, 0) and x =
36
as one of
5
the directrices.
Solution We have ae = 5,
a 36
5
=
which give a2 = 36 or a = 6. Therefore, e = .
e 5
6
2
Now b = a 1 − e = 6 1 −
25
x2 y2
= 11 . Thus, the equation of the ellipse is
+
= 1.
36
36 11
Example 5 For the hyperbola 9x2 – 16y2 = 144, find the vertices, foci and eccentricity.
Solution The equation of the hyperbola can be written as
x2 y2
–
= 1 , so a = 4, b = 3
16 9
9
25
5
+1=
, which gives e = . Vertices are (± a, 0) =
16
16
4
(± 4, 0) and foci are (± ae, 0) = (± 5, 0).
and 9 = 16 (e2 – 1), so that e2 =
Example 6 Find the equation of the hyperbola with vertices at (0, ± 6) and e =
5
.
3
Find its foci.
Solution Since the vertices are on the y-axes (with origin at the mid-point), the equation
is of the form
y2 x2
−
= 1.
a 2 b2
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CONIC SECTIONS
As vertices are (0, ± 6), a = 6, b2 = a2 (e2 – 1) = 36
equation of the hyperbola is
195
25
− 1 = 64 , so the required
9
y 2 x2
−
= 1 and the foci are (0, ± ae) = (0, ± 10).
36 64
Example 7 Find the equation of the circle which passes through the points (20, 3),
(19, 8) and (2, –9). Find its centre and radius.
Solution By substitution of coordinates in the general equation of the circle given by
x2 + y2 + 2gx + 2fy + c = 0, we have
40g + 6f + c = – 409
38g + 16 f + c = – 425
4g – 18 f + c = – 85
From these three equations, we get
g = – 7, f = – 3 and c = –111
Hence, the equation of the circle is
x2 + y2 – 14x – 6y – 111 = 0
or
(x – 7)2 + (y – 3)2 = 132
Therefore, the centre of the circle is (7, 3) and radius is 13.
Example 8 An equilateral triangle is inscribed in the parabola y2 = 4ax whose one
vertex is at the vertex of the parabola. Find the length of the side of the triangle.
Solution As shown in the figure APQ denotes the equilateral triangle with its equal
sides of length l (say).
Here
AP = l so AR = l cos30°
= l
Also,
3
2
PR = l sin 30° =
l
.
2
Thus l 3 , l are the coordinates of the point P lying on the
2 2
parabola y2 = 4ax.
Fig. 11.10
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196
EXEMPLAR PROBLEMS – MATHEMATICS
l2
= 4a
4
Therefore,
l 3
2
⇒l = 8 a 3.
Thus, 8 a 3 is the required length of the side of the equilateral triangle inscribed in
the parabola y2 = 4ax.
Example 9 Find the equation of the ellipse which passes through the point (–3, 1) and
has eccentricity
2
, with x-axis as its major axis and centre at the origin.
5
x2 y 2
Solution Let 2 + 2 = 1 be the equation of the ellipse passing through the point (–3, 1).
a
b
or
or
9
1
+ 2 =1.
2
a
b
9b2 + a2 = a2 b2
9a2 (a2 – e2) + a2 = a2 a2 (1 – e2)
or
a2 =
Again
b2 = a2 (1 – e2) =
Therefore, we have
(Using b2 = a2 (1 – e2)
32
3
32
3
1−
2
32
=
5
5
Hence, the required equation of the ellipse is
x2 y 2
+
= 1
32 32
3
5
or
3x2 + 5y2 = 32.
Example 10 Find the equation of the hyperbola whose vertices are (± 6, 0) and one of
the directrices is x = 4.
Solution As the vertices are on the x-axis and their middle point is the origin, the
equation is of the type
x2 y 2
−
= 1.
a 2 b2
Here b2 = a2 (e2 – 1), vertices are (± a, 0) and directrices are given by x = ±
a
.
e
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CONIC SECTIONS
Thus a = 6,
197
a
9
3
− 1 = 45
= 4 and so e = which gives b2 = 36
e
4
2
x2 y2
−
=1
Consequently, the required equation of the hyperbola is
36 45
Objective Type Questions
Each of the examples from 11 to 16, has four possible options, out of which one is
correct. Choose the correct answer from the given four options (M.C.Q.)
Example 11 The equation of the circle in the first quadrant touching each coordinate
axis at a distance of one unit from the origin is:
(A) x2 + y2 – 2x – 2y + 1= 0
(B) x2 + y2 – 2x – 2y – 1 = 0
2
2
(C) x + y – 2x – 2y = 0
(C) x2 + y2 – 2x + 2y – 1 = 0
Solution The correct choice is (A), since the equation can be written as (x – 1)2 +
(y – 1)2 = 1 which represents a circle touching both the axes with its centre (1, 1) and
Example 12 The equation of the circle having centre (1, –2) and passing through the
point of intersection of the lines 3x + y = 14 and 2x + 5y = 18 is
(A) x2 + y2 – 2x + 4y – 20 = 0
(B) x2 + y2 – 2x – 4y – 20 = 0
2
2
(C) x + y + 2x – 4y – 20 = 0
(D) x2 + y2 + 2x + 4y – 20 = 0
Solution The correct option is (A). The point of intersection of 3x + y – 14 = 0 and 2x
+ 5y – 18 = 0 are x = 4, y = 2, i.e., the point (4, 2)
9 + 16 = 5 and hence the equation of the circle is given by
2
(x – 1) + (y + 2)2 = 25
or
x + y2 – 2x + 4y – 20 = 0.
Example 13 The area of the triangle formed by the lines joining the vertex of the
parabola x2 = 12y to the ends of its latus rectum is
(A) 12 sq. units
(B) 16 sq. units
(C) 18 sq. units
(D) 24 sq. units
Solution The correct option is (C). From the
figure, OPQ represent the triangle whose area
is to be determined. The area of the triangle
2
=
1
1
PQ × OF =
(12 × 3) = 18
2
2
Fig. 11.11
18/04/18
198
EXEMPLAR PROBLEMS – MATHEMATICS
Example 14 The equations of the lines joining the vertex of the parabola y2 = 6x to the
points on it which have abscissa 24 are
(A) y ± 2x = 0
(B) 2y ± x = 0
(C) x ± 2y = 0
(D) 2x ± y = 0
Solution (B) is the correct choice. Let P and Q be
points on the parabola y2 = 6x and OP, OQ be the lines
joining the vertex O to the points P and Q whose
abscissa are 24.
Thus
y2 = 6 × 24 = 144
or
y = ± 12.
Therefore the coordinates of the points P and Q are
(24, 12) and (24, –12) respectively. Hence the lines are
y= ±
12
x
24
2y
x.
Fig. 11.12
Example 15 The equation of the ellipse whose centre is at the origin and the x-axis,
the major axis, which passes through the points (–3, 1) and (2, –2) is
(A) 5x2 + 3y2 32
(B) 3x2 + 5y2 = 32
(C) 5x2 – 3y2 = 32
(D) 3x2 + 5y2 + 32 = 0
x2 y 2
Solution (B) is the correct choice. Let 2 + 2 = 1 be the equation of the ellipse.
a
b
Then according to the given conditions, we have
which gives
9
1
+ 2 =1
2
a
b
and
1
1 1
+ 2 =
2
4
a
b
32
3
and
b2 =
a2 =
32
.
5
Hence, required equation of ellipse is 3x2 + 5y2 = 32.
Example 16 The length of the transverse axis along x-axis with centre at origin of a
hyperbola is 7 and it passes through the point (5, –2). The equation of the hyperbola is
18/04/18
CONIC SECTIONS
(A)
4 2 196 2
x −
y =1
49
51
(B)
49 2 51 2
x −
y =1
4
196
(C)
4 2 51 2
x −
y =1
49
196
(D)
none of these
Solution (C) is the correct choice. Let
199
x2 y 2
−
= 1 represent the hyperbola. Then
a 2 b2
according to the given condition, the length of transverse axis, i.e., 2a = 7 ⇒ a =
7
.
2
Also, the point (5, –2) lies on the hyperbola, so, we have
4
4
(25) – 2 = 1 which gives
49
b
b2 =
196
. Hence, the equation of the hyperbola is
51
4 2 51 2
x −
y =1
49
196
State whether the statements in Examples 17 and 18 are correct or not. Justify.
Example 17 Circle on which the coordinates of any point are (2 + 4 cosθ, –1 +
4 sinθ) where θ is parameter is given by (x – 2)2 + (y + 1)2 = 16.
Solution True. From given conditions, we have
x = 2 + 4 cosθ ⇒ (x – 2) = 4 cosθ
and
Squaring
y = –1 + 4 sinθ ⇒ y + 1 = 4 sinθ.
and adding, we get (x – 2)2 + (y + 1)2 = 16.
Example 18 A bar of given length moves with its
extremities on two fixed straight lines at right angles.
Any point of the bar describes an ellipse.
Solution True. Let P (x, y) be any point on the bar
such that PA = a and PB = b, clearly from the
Fig. 11.13.
Fig. 11.13
18/04/18
200
EXEMPLAR PROBLEMS – MATHEMATICS
x = OL = b cosθ and
y = PL = a sinθ
These give
x2 y 2
+
= 1, which is an ellipse.
b2 a2
Fill in the blanks in Examples 19 to 23.
Example 19 The equation of the circle which passes through the point (4, 5) and
has its centre at (2, 2) is ________________ .
Solution As the circle is passing through the point (4, 5) and its centre is (2, 2) so its
(4 − 2)2 + (5 − 2)2 = 13 . Therefore the required answer is
(x – 2)2 + (y – 2)2 = 13.
Example 20 A circle has radius 3 units and its centre lies on the line y = x – 1. If it
passes through the point (7, 3), its equation is ________________ .
Solution Let (h, k) be the centre of the circle. Then k = h – 1. Therefore, the equation
of the circle is given by (x – h)2 + [y – (h – 1)]2 = 9
... (1)
Given that the circle passes through the point (7, 3) and hence we get
(7 – h)2 + (3 – (h – 1))2 = 9
or
(7 – h)2 + (4 – h)2 = 9
or
h2 – 11h + 28 = 0
which gives (h – 7) (h – 4) = 0
⇒ h = 4 or h = 7
Therefore, the required equations of the circles are x2 + y2 – 8x – 6y + 16 = 0
or
x2 + y2 – 14x – 12y + 76 =
0
Example 21 If the latus rectum of an ellipse with axis along x-axis and centre at origin
is 10, distance between foci = length of minor axis, then the equation of the ellipse is
________________ .
Solution Given that
2b 2
= 10 and 2ae = 2b ⇒ b = ae
a
Again, we know that
b2 = a2 (1 – e2)
or
Thus
2a2e2 = a2 ⇒ e =
1
2
(using b = ae)
a= b 2
18/04/18
CONIC SECTIONS
2b2
= 10
a
or
b= 5 2.
Therefore, the required equation of the ellipse is
201
Again
Thus we get a = 10
x2
y2
+
= 1
100 50
Example 22 The equation of the parabola whose focus is the point (2, 3) and
directrix is the line x – 4y + 3 = 0 is ________________ .
Solution Using the definition of parabola, we have
( x − 2) 2 + ( y – 3) 2 =
x − 4y + 3
17
Squaring, we get
17 (x2 + y2 – 4x – 6y + 13) = x2 + 16y2 + 9 – 8xy – 24y + 6x
or 16x2 + y2 + 8xy – 74x – 78y + 212 = 0
Example 23 The eccentricity of the hyperbola
x2 y 2
−
= 1 which passes through
a 2 b2
the points (3, 0) and (3 2 , 2) is ________________ .
Solution Given that the hyperbola
x2 y 2
−
= 1 is passing through the points (3, 0) and
a 2 b2
(3 2, 2) , so we get a2 = 9 and b2 = 4.
Again, we know that b2 = a2 (e2 – 1). This gives
4 = 9 (e2 – 1)
or
e2 =
or
e=
13
9
13
.
3
18/04/18
202
EXEMPLAR PROBLEMS – MATHEMATICS
11.3 EXERCISE
1. Find the equation of the circle which touches the both axes in first quadrant and
2at
a (1 − t 2 )
y
=
and
lies on a circle
2. Show that the point (x, y) given by x =
1 + t2
1 + t2
3.
4.
5.
6.
7.
8.
for all real values of t such that –1 < t < 1 where a is any given real numbers.
If a circle passes through the point (0, 0) (a, 0), (0, b) then find the coordinates of
its centre.
Find the equation of the circle which touches x-axis and whose centre is (1, 2).
If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find
[Hint: Distance between given parallel lines gives the diameter of the circle.]
Find the equation of a circle which touches both the axes and the line 3x – 4y + 8 = 0
and lies in the third quadrant.
[Hint: Let a be the radius of the circle, then (– a, – a) will be centre and
perpendicular distance from the centre to the given line gives the radius of the
circle.]
If one end of a diameter of the circle x2 + y2 – 4x – 6y + 11 = 0 is (3, 4), then find
the coordinate of the other end of the diameter.
Find the equation of the circle having (1, –2) as its centre and passing through
3x + y = 14, 2x + 5y = 18
3 x + k touches the circle x2 + y2 = 16, then find the value of k.
[Hint: Equate perpendicular distance from the centre of the circle to its radius].
10. Find the equation of a circle concentric with the circle x2 + y2 – 6x + 12y + 15 = 0
and has double of its area.
[Hint: concentric circles have the same centre.]
11. If the latus rectum of an ellipse is equal to half of minor axis, then find its
eccentricity.
12. Given the ellipse with equation 9x2 + 25y2 = 225, find the eccentricity and foci.
9. If the line y =
13. If the eccentricity of an ellipse is
5
and the distance between its foci is 10, then
8
find latus rectum of the ellipse.
18/04/18
CONIC SECTIONS
14. Find the equation of ellipse whose eccentricity is
203
2
, latus rectum is 5 and the
3
centre is (0, 0).
x2 y2
+
= 1.
36 20
Find the coordinates of a point on the parabola y2 = 8x whose focal distance is 4.
Find the length of the line-segment joining the vertex of the parabola y2 = 4ax
and a point on the parabola where the line-segment makes an angle θ to the xaxis.
If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola,
then find the equation of the parabola.
If the line y = mx + 1 is tangent to the parabola y2 = 4x then find the value of m.
[Hint: Solving the equation of line and parabola, we obtain a quadratic equation
and then apply the tangency condition giving the value of m].
15. Find the distance between the directrices of the ellipse
16.
17.
18.
19.
20. If the distance between the foci of a hyperbola is 16 and its eccentricity is 2 ,
then obtain the equation of the hyperbola.
21. Find the eccentricity of the hyperbola 9y2 – 4x2 = 36.
3
22. Find the equation of the hyperbola with eccentricity
and foci at (± 2, 0).
2
23. If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154
square units, then obtain the equation of the circle.
24. Find the equation of the circle which passes through the points (2, 3) and (4, 5)
and the centre lies on the straight line y – 4x + 3 = 0.
25. Find the equation of a circle whose centre is (3, –1) and which cuts off a chord
of length 6 units on the line 2x – 5y + 18 = 0.
[Hint: To determine the radius of the circle, find the perpendicular distance
from the centre to the given line.]
26. Find the equation of a circle of radius 5 which is touching another circle
x2 + y2 – 2x – 4y – 20 = 0 at (5, 5).
27. Find the equation of a circle passing through the point (7, 3) having radius 3 units
and whose centre lies on the line y = x – 1.
28. Find the equation of each of the following parabolas
(a) Directrix x = 0, focus at (6, 0)
(b) Vertex at (0, 4), focus at (0, 2)
(c) Focus at (–1, –2), directrix x – 2y + 3 = 0
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204
EXEMPLAR PROBLEMS – MATHEMATICS
29. Find the equation of the set of all points the sum of whose distances from the
points (3, 0) and (9, 0) is 12.
30. Find the equation of the set of all points whose distance from (0, 4) are
2
of
3
their distance from the line y = 9.
31. Show that the set of all points such that the difference of their distances from (4, 0)
and (– 4, 0) is always equal to 2 represent a hyperbola.
32. Find the equation of the hyperbola with
(a) Vertices (± 5, 0), foci (± 7, 0)
(b) Vertices (0, ± 7), e =
4
3
(c) Foci (0, ± 10 ), passing through (2, 3)
Objective Type Questions
State Whether the statements in each of the Exercises from 33 to 40 are True or False.
Justify
33. The line x + 3y = 0 is a diameter of the circle x2 + y2 + 6x + 2y = 0.
34. The shortest distance from the point (2, –7) to the circle x2 + y2 – 14x – 10y – 151 = 0
is equal to 5.
[Hint: The shortest distance is equal to the difference of the radius and the
distance between the centre and the given point.]
35. If the line lx + my = 1 is a tangent to the circle x2 + y2 = a2, then the point (l, m)
lies on a circle.
[Hint: Use that distance from the centre of the circle to the given line is equal to
36. The point (1, 2) lies inside the circle x2 + y2 – 2x + 6y + 1 = 0.
37. The line lx + my + n = 0 will touch the parabola y2 = 4ax if ln = am2.
38. If P is a point on the ellipse
x2 y2
+
= 1 whose foci are S and S′, then PS + PS′ = 8.
16 25
39. The line 2x + 3y = 12 touches the ellipse
x2 y2
+
= 2 at the point (3, 2).
9
4
40. The locus of the point of intersection of lines
3 x − y − 4 3k = 0 and
18/04/18
CONIC SECTIONS
205
3kx + ky – 4 3 = 0 for different value of k is a hyperbola whose eccentricity
is 2.
[Hint:Eliminate k between the given equations]
Fill in the Blank in Exercises from 41 to 46.
41. The equation of the circle having centre at (3, – 4) and touching the line
5x + 12y – 12 = 0 is ________________ .
[Hint: To determine radius find the perpendicular distance from the centre of
the circle to the line.]
42. The equation of the circle circumscribing the triangle whose sides are the lines
y = x + 2, 3y = 4x, 2y = 3x is ________________ .
43. An ellipse is described by using an endless string which is passed over two pins.
If the axes are 6 cm and 4 cm, the length of the string and distance between the
pins are ____________.
44. The equation of the ellipse having foci (0, 1), (0, –1) and minor axis of length 1 is
________________ .
45. The equation of the parabola having focus at (–1, –2) and the directrix
x – 2y + 3 = 0 is ________________ .
46. The equation of the hyperbola with vertices at (0, ± 6) and eccentricity
5
is
3
______________ and its foci are ________________ .
Choose the correct answer out of the given four options (M.C.Q.) in Exercises 47 to 59.
47. The area of the circle centred at (1, 2) and passing through (4, 6) is
(A) 5π
(B) 10π
(C) 25π
(D) none of these
48. Equation of a circle which passes through (3, 6) and touches the axes is
(A) x2 + y2 + 6x + 6y + 3 = 0
(B) x 2 + y2 – 6x – 6y – 9 = 0
(C) x2 + y2 – 6x – 6y + 9 = 0
(D) none of these
49. Equation of the circle with centre on the y-axis and passing through the origin
and the point (2, 3) is
(A) x2 + y2 + 13y = 0
(B) 3x 2 + 3y2 + 13x + 3 = 0
(C) 6x2 + 6y2 – 13x = 0
(D) x 2 + y2 + 13x + 3 = 0
18/04/18
206
EXEMPLAR PROBLEMS – MATHEMATICS
50. The equation of a circle with origin as centre and passing through the vertices of
an equilateral triangle whose median is of length 3a is
(A) x2 + y2 = 9a2
(B) x 2 + y2 = 16a2
(C) x2 + y2 = 4a2
(D) x 2 + y2 = a2
[Hint: Centroid of the triangle coincides with the centre of the circle and the
2
of the length of the median]
3
51. If the focus of a parabola is (0, –3) and its directrix is y = 3, then its equation is
(A) x2 = –12y
(B) x 2 = 12y
(C) y2 = –12x (D) y2 = 12x
52. If the parabola y2 = 4ax passes through the point (3, 2), then the length of its latus
rectum is
2
4
1
(B)
(C)
(D) 4
3
3
3
53. If the vertex of the parabola is the point (–3, 0) and the directrix is the line x + 5 = 0,
then its equation is
(A) y2 = 8 (x + 3)
(B) x2 = 8 (y + 3)
(C) y2 = – 8 (x + 3)
(D) y2 = 8 ( x + 5)
54. The equation of the ellipse whose focus is (1, –1), the directrix the line x – y – 3
(A)
= 0 and eccentricity
1
is
2
(A) 7x2 + 2xy + 7y2 – 10x + 10y + 7 = 0
(B) 7x2 + 2xy + 7y2 + 7 = 0
(C) 7x2 + 2xy + 7y2 + 10x – 10y – 7 = 0
(D) none
55. The length of the latus rectum of the ellipse 3x2 + y2 = 12 is
(A) 4
(B) 3
56. If e is the eccentricity of the ellipse
(A) b2 = a2 (1 – e2)
(C) a2 = b2 (e2 – 1)
(C) 8
(D)
4
3
x2 y 2
+
= 1 (a < b), then
a 2 b2
(B) a2 = b2 (1 – e2)
(D) b2 = a2 (e2 – 1)
18/04/18
CONIC SECTIONS
207
57. The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is
equal to half of the distance between the foci is
(A)
4
3
(B)
4
3
(C)
2
3
(D) none of these
58. The distance between the foci of a hyperbola is 16 and its eccentricity is
equation is
x2 y2
−
=1
(A) x – y = 32 (B)
4
9
2
2
(C) 2x – 3y2 = 7 (D) none of these
59. Equation of the hyperbola with eccentricty
(A)
x2 y 2 4
x2 y 2 4
−
= (B)
−
=
4
5 9
9
9 9
2 . Its
3
and foci at (± 2, 0) is
2
x2 y2
−
= 1 (D) none of these
(C)
4
9
18/04/18
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At 149 seconds Isp the performance drops to about 130 tons per ton, which is close to the 160 optimum. Figure 1 also shows that when the tank fraction is appropriate for liquid hydrogen propellant (LH2), with value 1/10 ton tank per ton propellant, the performance is an order of magnitude lower. | 1,690 | 7,259 | {"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} | 3.15625 | 3 | CC-MAIN-2022-05 | latest | en | 0.877172 |
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Example 6 Write the component statements of the following compound statements and check whether the compound statement is true or false. (iii) All living things have two legs and two eyes. Compound statement are p : All living things have two legs. q : All living things have two eyes. We know that Plants are living beings which do not have legs and do not have eyes Hence p is false , & q is false Hence both the statement are false. The connection work is And So the compound statement is false. | 137 | 643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | longest | en | 0.91149 |
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Welcome to school 2001-2002!! This is the location to check the calender for homework assignments and tests, and to link to helpful lessons and practice games. Algebra2 3rd Six Weeks Calender 2001 Cline (periods 3,5,6) 11-05 Review 5.5-5.6 worksheet 11-06 Test 5.5-5.6 11-07 5.9 complex numbers (intro,simplifying) worksheet 11-08 5.9 add/subt complex numbers worksheet 11-09 5.9 mult/div complex numbers worksheet 11-12 5.9 mult/div complex numbers/Quiz worksheet 11-13 5.9 graphing complex numbers worksheet 11-14 review worksheet 11-15 review 11-16 Test 5.9 complex numbers Thanksgiving break 11-26 Parent functions y=x worksheet 111-27 Parent functions worksheet 11-28 Parent functions worksheet 11-29 Parent functions worksheet 11-30 ½ TEST (will be combined with matrix ½ test) 12-03 4.1 Intro to matrices worksheet 12-04 4.2 add/subt matrices worksheet 12-05 4.3 mult/div matrices worksheet 12-06 4.4 matrices and determinants worksheet 12-07 4.5 identity and inverse matrices worksheet 12-10 4.6 using matrices to solve systems of equations worksheet 12-11 review 12-12 ½ TEST (will be combined with parent functions ½ test) 12-13 review for semester finals, bring textbook 12-14 review for semester finals, bring textbook 12-17 review for semester finals, bring textbook 12-18 Semester finals 12-19 5emester finals 12-20 Semester finals 12-21 Semester finals
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https://www.quia.com/jg/68603.htmlMatch polynomials with factored form https://www.quia.com/jq/19732.htmlFactor given polynomials
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https://tasks.illustrativemathematics.org/7 | 1,680,208,480,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949387.98/warc/CC-MAIN-20230330194843-20230330224843-00068.warc.gz | 608,880,471 | 14,199 | #### 7.RP.A.2.d. Explain what a point $(x, y)$ on the graph of a proportional relationship means in terms of the situation, with special attention to the points $(0, 0)$ and $(1, r)$ where $r$ is the unit rate.
• No tasks yet illustrate this standard.
#### 7.NS.A.1.a. Describe situations in which opposite quantities combine to make 0. For example, a hydrogen atom has 0 charge because its two constituents are oppositely charged.
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#### 7.NS.A.1.d. Apply properties of operations as strategies to add and subtract rational numbers.
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#### 7.NS.A.2.c. Apply properties of operations as strategies to multiply and divide rational numbers.
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#### 7.G.B.5. Use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure.
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#### 7.SP.C.5. Understand that the probability of a chance event is a number between 0 and 1 that expresses the likelihood of the event occurring. Larger numbers indicate greater likelihood. A probability near 0 indicates an unlikely event, a probability around 1/2 indicates an event that is neither unlikely nor likely, and a probability near 1 indicates a likely event.
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#### 7.SP.C.7.b. Develop a probability model (which may not be uniform) by observing frequencies in data generated from a chance process. For example, find the approximate probability that a spinning penny will land heads up or that a tossed paper cup will land open-end down. Do the outcomes for the spinning penny appear to be equally likely based on the observed frequencies?
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#### 7.SP.C.8.c. Design and use a simulation to generate frequencies for compound events. For example, use random digits as a simulation tool to approximate the answer to the question: If 40% of donors have type A blood, what is the probability that it will take at least 4 donors to find one with type A blood?
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https://demo.formulasearchengine.com/wiki/Day_count_convention | 1,580,032,697,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251687958.71/warc/CC-MAIN-20200126074227-20200126104227-00364.warc.gz | 401,876,184 | 25,002 | # Day count convention
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In finance, a day count convention determines how interest accrues over time for a variety of investments, including bonds, notes, loans, mortgages, medium-term notes, swaps, and forward rate agreements (FRAs). This determines the number of days between two coupon payments; thus, calculating the amount transferred on payment dates and also the accrued interest for dates between payments.[1] The day count is also used to quantify periods of time when discounting a cash-flow to its present value. When a security such as a bond is sold between interest payment dates, the seller is eligible to some fraction of the coupon amount.
The day count convention is used in many other formulas in financial mathematics as well.
## Development
The need for day count conventions is a direct consequence of interest-earning investments. Different conventions were developed to address often conflicting requirements, including ease of calculation, constancy of time period (day, month, or year) and the needs of the Accounting department. This development occurred long before the advent of computers.
There is no central authority defining day count conventions, so there is no standard terminology, however the International Swaps and Derivatives Association (ISDA) and the International Capital Market Association (ICMA) have done work gathering and documenting conventions. Certain terms, such as "30/360", "Actual/Actual", and "money market basis" must be understood in the context of the particular market.
The conventions have evolved, and this is particularly true since the mid-1990s. Part of it has simply been providing for additional cases[2] or clarification.[3]
There has also been a move towards convergence in the marketplace, which has resulted in the number of conventions in use being reduced. Much of this has been driven by the introduction of the euro.[4][5]
## Definitions
Interest
Amount of interest accrued on an investment.
CouponFactor
The Factor to be used when determining the amount of interest paid by the issuer on coupon payment dates. The periods may be regular or irregular.
CouponRate
The interest rate on the security or loan-type agreement, e.g., 5.25%. In the formulas this would be expressed as 0.0525.
Date1 (Y1.M1.D1)
Starting date for the accrual. It is usually the coupon payment date preceding Date2.
Date2 (Y2.M2.D2)
Date through which interest is being accrued. You could word this as the "to" date, with Date1 as the "from" date. For a bond trade, it is the settlement date of the trade.
Date3 (Y3.M3.D3)
The coupon payment date following Date2. This would be the maturity date if there are no more interim payments to be made.
Days(StartDate, EndDate)
Function returning the number of days between StartDate and EndDate on a Julian basis (i.e., all days are counted). For instance, Days(15 October 2007, 15 November 2007) returns 31.
EOM
Indicates that the investment always pays interest on the last day of the month. If the investment is not EOM, it will always pay on the same day of the month (e.g., the 10th).
Factor
Figure representing the amount of the CouponRate to apply in calculating Interest. It is often expressed as "days in the accrual period / days in the year". If Date2 is a coupon payment date, Factor is zero.
Freq
The coupon payment frequency. 1 = annual, 2 = semi-annual, 4 = quarterly, 12 = monthly, etc.
Principal
Par value of the investment.
For all conventions, the Interest is calculated as:
${\displaystyle {\mathrm {Interest} }={\mathrm {Principal} }\times {\mathrm {CouponRate} }\times {\mathrm {Factor} }}$
## 30/360 methods
All conventions of this class calculate the Factor as:
${\displaystyle \mathrm {Factor} ={\frac {360\times (Y_{2}-Y_{1})+30\times (M_{2}-M_{1})+(D_{2}-D_{1})}{360}}}$
They calculate the CouponFactor as:
${\displaystyle {\mathrm {CouponFactor} }={\frac {360\times (Y_{3}-Y_{1})+30\times (M_{3}-M_{1})+(D_{3}-D_{1})}{360}}}$
This is the same as the Factor calculation, with Date2 replaced by Date3. In the case that it is a regular coupon period, this is equivalent to:
${\displaystyle {\mathrm {CouponFactor} }={\frac {1}{\mathrm {Freq} }}}$
The conventions are distinguished by the manner in which they adjust Date1 and/or Date2 for the end of the month. Each convention has a set of rules directing the adjustments.
Treating a month as 30 days and a year as 360 days was devised for its ease of calculation by hand compared with manually calculating the actual days between two dates. Also, because 360 is highly factorable, payment frequencies of semi-annual and quarterly and monthly will be 180, 90, and 30 days of a 360 day year, meaning the payment amount will not change between payment periods.
### 30/360 US
Date adjustment rules (more than one may take effect; apply them in order, and if a date is changed in one rule the changed value is used in the following rules):
• If the investment is EOM and (Date1 is the last day of February) and (Date2 is the last day of February), then change D2 to 30.
• If the investment is EOM and (Date1 is the last day of February), then change D1 to 30.
• If D2 is 31 and D1 is 30 or 31, then change D2 to 30.
• If D1 is 31, then change D1 to 30.
This convention is used for US corporate bonds and many US agency issues. It is most commonly referred to as "30/360", but the term "30/360" may also refer to any of the other conventions of this class, depending on the context.
Other names:
• 30U/360 - 30U/360 is not strictly the same as 30/360, it is used for the Euribor (Euro denominated Libor) curve and Euro denominated swaps, with the distinction that under 30/360, each day in a 31 day month accrues 30/31 of interest, whereas in 30U/360 payment occurs on the 30th and the 31st is considered to be part of the next month. - Bloomberg
• Bond basis
• 30/360
Sources:
• ISDA 2006 Section 4.16(f), though the first two rules are not included.[6]
• Template:Harv
### 30E/360
Date adjustment rules:
• If D1 is 31, then change D1 to 30.
• If D2 is 31, then change D2 to 30.
Other names:
• 30/360 ICMA
• 30S/360
• Eurobond basis (ISDA 2006)
• Special German
Sources:
• ICMA Rule 251.1(ii), 251.2.[7]
• ISDA 2006 Section 4.16(g).[6]
### 30E/360 ISDA
Date adjustment rules:
• If D1 is the last day of the month, then change D1 to 30.
• If D2 is the last day of the month (unless Date2 is the maturity date and M2 is February), then change D2 to 30.
Other names:
• 30E/360 ISDA
• Eurobond basis (ISDA 2000)
• German
Sources:
• ISDA 2006 Section 4.16(h).[6]
## Actual methods
The conventions of this class calculate the number of days between two dates (e.g., between Date1 and Date2) as the Julian difference. This is the function Days(StartDate, EndDate).
The conventions are distinguished primarily by the amount of the CouponRate they assign to each day of the accrual period.
### Actual/Actual ICMA
Formulas:
${\displaystyle {\mathrm {Factor} }={\frac {{\mathrm {Days} }({\mathrm {Date1} },{\mathrm {Date2} })}{{\mathrm {Freq} }\times {\mathrm {Days} }({\mathrm {Date1} },{\mathrm {Date3} })}}}$
For regular coupon periods:
${\displaystyle {\mathrm {CouponFactor} }={\frac {1}{\mathrm {Freq} }}}$
For irregular coupon periods, the period has to be divided into one or more quasi-coupon periods (also called notional periods) that match the normal frequency of payment dates. The interest in each such period (or partial period) is then computed, and then the amounts are summed over the number of quasi-coupon periods. For details, see Template:Harv or the ISDA paper.[4]
This method ensures that all coupon payments are always for the same amount.
It also ensures that all days in a coupon period are valued equally. However, the coupon periods themselves may be of different lengths; in the case of semi-annual payment on a 365 day year, one period can be 182 days and the other 183 days. In that case, all the days in one period will be valued 1/182nd of the payment amount and all the days in the other period will be valued 1/183rd of the payment amount.
This is the convention used for US Treasury bonds and notes, among other securities.
Other names:
• Actual/Actual
• Act/Act ICMA
• ISMA-99
• Act/Act ISMA
Sources:
• ICMA Rule 251.1(iii).[7]
• ISDA 2006 Section 4.16(c).[6]
• Template:Harv
• Actual/Actual comparison, EMU and Market Conventions: Recent Developments.[4]
### Actual/Actual ISDA
Formulas:
${\displaystyle {\mathrm {Factor} }={\frac {\mbox{Days not in leap year}}{365}}+{\frac {\mbox{Days in leap year}}{366}}}$
This convention accounts for days in the period based on the portion in a leap year and the portion in a non-leap year.
The days in the numerators are calculated on a Julian day difference basis. In this convention the first day of the period is included and the last day is excluded.
The CouponFactor uses the same formula, replacing Date2 by Date3. In general, coupon payments will vary from period to period, due to the differing number of days in the periods. The formula applies to both regular and irregular coupon periods.
Other names are:
• Actual/Actual
• Act/Act
• Actual/365
• Act/365
Sources:
• ISDA 2006 Section 4.16(b).[6]
### Actual/365 Fixed
Formulas:
${\displaystyle {\mathrm {Factor} }={\frac {{\mathrm {Days} }({\mathrm {Date1} },{\mathrm {Date2} })}{365}}}$
Each month is treated normally and the year is assumed to be 365 days. For example, in a period from February 1, 2005 to April 1, 2005, the Factor is considered to be 59 days divided by 365.
The CouponFactor uses the same formula, replacing Date2 by Date3. In general, coupon payments will vary from period to period, due to the differing number of days in the periods. The formula applies to both regular and irregular coupon periods.
Other names:
• Act/365 Fixed
• A/365 Fixed
• A/365F
• English
Sources:
### Actual/360
Formulas:
${\displaystyle {\mathrm {Factor} }={\frac {{\mathrm {Days} }({\mathrm {Date1} },{\mathrm {Date2} })}{360}}}$
This convention is used in money markets for short-term lending of currencies, including the US dollar and Euro, and is applied in ESCB monetary policy operations. It is the convention used with Repurchase agreements. Each month is treated normally and the year is assumed to be 360 days. For example, in a period from February 1, 2005 to April 1, 2005, the Factor is 59 days divided by 360 days.
The CouponFactor uses the same formula, replacing Date2 by Date3. In general, coupon payments will vary from period to period, due to the differing number of days in the periods. The formula applies to both regular and irregular coupon periods.
Other names:
• Act/360
• A/360
• French
Sources:
### Actual/364
Formulas:
${\displaystyle {\mathrm {Factor} }={\frac {{\mathrm {Days} }({\mathrm {Date1} },{\mathrm {Date2} })}{364}}}$
Each month is treated normally and the year is assumed to be 364 days. For example, in a period from February 1, 2005 to April 1, 2005, the Factor is considered to be 59 days divided by 364.
The CouponFactor uses the same formula, replacing Date2 by Date3. In general, coupon payments will vary from period to period, due to the differing number of days in the periods. The formula applies to both regular and irregular coupon periods.
### Actual/365L
Formulas:
${\displaystyle {\mathrm {Factor} }={\frac {{\mathrm {Days} }({\mathrm {Date1} },{\mathrm {Date2} })}{\mathrm {DiY} }}}$
This convention requires a set of rules in order to determine the days in the year (DiY).
• If Freq = 1 (annual coupons):
• If February 29 is in the range from Date1 (exclusive) to Date3 (inclusive), then DiY = 366, else DiY = 365.
• If Freq <> 1:
• If Date3 is in a leap year, then DiY = 366, else DiY = 365.
The CouponFactor uses the same formula, replacing Date2 by Date3. In general, coupon payments will vary from period to period, due to the differing number of days in the periods. The formula applies to both regular and irregular coupon periods.
Other names:
• ISMA-Year
Sources:
• ICMA Rule 251.1(i) (Euro-sterling floating-rate notes).[7]
### Actual/Actual AFB
Formulas:
${\displaystyle {\mathrm {Factor} }={\frac {{\mathrm {Days} }({\mathrm {Date1} },{\mathrm {Date2} })}{\mathrm {DiY} }}}$
This convention requires a set of rules in order to determine the days in the year (DiY).
The basic rule is that if February 29 is in the range from Date1 (inclusive) to Date2 (exclusive), then DiY = 366, else DiY = 365.
If the period from Date1 to Date2 is more than one year, the calculation is split into two parts:
• the number of complete years, counted back from the last day of the period
• the remaining initial stub, calculated using the basic rule.
As an example, a period from 1994-02-10 to 1997-06-30 is split as follows:
• 1994-06-30 to 1997-06-30 = 3 (whole years calculated backwards from the end)
• 1994-02-10 to 1994-06-30 = 140/365
Resulting in a total value of 3 + 140/365.
This convention was originally written in French and during translation the term "Période d'Application" was converted to "Calculation Period". As ISDA assigns a very specific meaning to "Calculation Period" (Date1 to Date3) confusion can ensue. Reading the original French, the period referred to is Date1 to Date2, not Date1 to Date3.[8]
The original French version of the convention contained no specific rules for counting back the years. A later ISDA paper [4] added an additional rule: "When counting backwards for this purpose, if the last day of the relevant period is 28 February, the full year should be counted back to the previous 28 February unless 29 February exists, in which case, 29 February should be used". No source can be found explaining the appearance or rationale of the extra rule. The table below compares the later ISDA count back rule to a simple count back rule (which would have been implied by the original French) for one of the few cases where they differ. The simple rule illustrated here is based on subtraction of n years from Date2, where subtracting whole years from a date goes back to the same day-of-month, except if starting on 29 February and going back to a non-leap year then 28 February results.
Date range ISDA count back rule Simple count back rule
From 2004-02-28 to 2008-02-27 3 + 365 / 366 3 + 365 / 366
From 2004-02-28 to 2008-02-28 4 + 1 / 366 4
From 2004-02-28 to 2008-02-29 4 + 1 / 366 4 + 1 / 366
Sources:
• "Definitions Communes plusieurs Additifs Techniques", by the Association Francaise des Banques in September 1994.[8]
• FBF Master Agreement for Financial Transactions, Supplement to the Derivatives Annex, Edition 2004, section 7i.[9]
• Actual/Actual comparison, EMU and Market Conventions: Recent Developments.[4]
### 1/1
This is used for inflation instruments and divides the overall 4 year period distributing the additional day across all 4 years i.e. giving 365.25 days to each year.
Sources:
• ISDA 2006 Section 4.16(a).[6]
• FBF Master Agreement for Financial Transactions, Supplement to the Derivatives Annex, Edition 2004, section 7a.[9]
## Discussion
### Comparison of 30/360 and Actual/360
The 30/360 methods assume every month has 30 days and each year has 360 days. The 30/360 calculation is listed on standard loan constant charts and is now typically used by a calculator or computer in determining mortgage payments. This method of treating a month as 30 days and a year as 360 days was originally devised for its ease of calculation by hand compared with the actual days between two dates. Because 360 is highly factorable, payment frequencies of semi-annual and quarterly and monthly will be 180, 90, and 30 days of a 360 day year, meaning the payment amount will not change between payment periods.
The Actual/360 method calls for the borrower for the actual number of days in a month. This effectively means that the borrower is paying interest for 5 or 6 additional days a year as compared to the 30/360 day count convention. Spreads and rates on Actual/360 transactions are typically lower, e.g., 9 basis points. Since monthly loan payments are the same for both methods and since the investor is being paid for an additional 5 or 6 days of interest with the Actual/360 year base, the loan’s principal is reduced at a slightly lower rate. This leaves the loan balance 1-2% higher than a 30/360 10-year loan with the same payment.
### Business date convention
Date rolling (business date) conventions are a common practice to adjust non-business days into business days.
## Footnotes
1. Template:Cite web
2. see the treatment of 30/360 in Template:Harv.
3. the ISDA 2006 vs. ISDA 2000 definitions, for instance.
4. Template:Cite web
5. Template:Cite web
6. Template:Cite web
7. Template:Cite web
8. Template:Cite web
9. Template:Cite web
## References
• {{#invoke:citation/CS1|citation
|CitationClass=citation }}. The standard reference for conventions applicable to US securities. For the 30/360 US convention, this edition adds the first two rules to those given in earlier editions.
• {{#invoke:citation/CS1|citation
|CitationClass=citation }}. ICMA's definition of certain day count conventions.
• {{#invoke:citation/CS1|citation
|CitationClass=citation }}. ISDA's definition of certain day count conventions. Note that these definitions differ in some cases from the ISDA's Annex to the 2000 Definitions.
• {{#invoke:citation/CS1|citation
|CitationClass=citation }}. ISDA's discussion of market convergence, including an extensive discussion of irregular coupon periods.
• {{#invoke:citation/CS1|citation
|CitationClass=citation }}. Definition of various day counts in section 7.
## Further reading
|CitationClass=citation }}. Web page on the history and context of day count conventions, including a cross-reference. | 4,544 | 17,804 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 12, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | longest | en | 0.950705 |
https://studyres.com/doc/154710/lecture-10-presentation | 1,627,602,614,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153899.14/warc/CC-MAIN-20210729234313-20210730024313-00300.warc.gz | 566,601,096 | 11,074 | Survey
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Transcript
```Physics 1161: Lecture 10
Magnetism
Textbook Sections 22-1 – 22-3, 22-8
Checkpoint
Magnets
1
Which drawing shows the correct
field lines for a bar magnet?
(1)
(2)
(3)
Magnetic field lines are continuous
S
N
S
N
S
N
2
Arrows go from N to S outside the
magnet (S to N inside).
3
• F=qvxB
Magnetic Field Units
• SI units: N-s/C-m = Tesla
• 1 Tesla = 10,000 Gauss
• Earth’s magnetic field is
approximately 0.5 Gauss
Nikola Tesla
1856- 1943
• Refrigerator magnets are
• Superconducting
electromagnets can be as
much as 40 Tesla
Carl Friedrich Gauss
1777-1855
Tesla Videos
vQ&feature=youtu.be
Direction of Magnetic Force on
Moving Charges
Velocity
out of screen
out of screen
out of screen
out of screen
B
right
left
up
down
Force
up
down
left
right
B
Right Hand Rule
• Thumb v, Fingers B, palm F
• Negative charge has opposite F!
v
F
Checkpoint
Particle in a Magnetic Field 1
Each chamber has a unique magnetic
field. A positively charged particle
enters chamber 1 with velocity 75 m/s
up, and follows the dashed trajectory.
What is the direction of the force on
the particle just as it enters region 1?
1) up
2) down
3) left
Particle is moving straight upwards then veers to the
4) right
right.
5) into page
6) out of page
Checkpoint
Particle in a Magnetic Field 2
Each chamber has a unique magnetic
field. A positively charged particle
enters chamber 1 with velocity 75 m/s
up, and follows the dashed trajectory.
What is the direction of the magnetic
field in region 1?
1) up
2) down
3) left
v (thumb) points up, F(palm) points right: so
4) right
B(fingers) must point out.
5) into page
6) out of page
Each chamber has a unique magnetic field. A positively
charged particle enters chamber 2 with velocity 75 m/s to
the right, and follows the dashed trajectory.
What is the direction of the
magnetic field in region 2?
1. up
2. down
3. left
4. right
5. into page
6. out of page
59%
34%
7%
0%
1
2
0%
0%
3
4
5
6
Each chamber has a unique magnetic field. A positively
charged particle enters chamber 2 with velocity 75 m/s to
the right, and follows the dashed trajectory.
What is the direction of the
magnetic field in region 2?
1. up
2. down
3. left
v (thumb) points right, F(palm) points
up, B(fingers) point in.
4. right
5. into page
6. out of page
69%
24%
7%
0%
1
2
0%
0%
3
4
5
6
Magnitude of Magnetic Force on
Moving Charges
• The magnetic force on a charge depends on
the magnitude of the charge, its velocity,
and the magnetic field.
V
F = q v B sin(q)
– Direction from RHR
v fingers (__),
B palm (__)
F
• Thumb (__),
– Note if v is parallel to B then F = 0
q
B
The three charges below have equal charge and speed,
but are traveling in different directions in a uniform
magnetic field.
Which particle experiences the greatest magnetic force?
B
1.
2.
3.
4.
1
2
3
All Same
3
2
1
52%
28%
14%
7%
1
2
3
4
The three charges below have equal charge and speed,
but are traveling in different directions in a uniform
magnetic field.
Which particle experiences the greatest magnetic force?
B
1.
2.
3.
4.
1
2
3
All Same
3
2
54%
1
46%
F = q v B sin(q)
0%
1
0%
2
3
4
The three charges below have equal charge and speed,
but are traveling in different directions in a uniform
magnetic field.
The force on particle 3 is in the same direction as the
force on particle 1.
1. True
2. False
B
3
62%
2
1
38%
1
2
The three charges below have equal charge and speed,
but are traveling in different directions in a uniform
magnetic field.
The force on particle 3 is in the same direction as the
force on particle 1.
1. True
2. False
B
3
2
1
Thumb (v), fingers (B), palm (F)
into page
Electric Force vs Magnetic Force
Electric
Source:
Charges
Acts on:
Charges
Magnitude:
F = qE
Direction:
Parallel to E
Magnetic
Moving
charges
Moving
charges
F = qvB sin
Perpendicular
to v & B
Velocity Selector
Determine magnitude and direction
of magnetic field such that a
v
positively charged particle with
initial velocity v travels straight
through and exits the other side.
Electric force is down, so need magnetic force up.
By RHR, B must be into page
For straight line, need |FE |= |FB |
q E= q v B sin(90)
FB
E
FE
B = E/v
What direction should B point if you want to select negative charges?
1) Into Page
2) Out of page
3) Left
4) Right
FE would be up so FB must be down.
Motion of Q in uniform B field
• Force is perpendicular to B,v
– B does no work! (W=F d cos q )
– Speed is constant (W=D K.E. )
– Circular motion
2
v
ac
R
Fnet mac
x x x x x x x
x x x x x x x
x x x x x x x
x x x x x x x
x x x x x x x
Uniform B into page
2
v
Fnet m
R
x x x x x x x
Fnet qvB
Trajectory in Constant B Field
• Suppose charge q enters B-field with velocity v as
shown below. What will be the path q follows?
x x x x x x x x x x x x
x x x x x x x x x x x vx B
x x x x x x x x x x x x
q
v
F
F
R
• Force is always ^ to velocity and B. What is path?
– Path will be circle. F will be the centripetal force
needed to keep the charge in its circular orbit.
– Calculate R:
• Lorentz force:
F qvB
x x x x x x x x x x x x
x x x x x x x x x x x
• centripetal acc:
v2
x
x
x
x
x
x
x
x
x
x
x
a
v
F
F
R
• Newton's 2nd Law:
R
F ma
R
qvB m
mv
qB
v2
R
an important result, with
useful experimental
consequences !
B
vx
x
q
Checkpoint
Particle in a Magnetic Field 4
Each chamber has a unique magnetic field. A
positively charged particle enters chamber 1 with
velocity v1= 75 m/s up, and follows the dashed
trajectory.
What is the speed of the particle when it
leaves chamber 2?
1) v2 < v1
2) v2 = v1
3) v2 > v1
Checkpoint
Particle in a Magnetic Field 4
Each chamber has a unique magnetic field. A
positively charged particle enters chamber 1 with
velocity v1= 75 m/s up, and follows the dashed
trajectory.
What is the speed of the particle when
it leaves chamber 2?
1) v2 < v1
2) v2 = v1
3) v2 > v1
Magnetic force is always perpendicular to velocity, so it changes direction, not
speed of particle.
43
Checkpoint
Particle in a Magnetic Field 6
Each chamber has a unique magnetic field. A
positively charged particle enters chamber 1 with
velocity v1= 75 m/s up, and follows the dashed
trajectory.
Compare the magnitude of the
magnetic field in chambers 1 and 2
1) B1 > B2
2) B1 = B2
3) B1 < B2
Checkpoint
Particle in a Magnetic Field 6
Each chamber has a unique magnetic field. A
positively charged particle enters chamber 1 with
velocity v1= 75 m/s up, and follows the dashed
trajectory.
Compare the magnitude of the magnetic field in
chambers 1 and 2
1) B1 > B2
2) B1 = B2.
3) B1 < B2
R
mv
qB
Larger B, greater force, smaller R
Checkpoint
Particle in a Magnetic Field 7
Each chamber has a unique magnetic field. A
positively charged particle enters chamber 1 with
velocity v1= 75 m/s up, and follows the dashed
trajectory.
A second particle with mass 2m enters the
chamber and follows the same path as the
particle with mass m and charge q=25 mC.
What is its charge?
1) Q = 12.5 mC
2) Q = 25 mC
3) Q = 50 mC
Checkpoint
Particle in a Magnetic Field 7
Each chamber has a unique magnetic
field. A positively charged particle enters
chamber 1 with velocity v1= 75 m/s up,
and follows the dashed trajectory.
A second particle with mass 2m enters the chamber and
follows the same path as the particle with mass m and charge
q=25 mC. What is its charge?
1) Q = 12.5 mC
2) Q = 25 mC
3) Q = 50 mC
mv
R
qB
```
Related documents | 2,516 | 8,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.720117 |
https://www.physicstutorials.org/exams-and-problem-solutions/rotational-motion-exams-and-problem-solutions/rotational-motion-exam1-and-problem-solutions/ | 1,669,914,018,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710829.5/warc/CC-MAIN-20221201153700-20221201183700-00802.warc.gz | 1,021,050,545 | 9,280 | # Rotational Motion Exam 1 and Problem Solutions
1. An object, attached to a 0,5m string, does 4 rotation in one second. Find
a) Period
b) Tangential velocity
c) Angular velocity of the object.
a) If the object does 4 rotation in one second, its frequency becomes; f=4s⁻¹
T=1/f=1/4s
b) Tangential velocity of the object;
V=2.π.f.r
V=2.3.4.0,5
V=12m/s
c) Angular velocity of the object
2. Find the relation between tangential and angular velocities of points X, Y and Z. X and Y rotate together, so if X does one rotation then Y also does one rotation. On the contrary, if Y does one revolution, Z does two revolutions. Angular velocities of the X, Y and Z are;
ωX=ωY=ωZ/2
3. An object hanged on a rope L=0,5m, does rotational motion. If the angle between rope and vertical is 37⁰, find the tangential velocity of the object. (g=10m/s², cos37⁰=0,8, sin37⁰=0,6) Free body diagram of system is given below; Horizontal component of tension on the rope makes object rotate.
Tx=mV²/r, Ty=m.g
Radius of the motion path is;
r=L.sin37⁰=0,5.0,6=0,3m
tan37⁰=Tx/Ty
3/4=mV²/r/m.g
3/4=V²/g.r
V=3/2m/s
4. An object having mass m does rotational motion. Its angular velocity is ω and radius of motion path is r. Find kinetic energy of the object in terms of r, ω, and m.
EK=1/2m.V²
V=ω.r
EK=1/2m(ω.r)²
EK=mω².r²/2
5. Stone having mass 0,5kg rotates in horizontal. It is hanged on 1m rope. If the tension on the rope is 80 N, find the frequency of the motion.
Fnet=80N=m.ω².r
80=m.4.π².f².r
80=0,5.4.3².f².1
f=2s⁻¹ | 530 | 1,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2022-49 | latest | en | 0.76907 |
https://circlecoder.com/construct-binary-tree-from-inorder-and-postorder-traversal/ | 1,680,390,790,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950363.89/warc/CC-MAIN-20230401221921-20230402011921-00605.warc.gz | 203,412,853 | 9,222 | # Construct Binary Tree From Inorder And Postorder Traversal Problem
## Description
LeetCode Problem 106.
Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of the same tree, construct and return the binary tree.
Example 1:
``````1
2
Input: inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
Output: [3,9,20,null,null,15,7]
``````
Example 2:
``````1
2
Input: inorder = [-1], postorder = [-1]
Output: [-1]
``````
Constraints:
• 1 <= inorder.length <= 3000
• postorder.length == inorder.length
• -3000 <= inorder[i], postorder[i] <= 3000
• inorder and postorder consist of unique values.
• Each value of postorder also appears in inorder.
• inorder is guaranteed to be the inorder traversal of the tree.
• postorder is guaranteed to be the postorder traversal of the tree.
## Sample C++ Code
``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
return create(inorder, postorder, 0, inorder.size() - 1, 0, postorder.size() - 1);
}
TreeNode* create(vector<int> &inorder, vector<int> &postorder, int is, int ie, int ps, int pe){
if(ps > pe){
return nullptr;
}
TreeNode* node = new TreeNode(postorder[pe]);
int pos;
for(int i = is; i <= ie; i++){
if(inorder[i] == node->val){
pos = i;
break;
}
}
node->left = create(inorder, postorder, is, pos - 1, ps, ps + pos - is - 1);
node->right = create(inorder, postorder, pos + 1, ie, pe - ie + pos, pe - 1);
return node;
}
`````` | 486 | 1,543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.424312 |
https://apprendre-forex.com/art-projects-with-dominoes/ | 1,718,451,652,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861586.40/warc/CC-MAIN-20240615093342-20240615123342-00419.warc.gz | 85,677,976 | 11,585 | # Art Projects With Dominoes
Dominoes are one of those toys that have stood the test of time. They are fun to play with, and you can line them up in many different ways, forming shapes like towers and pyramids. They can also be used to create interesting patterns, and they are often used in art projects. Dominoes are also a great tool for teaching children about simple physics concepts, such as gravity and force.
Most domino sets feature both a set of numbered tiles (the dominoes) and a set of blank or 0 tiles. The number of pips on a domino can vary from set to set, but most large domino sets use Arabic numerals on their pips. The most popular types of domino games fall into two categories: blocking games and scoring games. The first category involves placing tiles side-to-side so that the numbers match: a domino with a single number touches another, requiring the other to have a matching number on its exposed end. When all of the dominoes are matched up, the chain becomes a row of numbers that can be scored with one or more dice.
The second category of domino games involves arranging dominoes into a grid, with certain spaces left empty or “blank.” The goal of this type of game is to place additional dominoes in a way that increases the number of matches. The winner is the player who completes the grid with a number greater than all of his or her opponents. Dominos can be placed in a horizontal, vertical, or diagonal grid.
Hevesh is able to create her mind-blowing domino setups by using simple physics. When a domino falls, it has potential energy that can be converted into kinetic energy, which can then be transmitted to the next domino and knock it over as well. Hevesh says that the biggest domino setups take several nail-biting minutes to fall.
When Hevesh is creating an intricate domino art piece, she usually starts by sketching out a plan on paper before she begins construction. This allows her to make precise adjustments if something doesn’t work out as planned. She also tests each section of a layout, filming it in slow motion, to ensure that the entire system works as intended.
She’s been working on her current project for a year, and it will take another year before she can complete it. But she doesn’t want to stop there—she has plans for even more ambitious creations. She wants to build a domino set that’s a replica of the Taj Mahal, for example, and she’s also interested in developing a domino set that can be used in science classrooms to help students understand complex issues such as climate change. | 554 | 2,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-26 | latest | en | 0.957423 |
https://math.stackexchange.com/questions/2955809/spectral-measure-is-sigma-finite | 1,568,975,820,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573988.33/warc/CC-MAIN-20190920092800-20190920114800-00272.warc.gz | 563,914,118 | 30,121 | # Spectral measure is sigma finite
The spectral theorem states that for a self-adjoint linear operator $$A$$ with $$\left\|A\right\|=1$$ on a Hilbert space, there is a measure space $$(X,\mathcal M,\mu)$$ so that $$A$$ is unitarily equivalent to a multiplication operator on $$L^2(I,d\mu)$$ where $$I:=[-1,1]$$. If $$H$$ is separable, then $$\mu$$ can be taken to be a finite measure.
Now my question is: If $$H$$ is not separable, can we always take $$\mu$$ to be $$\sigma$$-finite?
• you should probably say "on a Hilbert space $H$" – mathworker21 Oct 14 '18 at 23:32
If every real $$x\in[0,1]$$ is an eigenvalue, then the $$\mu$$ cannot be sigma finite.
• For instance, let $H = \ell^2([0,1])$, i.e. $L^2$ of $[0,1]$ equipped with the discrete $\sigma$-algebra and counting measure. Then define $(Af)(x) = x f(x)$. – Nate Eldredge Oct 15 '18 at 4:30 | 283 | 856 | {"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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-39 | latest | en | 0.776519 |
http://exciting.wikidot.com/nitrogen-electronic-bandstructure-from-gw | 1,624,450,626,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488538041.86/warc/CC-MAIN-20210623103524-20210623133524-00141.warc.gz | 18,905,517 | 11,448 | Electronic Band Structure from GW
by Dmitrii Nabok for exciting nitrogen
Purpose: In this tutorial you will learn how to perform a basic G$_{0}$W$_{0}$ calculation. As an example, the electronic band structure of bulk Si is calculated. Notice that self-consistent GW is not yet implemented in exciting.
### 0. Define relevant environment variables
Read the following paragraphs before starting with the rest of this tutorial!
Before starting, be sure that relevant environment variables are already defined as specified in How to set environment variables for tutorials scripts. Here, you can find a list of the scripts which are relevant for this tutorial, with a short description.
• PLOT-compare-bands.py: Python visualization script for plotting and comparing energy bands.
• PLOT-gwdos.py: Python visualization script for plotting and comparing density of states.
From now on the symbol $will indicate the shell prompt. Requirements: Bash shell. Python numpy, lxml, matplotlib.pyplot, and sys libraries. Important note: All input parameters are in atomic units! ### 1. Theoretical background: G0W0 approximation. The quasiparticle energies in the G$_{0}$W$_{0}approximation are given as a solution of the linearized quasi-particle (QP) equation [Hybertsen1986] as (1) \begin{align} \epsilon_{n\mathbf{k}}^{QP} = \epsilon_{n\mathbf{k}} + Z_{n\mathbf{k}} \left[ {\rm Re}~\Sigma_{n\mathbf{k}}(\epsilon_{n\mathbf{k}}) - V_{n\mathbf{k}}^{xc} \right]\,, \end{align} where\epsilon_{n\mathbf{k}}$are the Kohn-Sham (KS) eigenvalues,$\Sigma_{n\mathbf{k}}$and$V_{n\mathbf{k}}^{xc}$are, respectively, the diagonal matrix elements of the self-energy and the exchange-correlation (xc) potential that is employed in the single-particle KS Hamiltonian. The QP renormalization factor$Z_{n\mathbf{k}}$accounts for the energy-dependence of the self-energy. The self-energy$\Sigma(\mathbf{r},\mathbf{r}';\omega)is given by [Hedin1965] (2) \begin{align} \Sigma(\mathbf{r},\mathbf{r}';\omega) = \frac{i}{2\pi} \int G_0(\mathbf{r},\mathbf{r}';\omega+\omega')~W_0(\mathbf{r},\mathbf{r}';\omega')~e^{i\omega'\eta}~d\omega', \end{align} whereG_0$is the noninteracting single-particle Green function obtained from the Kohn-Sham states and$W_0is the dynamically screened Coulomb potential (3) \begin{align} W_0(\mathbf{r},\mathbf{r}';\omega) = \int \epsilon^{-1}(\mathbf{r},\mathbf{r}'';\omega)~v_{\rm C}(\mathbf{r}'',\mathbf{r}') d\mathbf{r}''. \end{align} Here,v_{\rm C}(\mathbf{r},\mathbf{r}') = 1/|\mathbf{r}-\mathbf{r}'|$is the Coulomb potential and$\epsilon(\mathbf{r},\mathbf{r}';\omega)$is the dielectric function calculated in the random-phase approximation (RPA). ### 2. Ground-state calculation Create a working directory tutorial-GW and move inside it. $ mkdir tutorial-gw
$cd tutorial-gw Below is the input file (input.xml) for bulk Si, which we consider in this tutorial. <input> <title>Silicon</title> <structure speciespath="$EXCITINGROOT/species">
<crystal>
<basevect>5.13 5.13 0.00</basevect>
<basevect>5.13 0.00 5.13</basevect>
<basevect>0.00 5.13 5.13</basevect>
</crystal>
<species speciesfile="Si.xml" rmt="2.1">
<atom coord="0.00 0.00 0.00"></atom>
<atom coord="0.25 0.25 0.25"></atom>
</species>
</structure>
<groundstate
do="fromscratch"
rgkmax="7.0"
ngridk="4 4 4"
xctype="LDA_PW"
>
</groundstate>
</input>
To perform the actual calculation, copy and paste the text from above into a file called input.xml within the directory tutorial-gw. Make sure to set $EXCITINGROOT to the correct exciting root directory in the speciespath attribute using the command $ SETUP-excitingroot.sh
Now, you can start the groundstate calculation by invoking the exciting executable (serial version)
$time excitingser After the run, one can inspect the INFO.OUT file for checking the convergence behaviour of the SCF cycle and for other information related to the ground-state calculation. Please note that the quality of the groundstate calculations (the starting point) plays an important role when calculating G$_{0}$W$_{0}$QP corrections. Therefore one should always check the convergence of results with respect to the groundstate parameters (such as rgkmax, ngridk, etc.) in production calculations. ### 3. GW "single-shot" run Using the data from the previous run as an input, one performs a G$_{0}$W$_{0}$cycle to calculate quasiparticle QP energies. First, one should disable the execution of the groundstate part by changing the attribute do = "fromscratch" to do = "skip" in the groundstate element. ... <groundstate do="skip" ...> </groundstate> ... To initialize G$_{0}$W$_{0}$calculations, the following block should be present in your input file: ... <gw taskname="g0w0" ngridq="2 2 2" nempty="22" ibgw="1" nbgw="10" > <mixbasis lmaxmb="3" epsmb="1.d-4" gmb="1.0" ></mixbasis> <freqgrid nomeg="32" ></freqgrid> </gw> ... Let us look closer at the gw element. Attributes and elements used in this example are Parameter Description gw/taskname Specifies the task name. gw/ ngridq Specifies the number of both the k and the q-points used specifically in GW calculations. Notice that these values are independent from the one specified by groundstate/ ngridq. gw/nempty The number of empty states used as an input in GW calculations. Note that this value is independent from the one specified inside the groundstate element. gw/ibgw, gw/nbgw Give the range of resulting quasiparticle states to be calculated. ibgw (nbgw) is the lowest (highest) band index for which the GW correction is calculated. Notice that the numbering of the band indexes is such that the index 1 is the one for the lowest occupied valence band. Furthermore, nbgw must be chosen to be larger than the number of occupied bands. gw/ mixbasis Specifies the parameters for the mixed product basis. gw/ freqgrid Specifies the frequency grid for dynamical quantities. For all available parameters and options and their description, one should check the GW Section of Input Reference. It is interesting to notice that the computational setup given for the mixed-product basis and the frequency grid is rather universal and could be applied for the most of materials without modifications. Now, you can start the calculation by invoking the exciting executable (serial version) $ time excitingser
The main output of G$_{0}$W$_{0}$ run is stored in GW_INFO.OUT and EVALQP.DAT. The former file contains information on the GW program work-flow. In particular, energy band-gap values for KS and QP spectra can be found in the end of this file.
The text file EVALQP.DAT contains all important quantities that are calculated during G$_{0}$W$_{0}$ run. Below, we describe the structure of this file. The line "k-point # 1:" starts the block of data and contains the lattice coordinates of the irreducible k-point and the corresponding Brillouin-zone-integration weight. Then, for each electronic state (within the range determined by the ibgw and nbgw input attributes) one can find the following quantities:
• the starting KS eigenenergies (E_KS);
• the quasiparticle energies as obtained by considering the exchange part of the self-energy only (E_HF);
• the G$_{0}$W$_{0}$ quasiparticle energies (E_GW);
• the diagonal matrix elements of the exchange self-energy (Sx);
• the diagonal matrix elements of the correlation self-energy (Sc);
• the diagonal matrix elements of the underlying exchange-correlation potential (Vxc);
• the energy difference E_HF$-$E_KS (DE_HF);
• the energy difference E_GW$-$E_KS (DE_GW);
• the linearization prefactor $Z_{n\mathbf{k}}$ (Znk).
It is important to note that the zero of the E_HF and E_GW quasiparticle energies listed in EVALQP.DAT is the KS Fermi energy (that can be found, e.g., in GW_INFO.OUT). Therefore, in order to obtain absolute QP eigenenergies one has to add to the data listed in EVALQP.DAT the corresponding KS Fermi energy.
### 4. Quasiparticle band structure
##### i) Band structure plot
Now, we will visualize the results to compare the Kohn-Sham (KS) and quasiparticle (QP) band structures. To calculate them, one should introduce the following changes into the input.xml file.
• After the groundstate block, add the new element block properties as indicated below.
...
<properties>
<bandstructure>
<plot1d>
<path steps="100">
<point coord=" 0.750 0.500 0.250" label="W"/>
<point coord=" 0.500 0.500 0.500" label="L"/>
<point coord=" 0.000 0.000 0.000" label="GAMMA"/>
<point coord=" 0.500 0.500 0.000" label="X"/>
<point coord=" 0.750 0.500 0.250" label="W"/>
<point coord=" 0.750 0.375 0.375" label="K"/>
</path>
</plot1d>
</bandstructure>
<dos
winddos="-0.5 0.5"
nwdos="200"
></dos>
</properties>
...
• In the gw element replace taskname = "g0w0" by taskname = "band" to calculate the QP band structure on the basis of results obtained from the previous G$_{0}$W$_{0}$ run. In this case, Fourier interpolation is used to evaluate the band energies for the k-point path specified in the properties element.
...
<gw
... >
</gw>
...
Finally, run exciting one more time.
$time excitingser & The output files BAND.OUT and BAND-QP.OUT can now be used to plot the electronic band structure. This can be easily done using the script PLOT-compare-bands.py. To execute it, type $ PLOT-compare-bands.py -15 10 gw
Comparison for GW calculations
$ The script produces a file PLOT.png which you can visualize on the screen with standard tools. The result should look like the following. ##### ii) Density of states Now, we will compute and compare the Kohn-Sham (KS) and quasiparticle (QP) density of states. To calculate them, one should introduce the following changes into the input.xml file. • In the gw element replace taskname = "band" by taskname = "dos" to calculate the QP density of states on the basis of results obtained from the previous G$_{0}$W$_{0}$run. ... <gw taskname="dos" ... > </gw> ... Now, you can run exciting: $ time excitingser &
The results for the KS and G$_{0}$W$_{0}$ density of states are stored in TDOS.OUT and TDOS-QP.OUT correspondingly. To visualize them one can simply execute the script PLOT-gwdos.py. | 2,737 | 10,102 | {"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": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-25 | longest | en | 0.716797 |
https://www.experts-exchange.com/questions/28713025/Math-between-Excel-and-approximations.html | 1,531,975,028,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590493.28/warc/CC-MAIN-20180719031742-20180719051742-00198.warc.gz | 864,320,053 | 25,458 | # Math between Excel and approximations
I need to find 2 values X1 and X2 which must respect next inequations and constrains:
Inequations:
1) (X1+X2)/360>=2.8E-03 or better >=3.0E-03 , ideal would be 3.2E-03 or 3.5E-03
2) 440*X1/(X1+X2)<350
3) 440*X2/(X1+X2)<200
4) 440*440*X1/[(X1+X2)*(X1+X2)]<1
5) 440*440*X2/[(X1+X2)*(X1+X2)]<0.6
Ideal would be to get at the 1) inequation the value >= as high as possible on the right side and at 4) and 5) inequations to get the values as low as possible at the right side as for example: at 4) instead of <1 to be <=0.6 and at 5) instead of <0.6 to be <=0.4.
Important to find the best matching values, no matter what method of calculations or approximation is used.
Any suggestions?
LVL 21
Asked:
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IT Services ConsultantCommented:
Do you know the range that one/other/both of the values of X1 & X2 can be?
Sorry, not all of us use the terms "inequation" on a daily basis...
What is a "best matching value" (of either X1 or X2)?
Is that where the right side of the "inequations" are closest to the result of the corresponding left side without breaking the constraint imposed by the arithmetic comparison?
Do all "inequations" need to be "best matching", or a majority, or just one?
0
Author Commented:
1) X1 and X2 ranges.
X1 may be between 50000 and 150000
X2 may be between 15000 and 50000
Generally speaking we may consider X1 and X2 between 15000 up to 150000, but will be restricted to lower values due to inequality constrains imposed.
2) The best matching values for X1 and X2 I do not know, but I referred to best matching value for X1 and X2 considering that will respect the inequalities 1), 4) and 5) in this way:
a. At 1) instead of >=2.8E-03 to be higher, ideal >=3.5E-03, but I can also survive with >=3.0E-03
b. At 4) instead of <1 to be <0.6, but I ca survive also with <0.7 or <0.8, because the <1 is the worst case.
c. At 5) is similar as at 4) only that instead of <0.6 to get <0.4
In such way can be understood X1 and X2 best matching values.
3) Do all "inequations" need to be "best matching", or a majority, or just one?
Basically 1) and 4) and 5), but would be better of course for all, meaning that at 2) and 3) to have lower numbers on the right side as for example at 2) instead of <350 to have <320 and at 3) instead of <200 to have <180.
You may use inequality instead of inequation if you like.
Inequation is still used as per Wikipedia:
https://en.wikipedia.org/wiki/Inequation
0
IT Services ConsultantCommented:
Thank you.
...and, again, for anybody wishing to contribute:
Are X1 &/or X2 whole numbers (i.e. not decimals) in the ranges you stated above?
If they can be decimal values, what level of precision (how many decimal places) are possible/practical?
0
Author Commented:
X1 and X2 are natural numbers, without decimal value.
But if the constrains force the solution for decimal, then you may use 1 decimal point (0.0), because from practical reasons, due to tolerances, the decimal values will be ignored anyway. For the sake of calculations, ideal values, you may use 1 decimal point.
For example I found by trials in Excel, even if a graph I think will do better, solutions very close to the limits as X1=80000 and X2=45000. I am just wondering if I can get "better values", meaning to satisfy the inequalities from the right side for lower values to 4) and 5) and higher values at 1)
A graphic I found using next webpage:
http://goo.gl/xX1VXV
0
Author Commented:
Here is an additional note regrading the numbers X1 and X2.
X1 and X2 should match "preferred numbers" from electronics as described by IEC 60063 under E series.
X1 and X2 are in fact resistors values and the above inequalities are restrictions for current, voltage and power in a specific circuit.
https://en.wikipedia.org/wiki/Preferred_number
More exactly, X1 is a E48 type, meaning 2% tolerance and can have only next values, of course you can multiply with 10 or 100 or whatever, just add zeros at the end of the following numbers as you like:
E48 ( 2%): 100 121 147 178 215 261 316 383 464 562 681 825
105 127 154 187 226 274 332 402 487 590 715 866
110 133 162 196 237 287 348 422 511 619 750 909
115 140 169 205 249 301 365 442 536 649 787 953
And X2 is a E96 type, 1% tolerance and can have only next values, of course you can again just add zeros at the end of the following numbers as you like:
E96 ( 1%): 100 121 147 178 215 261 316 383 464 562 681 825
102 124 150 182 221 267 324 392 475 576 698 845
105 127 154 187 226 274 332 402 487 590 715 866
107 130 158 191 232 280 340 412 499 604 732 887
110 133 162 196 237 287 348 422 511 619 750 909
113 137 165 200 243 294 357 432 523 634 768 931
115 140 169 205 249 301 365 442 536 649 787 953
118 143 174 210 255 309 374 453 549 665 806 976
So far I found X1=78700 and X2=46400 matching the preferred numbers and respecting the above inequalities as best matching values.
Any other suggestion how to solve this and find a better pair solution?
0
Author Commented:
Oops, I made a mistake in the 1st inequality
Instead of “(X1+X2)/360>=2.8E-03 or better >=3.0E-03 , ideal would be 3.2E-03 or 3.5E-03”
It must be „360/(X1+X2)>=2.8E-03 or better >=3.0E-03 , ideal would be 3.2E-03 or 3.5E-03”
Under these circumstances the solutions are narrowed.
Here is the theory:
https://www.sophia.org/tutorials/graphing-systems-of-non-linear-inequalities
Using next online solver we see the region of the solutions:
https://www.desmos.com/calculator/tdtxtprhmn
0
Commented:
Hi
you can use the solver add-in for excel to handle linear programming problems and get simplex optimised responses to problems, the add-in is available through the 'add-ins' manager and a brief tutorial is here:
http://www.msubillings.edu/asc/resources/math/Tutorials/FiniteMathHelps/Lin%20Prog%20with%20Excel.pdf
edit: just re-read your last post so my assumption that you meant 360/(x1... is correct. When I plug these into excels solver I do get a local optimised result but not a global one - Simplex generates an error indicating that one of the values may have gone negative - which clearly doesn't make sense for resistance (unless you inadvertently discovered a super conductor). but a non-linear solve gives a result in line with your quoted numbers from the earlier post with x1 appox 70000
I'm not any kind of expert in this field but I think you need further constraints in order to guarantee an optimum result from a mathematical perspective but in an engineering sense would it be correct to assume that a) any pair which generate a result in the solution set would be acceptable and b) certain resistor types will have an optimal set of values which may vary according to physical attributes and the optimal pairing would therefore depend on non mathematical factors?
1
Author Commented:
My previous found values X1=78700 and X2=46400 using trials in Excel cells with equations and just observations from my side are confirmed by the intersection/common region in the graph above.
How do you get negative numbers?
0
Commented:
I phrased that badly; I was using the Solver add-in which has various linear programming methods to automatically trial values and converge on an optimum solution. When using a linear method (the Simplex equation method) it hit an error condition and said one of the constraints had generated an invalid value - it doesn't tell you which one or why - since the only obvious error is a division by 0 it would indicate that it was either trying both values at 0 or one was the negative of the other. So its an artifact of the linear programming approach rather than an actual value.
Using an alternate method it converged on values of X1 from 69800 and above but warned that these were not a guaranteed global optimal because non linear methods cannot be extrapolated; however it would seem to confirm your own observations and I thought that if you had some time to investigate Solver and its various options it might help you explore the solution space.
- As I said I am not an expert , I'm just interpreting the Solver add-in output. when I've used solver in the past it was working with a quants analyst who handled the higher math whilst I did the problem setup
0
Author Commented:
Thank you for explanations and Solver suggestion.
For the moment I will not try that, because the graphical approach narrows the results good enough in a small region where I can pick up the preferred number values.
Even more, if I change the restrictions values (=the number in the right side of the inequality) as:
- for inequality 1) 3.0E-03 or higher instead of 2.8E-3 then I can see that region disappears, becoming too small.
- for inequality 4) 0.9 or smaller instead of 1.0, then the same as above.
- for inequality 5) 0.5 or smaller instead of 0.6, then the same as above.
And I changed only one constrain at one moment of the time.
The above observations indicates that is not too much room to play and optimize the result and I should be happy with what I found, because they have to match also the real values of the resistors.
0
Author Commented:
As a matter of fact, now knowing the region of solutions and the preferred numbers are few discrete values which can fit in that region, the situation becomes bearable. It is enough to make an Excel table with all possible solutions, which seems not too many and they should be around the values already found. Or another way is too look at the x and y axes for the limits of the region and see what discrete preferred numbers fit for each of the axes.
Making zoom on the obtained region with possible solutions we see the limits:
https://www.desmos.com/calculator/tdtxtprhmn
- For x (which is X1) is 75500 minimum to 85500 maximum.
- For y (which is X2) is 43000 minimum to 51500 maximum.
Now, looking at preferred numbers we get next possible values close/around the ranges above:
- For X1: 75000, 78700, 82500. And from these 3 values only 78700 is in the range actually.
- For X2: 43200, 44200, 45300, 46400, 47500, 48700, 49900, and 51100.
Now the situation became quite simple:
X1=78700 and X2={43200, 44200, 45300, 46400, 47500, 48700, 49900, 51100}
With these few numbers we can easy see which one gives the optimal results.
Here it is the Excel table:
https://drive.google.com/open?id=1zm0FWmZjBzEeCBj24J_eYt8bxQsUJii2JDso71VSuJg
It seems that X1=78700 and X2=46400, found empirical before, give optimal result being in the middle of the solutions range.
0
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Commented:
Hi viki2000, I'm glad you found the information you needed and hope the work goes well.
For future reference I'd note that you specifically asked for 'suggestions', which were given, and you will find people much more willing to respond on EE if you treat the time they devote to your requests with a little respect.
0
Author Commented:
@ regmigrant
I am really sorry that you felt disrespect from my side.
I relay have no problem to assign the points to someone who provides help, in this case your answers.
From my point of view, you can request the points to a supervisor and gladly I can give them to you.
I did not assign the points from beginning to you only because, even if you provided suggestions, they did not lead to solutions of my system inequalities and I did not even need to follow that path. I was just imagined that if 10 more people would have answered with suggestions, then I would have assigned the points to the person(s) who provided the (closest) solution(s) to my needs. That’s why I considered that I solved the problem by myself. It is not a question of disrespect.
Do you think I see it wrong?
If yes, then please ask the Zone Supervisor and you can have the 500 points for the Excel suggestion from my side, which I would have been probably considered valuable under other conditions.
I thank you for the suggestion and sorry if you felt bad, was no intention from my side.
0
Commented:
I appreciate this wasn't done out of any malice and I'm not requesting points or feeling dis-respected; I meant the comment more for your own information as some experts will block you if you do this.
It is quite common for someone to solve their own problem without help from the experts, sometimes just having the discipline to phrase the question on here can help your thought process and allow a solution to become apparent. However once the question is posed the experts do spend time formulating responses and our only reward is the points; so even if you manage to figure it out on your own it's nice to either withdraw the question and say why, or split the points amongst those who responded as an acknowledgement that they took some time to help you.
Another point is that you were not asking for a 'solution', you asked only for suggestions so technically your question was answered even if it didn't resolve your 'problem' so its worth being specific about what you expect to award points for!.
Best of luck
Reg
0
Author Commented:
Thank you for suggestion, but the solution was found with graphical approach after my research.
0
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Have a better answer? Share it in a comment. | 3,833 | 14,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | longest | en | 0.892429 |
http://mathhelpforum.com/advanced-algebra/193493-these-linear-transformations.html | 1,481,185,008,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542455.45/warc/CC-MAIN-20161202170902-00448-ip-10-31-129-80.ec2.internal.warc.gz | 165,961,818 | 9,836 | ## Are these linear transformations?
I shoud show whether $A: V \rightarrow \mathbb{F}^m$ and for all $A(v)=\begin{bmatrix} f_1(v)\\ f_2(v)\\ \vdots\\ f_m(v)\end{bmatrix}$ and $f_i$ is a linear functional $f_i \in L(V, \mathbb{F}^1), i=1,2,\dots ,m$. is a linear transformation. I think no, because the image of the zero vector is not 0. To be honest I am not really sure about these lienar functional thing.
Second. $C \in L(V), \{v_1,v_2,v_3\}$ v1 v2 v3 is a basis of V.
$C(\alpha v_1+\alpha v_2+\alpha v_2)=(\alpha-\beta+2\gamma)v_1 + (2\alpha + \gamma)v_2 + (3\alpha + \beta)v_3$. About this I know it is a linear transformation. Is it invertible. I know it would be enough to show its injectivity but how can I do this in this case?
Thank you! | 262 | 751 | {"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": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.769073 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?p=134354 | 1,600,568,358,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400193087.0/warc/CC-MAIN-20200920000137-20200920030137-00742.warc.gz | 503,628,544 | 11,136 | ## Partial Pressures Definition
Mya Majewski 1L
Posts: 89
Joined: Fri Sep 25, 2015 3:00 am
### Partial Pressures Definition
Can someone give the definition of partial pressures and why its important to use partial pressures for gases and concentration for everything else?
Shubham Rai 2C
Posts: 64
Joined: Fri Sep 28, 2018 12:27 am
### Re: Partial Pressures Definition
Partial Pressure is basically the pressure exerted by one gas in a mixture of multiple in a specific volume. You don't have to use partial pressure for gases per se, but gases are often measured by their pressure instead of their concentration. Also, you can't measure pressure for other states thus you use concentration as it is more convenient.
Jayasuriya Senthilvelan 4I
Posts: 31
Joined: Thu Jan 10, 2019 12:17 am
### Re: Partial Pressures Definition
Suppose we have a mixture of gases in a box. The total pressure of the mixture agains the walls of the box is equal to the pressure exerted by each individual gas. The pressure exerted by each individual gas is the partial pressure.
We use partial pressures for gases because I think because the partial pressure of a gas is proportional to the concentration, and so it allows us to calculate a k-value that is different from Kc, yet gives the same information as Kc. | 311 | 1,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2020-40 | latest | en | 0.922553 |
https://educationexpert.net/chemistry/2663465.html | 1,660,227,867,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571472.69/warc/CC-MAIN-20220811133823-20220811163823-00368.warc.gz | 227,874,920 | 7,211 | 13 April, 08:03
# What mass of boron sulfide must be processed with 2.1 x 10 4g of carbon to yield 3.11 x 10 4 g of boron and 1.47 x 10 5 g of carbon sulfide?
+3
1. 13 April, 09:17
0
The reaction between boron sulfide and carbon is given as:
2B2S3 + 3C → 4B + 3CS2
As per the law of conservation of mass, for any chemical reaction the total mass of reactants must be equal to the total mass of the products.
Given dа ta:
Mass of C = 2.1 * 10^ 4 g
Mass of B = 3.11*10^4 g
Mass of CS2 = 1.47*10^5
Mass of B2S3 = ?
Now based on the law of conservation of mass:
Mass of B2S3 + mass C = mass of B + mass of CS2
Mass of B2S3 + 2.1 * 10^ 4 = 3.11*10^4 + 1.47*10^5
Mass of B2S3 = 15.7 * 10^4 g | 281 | 698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2022-33 | latest | en | 0.808174 |
https://sailboat.guide/lecomte/alc-35-le-comte | 1,638,179,879,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358702.43/warc/CC-MAIN-20211129074202-20211129104202-00318.warc.gz | 580,335,482 | 13,531 | Alc 35 (LE Comte)
1968
Designer
William H. Tripp Jr.
Builder
LeComte Yachts
Associations
?
# Built
?
Hull
Monohull
Keel
Fin
Rudder
Skeg
Construction
FG
Dimensions
Length Overall
35 0 / 10.7 m
Waterline Length
26 1 / 8 m
Beam
10 0 / 3.1 m
Draft
5 6 / 1.7 m
Displacement
12,000 lb / 5,443 kg
Ballast
5,800 lb / 2,631 kg
Rig and Sails
Type
Sloop
Reported Sail Area
569′² / 52.9 m²
Total Sail Area
569′² / 52.9 m²
Sail Area
272′² / 25.3 m²
P
37 6 / 11.4 m
E
14 6 / 4.4 m
Air Draft
?
Sail Area
298′² / 27.6 m²
I
42 5 / 13 m
J
14 0 / 4.3 m
Forestay Length
44 9 / 13.6 m
Make
Vetus
Model
?
HP
?
Fuel Type
?
Fuel Capacity
?
Accomodations
Water Capacity
?
Holding Tank Capacity
?
?
Cabins
?
Hull Speed
7.2 kn
Classic: 6.85 kn
Hull Speed
The theoretical maximum speed that a displacement hull can move efficiently through the water is determined by it's waterline length and displacement. It may be unable to reach this speed if the boat is underpowered or heavily loaded, though it may exceed this speed given enough power. Read more.
Formula
Classic hull speed formula:
Hull Speed = 1.34 x √LWL
A more accurate formula devised by Dave Gerr in The Propeller Handbook replaces the Speed/Length ratio constant of 1.34 with a calculation based on the Displacement/Length ratio.
Max Speed/Length ratio = 8.26 ÷ Displacement/Length ratio.311
Hull Speed = Max Speed/Length ratio x √LWL
7.17 knots
Classic formula: 6.85 knots
Sail Area/Displacement
17.4
16-20: good performance
Sail Area / Displacement Ratio
A measure of the power of the sails relative to the weight of the boat. The higher the number, the higher the performance, but the harder the boat will be to handle. This ratio is a "non-dimensional" value that facilitates comparisons between boats of different types and sizes. Read more.
Formula
SA/D = SA ÷ (D ÷ 64)2/3
• SA: Sail area in square feet, derived by adding the mainsail area to 100% of the foretriangle area (the lateral area above the deck between the mast and the forestay).
• D: Displacement in pounds.
17.37
<16: under powered
16-20: good performance
>20: high performance
Ballast/Displacement
48.3
>40: stiffer, more powerful
Ballast / Displacement Ratio
A measure of the stability of a boat's hull that suggests how well a monohull will stand up to its sails. The ballast displacement ratio indicates how much of the weight of a boat is placed for maximum stability against capsizing and is an indicator of stiffness and resistance to capsize.
Formula
Ballast / Displacement * 100
48.34
<40: less stiff, less powerful
>40: stiffer, more powerful
Displacement/Length
299.6
275-350: heavy
Displacement / Length Ratio
A measure of the weight of the boat relative to it's length at the waterline. The higher a boat’s D/L ratio, the more easily it will carry a load and the more comfortable its motion will be. The lower a boat's ratio is, the less power it takes to drive the boat to its nominal hull speed or beyond. Read more.
Formula
D/L = (D ÷ 2240) ÷ (0.01 x LWL)³
• D: Displacement of the boat in pounds.
• LWL: Waterline length in feet
299.64
<100: ultralight
100-200: light
200-300: moderate
300-400: heavy
>400: very heavy
Comfort Ratio
29.7
20-30: coastal cruiser
Comfort Ratio
This ratio assess how quickly and abruptly a boat’s hull reacts to waves in a significant seaway, these being the elements of a boat’s motion most likely to cause seasickness. Read more.
Formula
Comfort ratio = D ÷ (.65 x (.7 LWL + .3 LOA) x Beam1.33)
• D: Displacement of the boat in pounds
• LWL: Waterline length in feet
• LOA: Length overall in feet
• Beam: Width of boat at the widest point in feet
29.74
<20: lightweight racing boat
20-30: coastal cruiser
30-40: moderate bluewater cruising boat
40-50: heavy bluewater boat
>50: extremely heavy bluewater boat
Capsize Screening
1.8
<2.0: better suited for ocean passages
Capsize Screening Formula
This formula attempts to indicate whether a given boat might be too wide and light to readily right itself after being overturned in extreme conditions. Read more.
Formula
CSV = Beam ÷ ³√(D / 64)
• Beam: Width of boat at the widest point in feet
• D: Displacement of the boat in pounds
1.75
<2: better suited for ocean passages
>2: better suited for coastal cruising
For Sale
Have a sailboat to sell? | 1,261 | 4,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-49 | latest | en | 0.789731 |
http://www.longrangehunting.com/forums/f19/hypothetical-bc-question-69402/ | 1,496,117,487,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463613780.89/warc/CC-MAIN-20170530031818-20170530051818-00020.warc.gz | 696,457,290 | 12,821 | Long Range Hunting Online Magazine Hypothetical BC question
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Rifles, Bullets, Barrels & Ballistics
# Hypothetical BC question
#1
03-06-2011, 03:56 PM
Gold Member Join Date: Oct 2008 Location: Wherever the Marine Corps sends us. Posts: 676
Hypothetical BC question
Assuming two bullets have the same exterior measurements, are of the ULD type design with RBBT, and both balance well. If one bullet weighs say 215 gr and the other weighs 240 gr, which would have the higher BC? Let's say these are .30 cal bullets traveling 3200 fps for the 215 and 3150 for the 240. I'm not trying to start an argument, just curious as to everyone's opinions.
Thanks
__________________
ACTIVE DUTY MARINE CORPS '96-'04
SERGEANT (MED RET)
COMBAT ENGINEER, MOUT INSTRUCTOR, RIFLE/PISTOL INSTRUCTOR
"Be polite, be professional, but have a plan to kill everyone you meet."
"I come in peace. I didn't bring artillery. But I'm pleading with you, tears in my eyes: If you f*ck with me, I'll kill you all." Marine General Mattis
Custom built 7mm SAUM
#2
03-06-2011, 05:24 PM
Silver Member Join Date: Sep 2008 Posts: 327
Re: Hypothetical BC question
wikipedia - ... "BC is a function of mass, diameter, and drag coefficient." It is given by the mass of the object divided by the diameter squared that it presents to the airflow divided by a dimensionless constant i that relates to the aerodynamics of its shape. Ballistic coefficient has units of lb/inē or kg/mē. BCs for bullets are normally stated in lb/inē by their manufacturers without referring to this unit..."
or in other words,
BC = sectional density/form factor
__________________
#3
03-06-2011, 05:27 PM
Platinum Member Join Date: Feb 2004 Location: Pennsyltucky Posts: 2,630
Re: Hypothetical BC question
the heavier bullet will have a higher BC.
#4
03-06-2011, 05:29 PM
Official LRH Sponsor Join Date: Mar 2007 Posts: 634
Re: Hypothetical BC question
To put the above information into the context of your specific question...
If they're truly the same shape,
the 240 grain bullet will have a BC that's 1.116 times higher than the 215 at all speeds.
Because 240/215 = 1.116.
That's 11.6% higher BC for the 240 grain.
So if the 215 has a BC of .665 at a given speed (or averaged over a given speed range), then the 240 will have a BC of .665*1.116 = .742.
Bear in mind the 240 will suffer depressed MV, so the performance of the heavier bullet won't be as much better as the BC implies, but it will be a little better even at a reduced MV assuming equal chamber pressure for both bullets.
-Bryan
#5
03-07-2011, 12:18 AM
Platinum Member Join Date: May 2008 Location: South of Canada and North of Wyoming Posts: 6,068
Re: Hypothetical BC question
Quote:
Quote:
Originally Posted by JARHEAD1371 Assuming two bullets have the same exterior measurements, are of the ULD type design with RBBT, and both balance well. If one bullet weighs say 215 gr and the other weighs 240 gr, which would have the higher BC? Let's say these are .30 cal bullets traveling 3200 fps for the 215 and 3150 for the 240. I'm not trying to start an argument, just curious as to everyone's opinions. Thanks
Well it all depends.....
#6
03-07-2011, 03:07 AM
Gold Member Join Date: Oct 2008 Location: Wherever the Marine Corps sends us. Posts: 676
Re: Hypothetical BC question
Quote:
Originally Posted by bsl135 To put the above information into the context of your specific question... If they're truly the same shape, the 240 grain bullet will have a BC that's 1.116 times higher than the 215 at all speeds. Because 240/215 = 1.116. That's 11.6% higher BC for the 240 grain. So if the 215 has a BC of .665 at a given speed (or averaged over a given speed range), then the 240 will have a BC of .665*1.116 = .742. Bear in mind the 240 will suffer depressed MV, so the performance of the heavier bullet won't be as much better as the BC implies, but it will be a little better even at a reduced MV assuming equal chamber pressure for both bullets. -Bryan
Thanks for breaking it down for me. It makes more sense now.
__________________
ACTIVE DUTY MARINE CORPS '96-'04
SERGEANT (MED RET)
COMBAT ENGINEER, MOUT INSTRUCTOR, RIFLE/PISTOL INSTRUCTOR
"Be polite, be professional, but have a plan to kill everyone you meet."
"I come in peace. I didn't bring artillery. But I'm pleading with you, tears in my eyes: If you f*ck with me, I'll kill you all." Marine General Mattis
Custom built 7mm SAUM
#7
03-07-2011, 10:47 AM
Platinum Member Join Date: Apr 2010 Location: Allen, TX Posts: 2,608
Re: Hypothetical BC question
Also, they won't be exactly the same shape if they are made from the same material/construction. The heavier bullet will be longer.
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All times are GMT -5. The time now is 11:11 PM. | 1,415 | 5,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-22 | latest | en | 0.90507 |
http://www.jiskha.com/display.cgi?id=1235185666 | 1,498,616,064,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128322275.28/warc/CC-MAIN-20170628014207-20170628034207-00691.warc.gz | 589,286,789 | 3,890 | # Math
posted by .
Matt has twice as many stickers as David. If David has d stickers, and Matt gives David 10 stickers, how many stickers does each have in terms of d?
• Math -
m = 2 d (initially)
After Matt gives Dave 10, they have new amounts m' and d'.
m' = 2 d - 10
d' = d + 10
• Math -
David: d stickers
Matt: 2d stickers
after give-away ...
David : d+10
Matt : 2d - 10
The question is not clearly stated.
Since you used the present tense throughout, one is forced to wonder if Matt "has" twice a many as David before or after the transfer.
I actually interpreted it as | 162 | 585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-26 | latest | en | 0.962598 |
https://cs.stackexchange.com/questions/120176/is-there-a-name-for-a-graph-whose-vertices-consist-of-edges-of-an-existing-graph | 1,723,179,597,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640759711.49/warc/CC-MAIN-20240809044241-20240809074241-00250.warc.gz | 147,879,074 | 40,337 | # Is there a name for a graph whose vertices consist of edges of an existing graph connected by an edge if they shared a common vertex?
I was wondering if there is a name for this construction: Take a graph $$G$$ and construct a new graph $$G'$$ in which the edges of $$G$$ now become vertices, and "vertices become edges" in the sense that vertices are joined by edge in $$G'$$ if they as edges in $$G$$ shared a common vertex.
More precisely: Given a graph $$G=(V=\{v_1,v_2,\dots,v_n\},E=\{e_1,e_2,\dots,e_k\})$$ where each $$e_i=\{u,v\}\subset V$$ we define the graph $$G'=(V'=\{e_1,e_2,\dots,e_k\},E')$$ so that $$\{e_i,e_j\}\in E' \iff e_i\cap e_j\neq\emptyset$$.
I don't know if this can be used for anything (my guess would be it can't) it just came to my mind and thought that perhaps this has a name.
This is called the line graph of $$G$$. It actually has a wide variety of uses, as seen on that Wikipedia page, and the terminology is sufficiently standard that you should be able to mention it in a paper without defining it. | 295 | 1,040 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-33 | latest | en | 0.949582 |
https://didactalia.net/pt/comunidade/materialeducativo/recursos/tag/denominator?skos:ConceptID=gnoss:2CA24DEC-226A-4D18-ABA6-136364277638 | 1,586,027,866,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370524604.46/warc/CC-MAIN-20200404165658-20200404195658-00280.warc.gz | 417,612,812 | 23,095 | Que pode fazer?
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## Adding and Subtracting Rational Expressions with U...
This lesson covers adding and subtracting rational expressions where the denominators are different.
It is an educational content of CK-12 Foundation (to access some of the CK-12 contents you must be l ...
Hipervinculo Matemática
## Adding and Subtracting Rational Expressions with L...
This lesson covers adding and subtracting rational expressions when the denominators are the same.
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Hipervinculo Matemática
## Adding and Subtracting Rational Expressions where ...
This lesson covers adding and subtracting rational expressions when one of the denominators is the LCD.
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## Write, Compare, and Order Ratio? At grade
Write ratios, order ratios from least to greatest and greatest to least and compare ratio
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## Use Unit Rates and Equivalent Rate? At grade
Define and identify unit rates, then find equivalent rates
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## Properties of Rational Number? Basic
Learn to simplify and order rational numbers.
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Identify rational numbers as the ratio of two integers
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## Positive and Negative Fraction and Decimal Compari...
Compare and order positive and negative fractions and decimals.
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## Mixed Number and Fraction Estimation, grade ? At g...
Approximate fractions and mixed numbers by using common benchmarks
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## Identification and Writing of Equivalent Rates, gr...
Identify and write equivalent rates
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Ir a Odite | 731 | 3,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-16 | latest | en | 0.841556 |
https://crypto.stackexchange.com/questions/65965/determine-rsa-modulus-from-encryption-oracle | 1,716,523,996,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00396.warc.gz | 149,333,978 | 39,939 | # Determine RSA modulus from encryption oracle
Suppose we have an RSA encryption oracle $$E(m)$$ which basically just calculates $$m^e \mod n$$ for a given message $$m$$. Here $$e=65537$$ is known but $$n$$ is not. Can we determine the value of $$n$$ without trying all values below $$n$$, assuming $$2^{1023}\leq n <2^{1024}$$?
I thought of using $$E(-1)$$ but sadly only positive messages are allowed. Another idea that i had was to just use $$E(2)$$ and if $$e$$ was small it would wrap the modulus only a few times and we would be able to determine $$n$$. Sadly $$2^{65537}$$ is many times bigger than $$n$$ so it doesn't work either.
• In the present question $e$ is known and $m$ can be chosen. It can only make finding $n$ easier than in said similar question.
– fgrieu
Dec 18, 2018 at 18:52
• Its for a private CTF challenge Dec 18, 2018 at 19:09
## 1 Answer
We have $$E(x) = x^{65537} - k \cdot n$$, for some integer $$k$$ (which will be different for different values of $$x$$), and the unknown modulus $$n$$, and hence $$x^{65537} - E(x)$$ will always be a multiple of $$n$$.
So, compute:
$$\gcd( 2^{65537} - E(2), 3^{65537} - E(3) )$$
That will be $$n$$ multiplied by some integer which is likely to be small...
As similar method (that doesn't involve computing on such large integers, and even works even if you don't know $$e$$) is to compute:
$$\gcd( E(2)^2 - E(4), E(3)^2 - E(9) )$$
How this works should be fairly obvious... | 440 | 1,452 | {"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": 24, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-22 | latest | en | 0.913663 |
https://www.got-it.ai/solutions/excel-chat/excel-problem/34633-column-function | 1,722,770,900,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640398413.11/warc/CC-MAIN-20240804102507-20240804132507-00009.warc.gz | 642,150,104 | 13,034 | # Excel - COLUMN Function Problem - Expert Solution
Question description:
This user has given permission to use the problem statement for this blog.
Two columns of data, trying to fill in the blanks in the second column. Example: A = B A = Blank A = B C = D C = D C = Blank C = D
Solved by S. Q. in 11 mins
This is the chat thread from the real Excelchat help session. It contains no private user information.
Excelchat Expert 07/06/2018 - 05:13
Hello
User 07/06/2018 - 05:13
Hi
Excelchat Expert 07/06/2018 - 05:13
Welcome to excelchat
Excelchat Expert 07/06/2018 - 05:14
I want to understand the problem clearly, can you please explain it once again by showing some sample data in the preview window?
User 07/06/2018 - 05:14
I can't show example due to the confidential nature of the data, but I will try to explain using the window
User 07/06/2018 - 05:15
Trying to fill in the blanks with the missing data
Excelchat Expert 07/06/2018 - 05:16
Meaning Bob Smith will be filled with 12345 in B3?
User 07/06/2018 - 05:16
Right
User 07/06/2018 - 05:16
Susan will be filled in with 34567
Excelchat Expert 07/06/2018 - 05:17
Okay, understood.
Excelchat Expert 07/06/2018 - 05:17
For this, the data need to be in sorted order.
User 07/06/2018 - 05:17
Column A is already sorted in alphabetical order
Excelchat Expert 07/06/2018 - 05:18
Great.
Excelchat Expert 07/06/2018 - 05:18
Excelchat Expert 07/06/2018 - 05:19
In C1, the formula is =B1. This is the first row.
Excelchat Expert 07/06/2018 - 05:19
Then in C2 the formula is =IF(B2="",C1,B2)
Excelchat Expert 07/06/2018 - 05:19
We can drag this formula down to fill in the empty cells.
Excelchat Expert 07/06/2018 - 05:20
Please let me know if you have any question.
User 07/06/2018 - 05:21
Trying it now
Excelchat Expert 07/06/2018 - 05:21
Okay, I'm here if you need me.
User 07/06/2018 - 05:22
Wow, I think that worked!
Excelchat Expert 07/06/2018 - 05:22
Great.
Excelchat Expert 07/06/2018 - 05:22
Anything else?
User 07/06/2018 - 05:23
I don't think so. Thank you so much
Excelchat Expert 07/06/2018 - 05:23
In that case, the session will end in 9 minutes unless you opt to end it now.
User 07/06/2018 - 05:24
I think we're good. Thanks again
Excelchat Expert 07/06/2018 - 05:24
Thanks for visiting excelchat, have a great day.
User 07/06/2018 - 05:24
You too
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Topic: Non-Euclidean Arithmetic
Replies: 108 Last Post: Sep 13, 2012 3:39 PM
Search Thread: Advanced Search
Messages: [ Previous | Next ]
Paul A. Tanner III Posts: 5,920 Registered: 12/6/04
Re: Non-Euclidean Arithmetic
Posted: Sep 8, 2012 6:56 PM
Plain Text Reply
On Sat, Sep 8, 2012 at 4:35 PM, kirby urner <kirby.urner@gmail.com> wrote:
> On Fri, Sep 7, 2012 at 4:14 PM, Jonathan Crabtree
> <sendtojonathan@yahoo.com.au> wrote:
>> Thank you Joe!
>>
>> That is the 'what and why' of ASAPANS (as simple as possible and no simpler).
>>
>> Multiplication should have been clarified by Einstein.
>>
>> Multiplication is NOT scaling.
>>
>
> I think a fairly common mistake among mathematicians is to think their
> job is to develop "global meanings" in the sense of their being a top
> level namespace that anchors all the others, a fixed firmament.
>
I think you are making a mistake of confusing the drive to generalize things with this attempt to impose as you say.
First, I don't think mathematicians do this imposing. It's this imposing that mathematicians are actually against. Sure, the drive to generalize may result in trying to see whether there is any "common denominator", but that is not this imposing.
This type of imposing is what Devlin I think is arguing against, that the operation of binary multiplication is - where "is" means the sense of "is identical to" - the operation of repeated addition. Sure, these different operations sometimes give identical results in some contexts, but that is different than saying that they are identical operations, that one operation is the other operation.
It seems to me that mathematicians say that modeling an operation is an OK thing, but if there are going to be models then please don't exclude the ones that have more generality, as the scaling model has more generality than the repeated addition model, as general as any set on which we can define magnitude or absolute value or if we wish, the geometric language of distance or length from 0.
Side note: No one is saying that scaling is what multiplication is - only that scaling is a model for multiplication.
With repeated addition as a model we have to keep changing the definition of repeated addition as we go to more and more general number systems and to the breaking point when we hit multiplication of two irrationals. But with scaling we don't have to change the definition of scaling at all. Even for multiplying two irrationals e and pi, where 1, e, pi, and e(pi) are viewed as distances from 0, we use the same exact model of scaling we use for multiplying two naturals: We have these ratios as a model: pi is to 1 as e(pi) is to e, and via commutativity we have e is to 1 as e(pi) is to pi. Or if we wish to use equality on fractions, pi/1 = e(pi)/e, and via commutativity we have e/1 = e(pi)/pi. And we can model this by construction on a graph: Connect 1 on the x-axis to pi on the y-axis, and, parallel to that drawn line segment, connect e on the x-axis to the y-axis, and this point on the y-axis is e(pi). And via commutativity we have the other way: Connect 1 on the x-axis to e on the y-axis, and, parallel to that drawn line segment, connect pi on the x-axis to the y-axis, and this point on the y-axis is e(pi).
No one is saying don't use repeated addition *as a model* in the naturals, and no is saying don't use the repeated idea later with the redefinitions. But just don't exclude the one and only model that needs no redefinition all the way up through multiplying two irrationals, especially the one and only model that has the added and probably vastly more important benefit of teaching proportional reasoning from day one via reasoning on equal or equivalent ratios or fractions (which is a great foundation for working with percentages - and I mean using the equivalent ratios or fractions x/100 = y/z.)
The lack of fluent proportional reasoning in a large percentage of students all the way up through adulthood is one of big problems that must be addressed, and a least including this scaling way of modeling multiplication using equivalent or equal ratios or fractions from day one is a ready-made way of attacking this problem.
Date Subject Author
9/1/12 Jonathan J. Crabtree
9/1/12 Paul A. Tanner III
9/2/12 kirby urner
9/3/12 Paul A. Tanner III
9/3/12 kirby urner
9/3/12 Paul A. Tanner III
9/4/12 kirby urner
9/4/12 Paul A. Tanner III
9/4/12 kirby urner
9/5/12 Paul A. Tanner III
9/5/12 Robert Hansen
9/6/12 kirby urner
9/1/12 kirby urner
9/1/12 Joe Niederberger
9/1/12 Wayne Bishop
9/1/12 Joe Niederberger
9/2/12 Robert Hansen
9/3/12 Paul A. Tanner III
9/3/12 Robert Hansen
9/5/12 Paul A. Tanner III
9/3/12 Joe Niederberger
9/3/12 Robert Hansen
9/5/12 Paul A. Tanner III
9/3/12 Joe Niederberger
9/3/12 Paul A. Tanner III
9/4/12 Joe Niederberger
9/5/12 Paul A. Tanner III
9/5/12 Joe Niederberger
9/5/12 Robert Hansen
9/5/12 Paul A. Tanner III
9/5/12 Joe Niederberger
9/5/12 kirby urner
9/5/12 Joe Niederberger
9/5/12 Robert Hansen
9/5/12 Joe Niederberger
9/6/12 Joe Niederberger
9/8/12 Robert Hansen
9/7/12 Jonathan J. Crabtree
9/8/12 kirby urner
9/8/12 Paul A. Tanner III
9/10/12 kirby urner
9/10/12 Paul A. Tanner III
9/10/12 kirby urner
9/10/12 Paul A. Tanner III
9/10/12 kirby urner
9/8/12 Robert Hansen
9/8/12 kirby urner
9/8/12 Robert Hansen
9/8/12 kirby urner
9/8/12 Joe Niederberger
9/8/12 Jonathan J. Crabtree
9/9/12 kirby urner
9/8/12 Clyde Greeno @ MALEI
9/8/12 Jonathan J. Crabtree
9/8/12 Jonathan J. Crabtree
9/8/12 Joe Niederberger
9/8/12 Joe Niederberger
9/9/12 Paul A. Tanner III
9/9/12 Robert Hansen
9/9/12 Paul A. Tanner III
9/9/12 Robert Hansen
9/9/12 Paul A. Tanner III
9/9/12 Robert Hansen
9/10/12 Paul A. Tanner III
9/10/12 Wayne Bishop
9/10/12 Paul A. Tanner III
9/9/12 Joe Niederberger
9/10/12 Clyde Greeno @ MALEI
9/9/12 Joe Niederberger
9/9/12 Paul A. Tanner III
9/9/12 Wayne Bishop
9/9/12 Paul A. Tanner III
9/10/12 Wayne Bishop
9/10/12 Paul A. Tanner III
9/9/12 Paul A. Tanner III
9/9/12 Joe Niederberger
9/10/12 Clyde Greeno @ MALEI
9/10/12 Joe Niederberger
9/10/12 Paul A. Tanner III
9/10/12 Joe Niederberger
9/10/12 Clyde Greeno @ MALEI
9/10/12 Joe Niederberger
9/11/12 Joe Niederberger
9/11/12 Paul A. Tanner III
9/11/12 kirby urner
9/11/12 Paul A. Tanner III
9/11/12 kirby urner
9/11/12 Paul A. Tanner III
9/11/12 kirby urner
9/12/12 Paul A. Tanner III
9/12/12 kirby urner
9/12/12 Paul A. Tanner III
9/12/12 kirby urner
9/13/12 Paul A. Tanner III
9/13/12 kirby urner
9/13/12 Paul A. Tanner III
9/11/12 Joe Niederberger
9/11/12 Joe Niederberger
9/11/12 kirby urner
9/11/12 Joe Niederberger
9/11/12 Joe Niederberger
9/11/12 Paul A. Tanner III
9/11/12 israeliteknight
9/11/12 Joe Niederberger
9/12/12 Paul A. Tanner III
9/12/12 kirby urner
9/12/12 Paul A. Tanner III
9/12/12 kirby urner
9/11/12 Jonathan J. Crabtree
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5 Answers · Science & Mathematics · 31/01/2011 | 1,168 | 2,954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | longest | en | 0.670813 |
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Write the following numbers as an $(\alpha + \beta i)$ which means as an algebraic expression : $[2(\cos5 + i\sin5)]^{12}$ and also $(1+i)^8$ . So,as for the first one, I tried writing $2^{12}(\cos5 + i\sin5)^{12}$ but could not take it further. As for the second one, I took the modulus of $z$, which is $1$ but I don't know how to find $\alpha$ or $\beta$.
-
What's for you "an algebraic expression"? Because both expressions you wrote are "algebraic", so what did you actually meant? Perhaps Polar something? – DonAntonio Dec 9 '12 at 14:26
No the first one is trigonometric expression...an algebric expression has the form : alfa +beta*i.. – einsteinwannabe Dec 9 '12 at 14:27
@Beyondhere: you are missing the point. Any number $\alpha+i\beta$ can be written using sines and cosines. The point of the question is to get rid of the power. – Martin Argerami Dec 9 '12 at 14:52
Using de Moivre's formula, $(\cos 5^{\circ}+i\sin 5^{\circ})^{12}=\cos 60^{\circ}+i\sin60^{\circ}=\frac12+i\frac{\sqrt3}2$
So, $$\{2(\cos 5^{\circ}+i\sin 5^{\circ})\}^{12}$$ $$=2^{12}(\cos 5^{\circ}+i\sin 5^{\circ})^{12}=4096\left(\frac12+i\frac{\sqrt3}2\right)=2048+i2048\sqrt3$$
$1+i=r(\cos \theta+i\sin\theta)$
Equating the real & the imaginary parts, $r\cos \theta=r\sin\theta=1$ where $r>0$
Squaring & adding we get, $r=\sqrt 2$
On division, $\tan \theta =1$ so $\theta=45^{\circ}$ as $\cos \theta,\sin\theta>0$
So, $1+i=\sqrt2(\cos45^{\circ}+i\sin45^{\circ})$
Hence, $(1+i)^8=(\sqrt2)^8\{\cos45^{\circ}+i\sin45^{\circ}\}^8=2^4(\cos360^{\circ}+i\sin360^{\circ})=16$
Alternatively, $(1+i)^2=1+i^2+2i=2i,(1+i)^4=(2i)^2=-4,(1+i)^8=(-4)^2=16$
-
In the first one you have 2*(cos5..... but thank you anyway :) – einsteinwannabe Dec 9 '12 at 14:29
I've just shown the power part – lab bhattacharjee Dec 9 '12 at 14:30
If incompleteness is the reason behind the down-voting, I've just completed it. – lab bhattacharjee Dec 9 '12 at 16:40
Use the standard formula: $r(\cos\theta + i \sin\theta) = re^{i\theta}.$ From this, we can conclude that $$[r(\cos\theta + i \sin\theta)]^n = [re^{i\theta}]^n = r^ne^{in\theta} = r^n(\cos(n\theta)+i\sin(n\theta)).$$
In the first case, $r=2$, $\theta =5$ and $n=12$. Putting this into the line above:
$$[2(\cos5+i\sin5)]^{12} = 2^{12}(\cos60+i\sin60).$$
Assuming that you are working in degrees, $\cos60 = 1/2$ while $\sin 60 = \sqrt{3}/2.$ Thus:
$$[2(\cos5+i\sin5)]^{12} = 4096\left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = 2048 + i \, 2048\sqrt{3} \, .$$
In the second case, note that $|1+i| = \sqrt{2}$ while $\arg(1+i) = 45^{\circ}.$ Thus:
$$1+i = \sqrt{2}(\cos45 + i \sin45) \, .$$
In this case, $r=\sqrt{2}$, $\theta = 45$ and $n = 8$, thus:
$$(1+i)^8 = 16(\cos360 + i \sin360) = 16 \, .$$
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### Author Topic: Large circle problem (Read 3879 times)
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#### Agwelder
• 4
##### Large circle problem
« on: November 01, 2007, 01:02:26 PM »
Howdy,
We are having a problem cutting a circle with approximately 12" radius. The item is a circular tool that needs to be reduced in diameter for better fit. Our table has a y-axis travel limit of 15", so we are cutting a little more than half of the part at a time, then rotating 180deg and running the toolpath again.
The toolpath begins at the proper radius (12.24"), but as reaches the midpoint (X0) the radius is reduced to 12.2137", as it reaches the end of the toolpath the radius is back to 12.24". We have tried this with backlash comp. on and off, CV on and off, and even ran a sim on other computers. The mechanical counters on our machine confirm our DRO's so we don't believe we are completely crazy. (yet)
%
N100 G20
N102 G0 G17 G40 G49 G80 G90
N104 T239 M6
N106 G0 G90 X12.234 Y-.3823 S1069 M3
N108 H239 Z2.
N110 Z.35
N112 G1 Z-.005 F6.42
N114 G3 X-12.234 Y.3823 R12.24 F10.
N116 Y-.3825 R12.24
N118 G1 Z.095 F6.42
N120 G0 Z2.
N122 M5
N124 G0 X0. Y0.
N128 M30
%
Thanks in advance for any help.
It's all fun and games till the Geckos die.
#### vmax549
##### Re: Large circle problem
« Reply #1 on: November 02, 2007, 03:41:24 PM »
Ok a couple of days ago you determened that your machine cut a circle to .0005" tolerance so we can assume that the machine can cut a good circle.
Are you cutting an already existing part?? Are you SURE you have it fixtured so that you are acutally cutting from the true center of the arc radius and are not slightly off center???
Did you measure the finished part to determine the error??
Just a thought (;-) TP
« Last Edit: November 02, 2007, 03:43:13 PM by vmax549 »
#### Agwelder
• 4
##### Re: Large circle problem
« Reply #2 on: November 02, 2007, 07:12:30 PM »
Yes, we are cutting an existing part. It is used to locate multiple components, one of which is mounted in the center via a 1" hole. We used this hole to center the part in our mill setup and the only time the part was moved, it was rotated at this point. There was no more than a .001" of play at the center. The finished part was measured 180 deg apart on the x-axis. There was only .0006" error at these points. At the points on the y-axis the error was nearly .030". There was no re-zeroing and the part was never removed from the mill. To measure the part we use an edge finder and jog the table in .0001" increments.
It's all fun and games till the Geckos die.
#### vmax549
##### Re: Large circle problem
« Reply #3 on: November 02, 2007, 11:46:02 PM »
Ok the next thing I would do is to reprogram the code from a radius format arc to a center format arc to see if it changes anything.
Can you mount an indicator and spin the part slowly to double check runout?
(;-) TP
(;-) TP
#### Graham Waterworth
• 2,437
• Yorkshire Dales, England
##### Re: Large circle problem
« Reply #4 on: November 03, 2007, 04:58:40 AM »
Hi,
sorry to say its your code that,s wrong, your machine is probably fine, try this code.
%
N100 G20
N102 G0 G17 G40 G49 G80 G90
N104 T239 M6
N106 G0 G90 X12.234 Y-.3823 S1069 M3
N108 H239 Z2.
N110 Z.35
N112 G01 Z-0.005 F6.42
N114 G03 X0. Y12.24 R12.24 F10.
N116 X-12.234 Y-0.383 R12.24
N118 G1 Z.095 F6.42
N120 G0 Z2.
N122 M5
N124 G0 X0. Y0.
N128 M30
%
Graham.
Without engineers the world stops
#### vmax549
##### Re: Large circle problem
« Reply #5 on: November 03, 2007, 10:58:15 AM »
Graham did you ever publish your Gcode help manual. I have some students that I help from time to time to understand CNC (local High school)
My manual Gcoding is way too rusty (CAM) Your book would be helpfull.
(;-) TP
#### Graham Waterworth
• 2,437
• Yorkshire Dales, England
##### Re: Large circle problem
« Reply #6 on: November 03, 2007, 11:06:12 AM »
I am just in the middle of a re-write, I was asked by one of the model engineering mags to do a 12 to 24 month CNC set up and programming tutorial, it has to cover milling and lathe work with working examples for each issue. Starting off simple and working up to complex parts. It is going to be much better than the original.
Graham.
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# Direct product
In mathematics, one can often define a direct product of objects already known, giving a new one. This generalizes the Cartesian product of the underlying sets, together with a suitably defined structure on the product set. More abstractly, one talks about the product in category theory, which formalizes these notions.
Examples are the product of sets, groups (described below), rings, and other algebraic structures. The product of topological spaces is another instance.[dubious ]
There is also the direct sum – in some areas this is used interchangeably, while in others it is a different concept.
## Examples
• If we think of ${\displaystyle \mathbb {R} }$ as the set of real numbers, then the direct product ${\displaystyle \mathbb {R} \times \mathbb {R} }$ is just the Cartesian product ${\displaystyle \{(x,y):x,y\in \mathbb {R} \}.}$
• If we think of ${\displaystyle \mathbb {R} }$ as the group of real numbers under addition, then the direct product ${\displaystyle \mathbb {R} \times \mathbb {R} }$ still has ${\displaystyle \{(x,y):x,y\in \mathbb {R} \}}$ as its underlying set. The difference between this and the preceding example is that ${\displaystyle \mathbb {R} \times \mathbb {R} }$ is now a group, and so we have to also say how to add their elements. This is done by defining ${\displaystyle (a,b)+(c,d)=(a+c,b+d).}$
• If we think of ${\displaystyle \mathbb {R} }$ as the ring of real numbers, then the direct product ${\displaystyle \mathbb {R} \times \mathbb {R} }$ again has ${\displaystyle \{(x,y):x,y\in \mathbb {R} \}}$ as its underlying set. The ring structure consists of addition defined by ${\displaystyle (a,b)+(c,d)=(a+c,b+d)}$ and multiplication defined by ${\displaystyle (a,b)(c,d)=(ac,bd).}$
• However, if we think of ${\displaystyle \mathbb {R} }$ as the field of real numbers, then the direct product ${\displaystyle \mathbb {R} \times \mathbb {R} }$ does not exist – naively defining addition and multiplication componentwise as in the above example would not result in a field since the element ${\displaystyle (1,0)}$ does not have a multiplicative inverse.
In a similar manner, we can talk about the direct product of finitely many algebraic structures, for example, ${\displaystyle \mathbb {R} \times \mathbb {R} \times \mathbb {R} \times \mathbb {R} .}$ This relies on the fact that the direct product is associative up to isomorphism. That is, ${\displaystyle (A\times B)\times C\cong A\times (B\times C)}$ for any algebraic structures ${\displaystyle A,}$ ${\displaystyle B,}$ and ${\displaystyle C}$ of the same kind. The direct product is also commutative up to isomorphism, that is, ${\displaystyle A\times B\cong B\times A}$ for any algebraic structures ${\displaystyle A}$ and ${\displaystyle B}$ of the same kind. We can even talk about the direct product of infinitely many algebraic structures; for example we can take the direct product of countably many copies of ${\displaystyle \mathbb {R} ,}$ which we write as ${\displaystyle \mathbb {R} \times \mathbb {R} \times \mathbb {R} \times \dotsb .}$
## Group direct product
In group theory one can define the direct product of two groups ${\displaystyle (G,\circ )}$ and ${\displaystyle (H,\cdot ),}$ denoted by ${\displaystyle G\times H.}$ For abelian groups which are written additively, it may also be called the direct sum of two groups, denoted by ${\displaystyle G\oplus H.}$
It is defined as follows:
• the set of the elements of the new group is the Cartesian product of the sets of elements of ${\displaystyle G{\text{ and }}H,}$ that is ${\displaystyle \{(g,h):g\in G,h\in H\};}$
• on these elements put an operation, defined element-wise:
${\displaystyle (g,h)\times \left(g',h'\right)=\left(g\circ g',h\cdot h'\right)}$
Note that ${\displaystyle (G,\circ )}$ may be the same as ${\displaystyle (H,\cdot ).}$
This construction gives a new group. It has a normal subgroup isomorphic to ${\displaystyle G}$ (given by the elements of the form ${\displaystyle (g,1)}$), and one isomorphic to ${\displaystyle H}$ (comprising the elements ${\displaystyle (1,h)}$).
The reverse also holds. There is the following recognition theorem: If a group ${\displaystyle K}$ contains two normal subgroups ${\displaystyle G{\text{ and }}H,}$ such that ${\displaystyle K=GH}$ and the intersection of ${\displaystyle G{\text{ and }}H}$ contains only the identity, then ${\displaystyle K}$ is isomorphic to ${\displaystyle G\times H.}$ A relaxation of these conditions, requiring only one subgroup to be normal, gives the semidirect product.
As an example, take as ${\displaystyle G{\text{ and }}H}$ two copies of the unique (up to isomorphisms) group of order 2, ${\displaystyle C^{2}:}$ say ${\displaystyle \{1,a\}{\text{ and }}\{1,b\}.}$ Then ${\displaystyle C_{2}\times C_{2}=\{(1,1),(1,b),(a,1),(a,b)\},}$ with the operation element by element. For instance, ${\displaystyle (1,b)^{*}(a,1)=\left(1^{*}a,b^{*}1\right)=(a,b),}$ and${\displaystyle (1,b)^{*}(1,b)=\left(1,b^{2}\right)=(1,1).}$
With a direct product, we get some natural group homomorphisms for free: the projection maps defined by
{\displaystyle {\begin{aligned}\pi _{1}:G\times H\to G,\ \ \pi _{1}(g,h)&=g\\\pi _{2}:G\times H\to H,\ \ \pi _{2}(g,h)&=h\end{aligned}}}
are called the coordinate functions.
Also, every homomorphism ${\displaystyle f}$ to the direct product is totally determined by its component functions ${\displaystyle f_{i}=\pi _{i}\circ f.}$
For any group ${\displaystyle (G,\circ )}$ and any integer ${\displaystyle n\geq 0,}$ repeated application of the direct product gives the group of all ${\displaystyle n}$-tuples ${\displaystyle G^{n}}$ (for ${\displaystyle n=0,}$ this is the trivial group), for example ${\displaystyle \mathbb {Z} ^{n}}$ and ${\displaystyle \mathbb {R} ^{n}.}$
## Direct product of modules
The direct product for modules (not to be confused with the tensor product) is very similar to the one defined for groups above, using the Cartesian product with the operation of addition being componentwise, and the scalar multiplication just distributing over all the components. Starting from ${\displaystyle \mathbb {R} }$ we get Euclidean space ${\displaystyle \mathbb {R} ^{n},}$ the prototypical example of a real ${\displaystyle n}$-dimensional vector space. The direct product of ${\displaystyle \mathbb {R} ^{m}}$ and ${\displaystyle \mathbb {R} ^{n}}$ is${\displaystyle \mathbb {R} ^{m+n}.}$
Note that a direct product for a finite index ${\displaystyle \prod _{i=1}^{n}X_{i}}$ is canonically isomorphic to the direct sum ${\displaystyle \bigoplus _{i=1}^{n}X_{i}.}$ The direct sum and direct product are not isomorphic for infinite indices, where the elements of a direct sum are zero for all but for a finite number of entries. They are dual in the sense of category theory: the direct sum is the coproduct, while the direct product is the product.
For example, consider ${\displaystyle X=\prod _{i=1}^{\infty }\mathbb {R} }$ and ${\displaystyle Y=\bigoplus _{i=1}^{\infty }\mathbb {R} ,}$ the infinite direct product and direct sum of the real numbers. Only sequences with a finite number of non-zero elements are in ${\displaystyle Y.}$ For example, ${\displaystyle (1,0,0,0,\ldots )}$ is in ${\displaystyle Y}$ but ${\displaystyle (1,1,1,1,\ldots )}$ is not. Both of these sequences are in the direct product ${\displaystyle X;}$ in fact, ${\displaystyle Y}$ is a proper subset of ${\displaystyle X}$ (that is, ${\displaystyle Y\subset X}$).[1][2]
## Topological space direct product
The direct product for a collection of topological spaces ${\displaystyle X_{i}}$ for ${\displaystyle i}$ in ${\displaystyle I,}$ some index set, once again makes use of the Cartesian product
${\displaystyle \prod _{i\in I}X_{i}.}$
Defining the topology is a little tricky. For finitely many factors, this is the obvious and natural thing to do: simply take as a basis of open sets to be the collection of all Cartesian products of open subsets from each factor:
${\displaystyle {\mathcal {B}}=\left\{U_{1}\times \cdots \times U_{n}\ :\ U_{i}\ \mathrm {open\ in} \ X_{i}\right\}.}$
This topology is called the product topology. For example, directly defining the product topology on ${\displaystyle \mathbb {R} ^{2}}$ by the open sets of ${\displaystyle \mathbb {R} }$ (disjoint unions of open intervals), the basis for this topology would consist of all disjoint unions of open rectangles in the plane (as it turns out, it coincides with the usual metric topology).
The product topology for infinite products has a twist, and this has to do with being able to make all the projection maps continuous and to make all functions into the product continuous if and only if all its component functions are continuous (that is, to satisfy the categorical definition of product: the morphisms here are continuous functions): we take as a basis of open sets to be the collection of all Cartesian products of open subsets from each factor, as before, with the proviso that all but finitely many of the open subsets are the entire factor:
${\displaystyle {\mathcal {B}}=\left\{\prod _{i\in I}U_{i}\ :\ (\exists j_{1},\ldots ,j_{n})(U_{j_{i}}\ \mathrm {open\ in} \ X_{j_{i}})\ \mathrm {and} \ (\forall i\neq j_{1},\ldots ,j_{n})(U_{i}=X_{i})\right\}.}$
The more natural-sounding topology would be, in this case, to take products of infinitely many open subsets as before, and this does yield a somewhat interesting topology, the box topology. However it is not too difficult to find an example of bunch of continuous component functions whose product function is not continuous (see the separate entry box topology for an example and more). The problem which makes the twist necessary is ultimately rooted in the fact that the intersection of open sets is only guaranteed to be open for finitely many sets in the definition of topology.
Products (with the product topology) are nice with respect to preserving properties of their factors; for example, the product of Hausdorff spaces is Hausdorff; the product of connected spaces is connected, and the product of compact spaces is compact. That last one, called Tychonoff's theorem, is yet another equivalence to the axiom of choice.
For more properties and equivalent formulations, see the separate entry product topology.
## Direct product of binary relations
On the Cartesian product of two sets with binary relations ${\displaystyle R{\text{ and }}S,}$ define ${\displaystyle (a,b)T(c,d)}$ as ${\displaystyle aRc{\text{ and }}bSd.}$ If ${\displaystyle R{\text{ and }}S}$ are both reflexive, irreflexive, transitive, symmetric, or antisymmetric, then ${\displaystyle T}$ will be also.[3] Similarly, seriality of ${\displaystyle T}$ is inherited from ${\displaystyle R{\text{ and }}S.}$ Combining properties it follows that this also applies for being a preorder and being an equivalence relation. However if ${\displaystyle R{\text{ and }}S}$ are connected relations, ${\displaystyle T}$ need not be connected; for example, the direct product of ${\displaystyle \,\leq \,}$ on ${\displaystyle \mathbb {N} }$ with itself does not relate ${\displaystyle (1,2){\text{ and }}(2,1).}$
## Direct product in universal algebra
If ${\displaystyle \Sigma }$ is a fixed signature, ${\displaystyle I}$ is an arbitrary (possibly infinite) index set, and ${\displaystyle \left(\mathbf {A} _{i}\right)_{i\in I}}$ is an indexed family of ${\displaystyle \Sigma }$ algebras, the direct product ${\displaystyle \mathbf {A} =\prod _{i\in I}\mathbf {A} _{i}}$ is a ${\displaystyle \Sigma }$ algebra defined as follows:
• The universe set ${\displaystyle A}$ of ${\displaystyle \mathbf {A} }$ is the Cartesian product of the universe sets ${\displaystyle A_{i}}$ of ${\displaystyle \mathbf {A} _{i},}$ formally: ${\displaystyle A=\prod _{i\in I}A_{i}.}$
• For each ${\displaystyle n}$ and each ${\displaystyle n}$-ary operation symbol ${\displaystyle f\in \Sigma ,}$ its interpretation ${\displaystyle f^{\mathbf {A} }}$ in ${\displaystyle \mathbf {A} }$ is defined componentwise, formally: for all ${\displaystyle a_{1},\dotsc ,a_{n}\in A}$ and each ${\displaystyle i\in I,}$ the ${\displaystyle i}$th component of ${\displaystyle f^{\mathbf {A} }\!\left(a_{1},\dotsc ,a_{n}\right)}$ is defined as ${\displaystyle f^{\mathbf {A} _{i}}\!\left(a_{1}(i),\dotsc ,a_{n}(i)\right).}$
For each ${\displaystyle i\in I,}$ the ${\displaystyle i}$th projection ${\displaystyle \pi _{i}:A\to A_{i}}$ is defined by ${\displaystyle \pi _{i}(a)=a(i).}$ It is a surjective homomorphism between the ${\displaystyle \Sigma }$ algebras ${\displaystyle \mathbf {A} {\text{ and }}\mathbf {A} _{i}.}$[4]
As a special case, if the index set ${\displaystyle I=\{1,2\},}$ the direct product of two ${\displaystyle \Sigma }$ algebras ${\displaystyle \mathbf {A} _{1}{\text{ and }}\mathbf {A} _{2}}$ is obtained, written as ${\displaystyle \mathbf {A} =\mathbf {A} _{1}\times \mathbf {A} _{2}.}$ If ${\displaystyle \Sigma }$ just contains one binary operation ${\displaystyle f,}$ the above definition of the direct product of groups is obtained, using the notation ${\displaystyle A_{1}=G,A_{2}=H,}$ ${\displaystyle f^{A_{1}}=\circ ,\ f^{A_{2}}=\cdot ,\ {\text{ and }}f^{A}=\times .}$ Similarly, the definition of the direct product of modules is subsumed here.
## Categorical product
The direct product can be abstracted to an arbitrary category. In a general category, given a collection of objects ${\displaystyle A_{i}}$ and a collection of morphisms ${\displaystyle p_{i}}$ from ${\displaystyle A}$ to ${\displaystyle A_{i}}$[clarification needed] with ${\displaystyle i}$ ranging in some index set ${\displaystyle I,}$ an object ${\displaystyle A}$ is said to be a categorical product in the category if, for any object ${\displaystyle B}$ and any collection of morphisms ${\displaystyle f_{i}}$ from ${\displaystyle B}$ to ${\displaystyle A_{i},}$ there exists a unique morphism ${\displaystyle f}$ from ${\displaystyle B}$ to ${\displaystyle A}$ such that ${\displaystyle f_{i}=p_{i}f}$ and this object ${\displaystyle A}$ is unique. This not only works for two factors, but arbitrarily (even infinitely) many.
For groups we similarly define the direct product of a more general, arbitrary collection of groups ${\displaystyle G_{i}}$ for ${\displaystyle i\in I,I}$ an index set. Denoting the Cartesian product of the groups by ${\displaystyle G}$ we define multiplication on ${\displaystyle G}$ with the operation of componentwise multiplication; and corresponding to the ${\displaystyle p_{i}}$ in the definition above are the projection maps
${\displaystyle \pi _{i}\colon G\to G_{i}\quad \mathrm {by} \quad \pi _{i}(g)=g_{i},}$
the functions that take ${\displaystyle \left(g_{j}\right)_{j\in I}}$ to its ${\displaystyle i}$th component ${\displaystyle g_{i}.}$
## Internal and external direct product
Some authors draw a distinction between an internal direct product and an external direct product. If ${\displaystyle A,B\subseteq X}$ and ${\displaystyle A\times B\cong X,}$ then we say that ${\displaystyle X}$ is an internal direct product of ${\displaystyle A{\text{ and }}B,}$ while if ${\displaystyle A{\text{ and }}B}$ are not subobjects then we say that this is an external direct product. | 4,424 | 15,933 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 164, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-05 | longest | en | 0.888782 |
https://ch.mathworks.com/matlabcentral/cody/problems/7-column-removal/solutions/1766227 | 1,571,306,996,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986673538.21/warc/CC-MAIN-20191017095726-20191017123226-00023.warc.gz | 418,063,411 | 15,384 | Cody
Problem 7. Column Removal
Solution 1766227
Submitted on 29 Mar 2019 by Rajani Mishra
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
A = [1 2 3; 4 5 6]; n = 2; B_correct = [1 3; 4 6]; assert(isequal(column_removal(A,n),B_correct))
2 Pass
A = magic(4); n = 3; B = [16 2 13; 5 11 8; 9 7 12; 4 14 1]; B_correct = B; assert(isequal(column_removal(A,n),B_correct))
3 Pass
A = 1:10; n = 7; B_correct = [1 2 3 4 5 6 8 9 10]; assert(isequal(column_removal(A,n),B_correct)) | 232 | 594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-43 | latest | en | 0.590967 |
https://www.jiskha.com/search/index.cgi?query=A+Ferris+Wheel%2C+24.0+m+in+diameter%2C+rotates+once+every+12.5s.+What+is+the+fractional+change+in+a+person%27s+apparent+weight%3A+a%29+at+the+top%2C+and+b%29+at+the+bottom%3F&page=17 | 1,521,414,844,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257646178.24/warc/CC-MAIN-20180318224057-20180319004057-00073.warc.gz | 815,063,917 | 13,254 | A Ferris Wheel, 24.0 m in diameter, rotates once every 12.5s. What is the fractional change in a person's apparent weight: a) at the top, and b) at the bottom?
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44. Chemistry
What order do i arrange the units so they can be from largest to smallest? m superscript 3, mL, cL, UL, L, dL. kilo hecto deca unit deci centi milli ----------- All of the above are by a factor of 10 from top to bottom (becoming smaller); 1 kilogram = 1000 grams...
45. PHYSICS
A 20.0 kg solid cylinder, 0.25m in radius, and a 5.0 kg thin hollow sphere, 0.50 m in radius, are released from rest at the exact same time at the top of a 10m long ramp inclined at 30º. a) What is the final velocity of the cylinder at the bottom of the ramp? b) What is the ...
46. Trigonometry
A bicycle is traveling at 20 feet/second. If the diameter of the wheel is 24 inches, what is the angular velocity in radians/second?
47. Physics
A constant retarding torque of 12 N*m stops a rolling 0.80 m diameter wheel in a distance of 15 m. How much work does the torque do?
48. Physics (repost)
(I'll try to describe this the best I can, it's a little hard without showing a picture) This question deals with projectile motion: An archer is standing 150 meters above another archer 450 meters away [so visually the 150m and 450m make the right angle]. The archer at the ...
49. math
A landscaping company sells 40-pound bags of top soil. The actual weight x of a bag, however, may differ from the advertised weight by as much as 0.75 pound. Write an inequality involving absolute value that expresses the relationship between the actual weight x of a bag and ...
50. Algebra
A landscaping company sells 40-pound bags of top soil. The actual weight x of a bag, however, may differ from the advertised weight by as much as 0.75 pound. Write an inequality involving absolute value that expresses the relationship between the actual weight x of a bag and ...
51. Statistics
T-Test The article “Does Smoking Cessation Lead to Weight Gain?” described an experiment in which 322 subjects, selected at random from those show successfully participated in a program to quit smoking were weighed at the beginning of the program and again one year later...
52. physics
A graduated cylinder is inverted in a very large bucket of water. The depth of the water in the container is 6 m and the top of the bucket is open to the air in the room. The mouth (bottom) of the graduated cylinder is 4 m above the bottom of the container. Due to a pressure ...
53. Physics
A hypothesis says the upper limit in frequency a human ear can hear can be determined by the diameter of the eardrum, which should have approximately the same diameter as the wavelength at the upper limit. Using the hypothesis what would be the radius of the eardrumfor a ...
54. English
1. A person is drifting on a deserted island. Change this person's thoughts into other words and say them to your friend. 2. The following are about some conditions. Fill in the blanks, and then say as in the example. 3. Think of your life habits which you want to change,and ...
55. Trigonometry
a flagpole 25 feet tall stands on a top of a building. from a point in the same horizontal plane with the base of the building. The angle of elevation of the top and the bottom are 61°30' and 56°20' respectively. How high is the building?
56. Trigonometry
a flagpole 25 feet tall stands on a top of a building. from a point in the same horizontal plane with the base of the building. The angle of elevation of the top and the bottom are 61°30' and 56°20' respectively. How high is the building?
57. Physics
AN astronaut on the Moon has a mass (including his spacesuit and equipment) of 180 kg. the acceleration due to gravity on the Moon is 1.6 m/s2 (seconds squared). a) calculate his weight on the moon. the astronaut climbs 100m to the top of the crater . b) by how much does his ...
58. Math
Peter has a cylindrical block of wood od diameter 6 cm and height 18 cm. He glues the base to a board, but wants to paint the rest of the block. He wants to paint the bottom half of the wood black and the top half white. a) Calculate the surface area of the block that Peter ...
59. math
Peter has a cylindrical block of wood od diameter 6 cm and height 18 cm. He glues the base to a board, but wants to paint the rest of the block. He wants to paint the bottom half of the wood black and the top half white. a) Calculate the surface area of the block that Peter ...
60. Physics
A cord is wrapped around the rim of a wheel .250 m in radius, and a steady pull of 40.0 N is exerted on the cord. The wheel is mounted on frictionless bearings on a horizontal shaft through its center. The moment of inertia of the wheel about this shaft is 5.00kg*m^2. Compute ...
61. physics
A bubble located 0.200m beneath the surface of a glass of beer rises to the top.the air pressure at the top of the glass is 1.01(10^5)pa.assume that the density of beer is the same as that of fresh water at 4 degrees celcius.if the temperature and number of moles of CO2 remain...
62. functions
A person’s weight, w, on a planet of radius r is given by: w= kr^-2 k < 0 where the constant, k, depends on the masses of the person and the planet.What fraction of the Earth’s radius must an equally massive planet have for a person to weigh ten times more on that ...
63. Dynamics
a flywheel has an angular speed of 1200rev/min when its motor is turned off.the wheel attains constant decelerations of 1.5rad/s2 due to friction in its bearing.determine the time required for the wheel to come to rest and the numer of revolutions the wheel makes before it ...
64. Math
James uses a measuring wheel to find the length of a cross-country race course he holds the handle and rolls the wheel along the course.The wheel has a radius of 1 foot and it turns 2,000 times from the beginning to the end of the race course, how long is the course?Explain ...
65. Calculus
a ladder is 20 feet long and its top is slidding at a constant rate of 2 feet per second. How fast is the bottom of the ladder moving along the ground when the bottom is 16 feet from the wall?
66. physics
After 10.0 s, a spinning roulette wheel at a casino has slowed down to an angular velocity of +1.56 rad/s. During this time, the wheel has an angular acceleration of -5.45 rad/s2. Determine the angular displacement of the wheel.
67. physics
After 10.0 s, a spinning roulette wheel at a casino has slowed down to an angular velocity of +1.74 rad/s. During this time, the wheel has an angular acceleration of -5.1 rad/s2. Determine the angular displacement of the wheel.
68. Maths
The top of a building 24cm high is observed from the top and from the bottom of a vertical tree. The angles of elevation are found to be 45degree and 60degree respectively . Find the height of the tree. ( leave your answer in surd form)
69. Maths
A ship is sighted from the top of a lighthouse. The angle of depression from the lighthouse to the top of the ship is 45 degrees. The distance from the top of the lighthouse directly to the ship is 4 miles. Calculate the horizontal distance of the ship from the bottom of the ...
70. english
sat skills. write the word in the space provided. 1-how sad it to see such beautiful flowers ----------- and die! 2-a person can usually tell how popular a new movie is by the length of the ----------- in front of the box office. 3-the desire to be the world's top tennis ...
71. Science
Compare an astronauts weight in orbit with the astronauts weight on Earth, assuming the mass of the astronaut doesn't change.
72. Science
Compare an astronauts weight in orbit with the astronauts weight on Earth, assuming the mass of the astronaut doesn't change.
73. bibb
Consider a rider on a Ferris wheel which is turning at constant speed. (a) At what point is the upward component of the force of the seat on the rider minimum? (b) At what point is the upward component of the force of the seat on the rider maximum? Justify your answers.
74. angular accleration
a wheel if a vehical rotates uniformly at the rate if 120 rotations per mint suddenly it goes faster and reaches 150 rotations per mint after 30s find the angular acc
75. Physics
A 1550 kg automobile has a wheel base (the distance between the axles) of 3.00 m. The center of mass of the automobile is on the center line at a point 1.10 m behind the front axle. Find the force exerted by the ground on each wheel. force on each rear wheel ___ kN (upward) ...
76. chemistry
A pound of body fat stores an amount of chemical energy equivalent to 3500 Cal. When sleeping, the average adult burns or expends about 0.45 Cal/h for every pound of body weight. How many Calories would a 135 lb person burn during 7 hours of sleep? (Note that 1 Cal = 1 kcal.)
77. science
A pound of body fat stores an amount of chemical energy equivalent to 3500 Cal. When sleeping, the average adult burns or expends about 0.45 Cal/h for every pound of body weight. How many Calories would a 126 lb person burn during 7 hours of sleep? (Note that 1 Cal = 1 kcal.)
78. physics
A pound of body fat stores an amount of chemical energy equivalent to 3500 Cal. When sleeping, the average adult burns or expends about 0.45 Cal/h for every pound of body weight. How many Calories would a 196 lb person burn during 7 hours of sleep? (Note that 1 Cal = 1 kcal.)
79. science
A pound of body fat stores an amount of chemical energy equivalent to 3500 Cal. When sleeping, the average adult burns or expends about 0.45 Cal/h for every pound of body weight. How many Calories would a 170 lb person burn during 10 hours of sleep? (Note that 1 Cal = 1 kcal.)
80. Health
What are warning signs that someone may have a drinking problem? A. A person continued to drink even though it was causing trouble with their family or friends B. A person found that when the effects of alcohol were wearing off, they had withdrawal symptoms, such as trouble ...
81. PHYSICS! -need help!-
A combination of an applied force and a frictional force produce a constant total torque of 38 N-m on a wheel rotating about a fixed axis. This combined force acts for 8 s, during which the wheel’s angular speed increases from 0 to 12 rad/s. Find the moment of inertia of the...
82. physics
You are pulling your sister on a sled to the top of a 20.0 m high, frictionless hill with a 10.0° incline. Your sister and the sled have a total mass of 50.0 kg. You pull the sled, starting from rest, with a constant force of 127 N at an angle of 45.0° to the hill. If you ...
83. PHYSICS
An 800.0 kg roller coaster car is at rest at the top of a 95 m hill. It rolls down the first drop to a height of 31 m. When it travels to the top of the second hill, it is moving at 28 m/s. It then rolls down the second hill until it is at ground level. What is the kinetic and...
84. Physics
A circular-motion addict of mass 85 kg rides a Ferris wheel around in a vertical circle of radius 10 m at a constant speed of 6.1 m/s. (a) What is the period of the motion? __________s (b) What is the magnitude of the normal force on the addict from the seat when both go ...
85. Algebra 2
I got no response so I guess I'll try again The force needed to overcome friction and drag a person across the floor depends on the person's weight. Manuel Dexterity can drag Gil O' Teen across the floor by pulling with a force of 51 pounds. Gil's brother, Nick who weighs ...
86. Physics
The combination of an applied force produces a constant total torque of 39.6 N.m on a wheel rotating about a fixed axis. The applied force acts for 5.3 s. During this time, the angular speed of the wheel increases from 0 to 9.8 rad/s. Determine the moment of inertia of the wheel.
87. physics
A stationary bicycle is raised off the ground, and its front wheel (m = 1.3 kg) is rotating at an angular velocity of 12.3 rad/s (see the drawing). The front brake is then applied for 3.0 s, and the wheel slows down to 3.3 rad/s. Assume that all the mass of the wheel is ...
88. physics
A 1.5 kg snowball is fired from a cliff 12.5m high with an inital velocity of 14 m/s directed 41 degrees above the horizontal. a) how much work is done on the snowball by the gravitaitonal force during its flight to the flat ground below the cliff? b) what is the change in the...
89. trigonometry
an automobile tire has a diameter of 36 inch.how many revolution will the wheel make an arc the automobile travels 1 mile (5280 A)?
90. Math
Sue's bike has whheels that are 60cm in diameter. 1.1 How far does the bike move in one full turn of the wheel?
91. algebra
there are approximately 4lbs of muscle for every 5lbs of body weight of a certain animal for a 115-lbs animal, how much of the weight is muscle
92. Physics
Archimedes principle. The weight of the bouy is given, so the weight of water displaced by the submerged part of the bouy equals that. Figure the volume that weight of water displaces, then figure the height of a .33m diameter column of water will have of that volume. A large ...
93. Math
1.If a person jogs at a rate of 5 1/2mi/h for 3 1/2 h in a week, how many calories do they burn? 2. If a person runs in place for 15 minutes, how many calories will be burned? 3. If a person cross-country skis for35minutes, how many calories will be burned? 4. How many hours ...
94. Resources and references
Once a manual is written, it may change over time as procedures change. What is recommended format for a manual to incorporate these changes?
95. PHYSICS
1. A 15 kg child sits in a swing that is attached in ropes 1.6 m long . Find the potential energy of the child relative to his lower position when the ropes (a) are horizontal (b) vertical and (c) make 30° with the horizontal. 2. A brick of mass 1.50 kg is placed on top of a ...
96. Physics
A quartz sphere is 14.0 cm in diameter. What will be its change in volume if its temperature is increased by 305 °F? (the coefficient of volume expansion of quartz is (1.50 × 10^-6) °C-1) Answer in cm^3 I tried this twice and got 0.657 cm^3 and 0.18cm^3 but they are both ...
97. Physics!
One of the new events in the 2002 Winter Olympics was the sport of skeleton (see the photo). Starting at the top of a steep, icy track, a rider jumps onto a sled (known as a skeleton) and proceeds - belly down and head first - to slide down the track. The track has fifteen ...
98. math
There are approximately 4 lbsof muscle for every 5 lbs of body weight of a certain animal. For a 125-lb animal, how much of the weight is muscle?
99. algebra
there are approximately 4lb of muscle for every 5lbs of body weight of a certain animal for a 115-lb animal how much of the weight is muscle
100. math
A cone shaped container with its vertex at the top and its circular base on the ground is filled to 3/4 of its height with water. What fractional part of the container is filled with water? | 6,387 | 25,081 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | longest | en | 0.932692 |
www.circuit-hut.co.uk | 1,368,960,187,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368697420704/warc/CC-MAIN-20130516094340-00027-ip-10-60-113-184.ec2.internal.warc.gz | 394,694,734 | 5,340 | ohms law calculations
Ohms Law Explained
Ohms Law
There is a post regarding ohms lawi you can see more of that by following: http://circuit-hut.co.uk/Ohms-Law<
This post aims to explain Ohms law in more detail.
Calulate a resistor value uising Ohms Law
I thought users would find this easy to understand
First of all we need to collect the following information:
• The voltage of the supply
• The forward voltage of the component eg L.E.Di<
• The current the component uses
• We can then apply Ohms law to calculate the value of resistori needed.
• For calculating resistance we use the formula R= V ÷ I lets take the following as an example: Blue LED = 3.4V Forward voltage and 30mA maximum current, The supply voltage is 12V DC.
• Step 1 We first need to work out the difference in the voltage between the source and the component this is 12V - 3.4V = 8.6V
•
Step 2 We convert the current rating from mA into A we simply divide by 1000. | 244 | 949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2013-20 | latest | en | 0.896279 |
https://view.joomag.com/fin-486-cart-become-exceptional-fin486cartcom-fin-486-entire-course/0563858001548153096?page=19 | 1,579,891,233,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250624328.55/warc/CC-MAIN-20200124161014-20200124190014-00347.warc.gz | 732,954,275 | 12,052 | FIN 486 CART Become Exceptional--fin486cart.com FIN 486 Entire Course - Page 19
Explain which income statement and balance sheet items you assumed were variable instead of fixed. Format your paper consistent with APA guidelines. Click the Assignment Files tab to submit your assignment. ============================================== FIN 486 Week 3 Team Assignment FOR MORE CLASSES VISIT www.fin486cart.com a) Calculate the after-tax cost of debt b) Calculate the cost of preferred stock c) Calculate the cost of retained earnings d) calculate the cost of common stock e) calculate the firms weighted average cost of capital using retained earnings and the capital structure weights f) calculate the firms weighted average cost of capital using new common stock and the capital structure weights a. Cost of debt: Proceeds from sale of \$1,000 par value bond: b. Cost of preferred stock: rp = Dp ÷ Np c. Cost of common stock: rj = RF + [bj × (rm − RF)] d. Weighted average cost of capital: ra = (wi × ri) + (wp × rp) + (ws × rn) e. 1. Change in risk Premium: Change in beta × market risk premium Note: 17.5% − 15.7% = 1.8% 2. Revised weighted average cost of capital: ra = (wi x ri) + (ws x rn) ============================================== FIN 486 Week 3 Video Summary Short Term Finance FOR MORE CLASSES VISIT www.fin486cart.com | 314 | 1,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.751962 |
http://mathhelpforum.com/advanced-algebra/141167-group-homomorphism-print.html | 1,516,795,628,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084894125.99/warc/CC-MAIN-20180124105939-20180124125939-00219.warc.gz | 196,646,775 | 3,014 | # group homomorphism
Printable View
• Apr 24th 2010, 06:56 PM
nhk
group homomorphism
24. Suppose that ƒ:Z50 to Z15 (co-domain is Z 15) is a group homomorphism with ƒ(7)=6.
a. Determine ƒ(x).
b. Determine ƒ(Z50).
c. Determine ker ƒ.
d. Determine ƒ–1({3}).
I am lost particularly on b, a,c and d I think I will be able to get by myself. Any suggestions to help me for b?
Thanks
• Apr 24th 2010, 07:19 PM
tonio
Quote:
Originally Posted by nhk
24. Suppose that ƒ:Z50 to Z15 (co-domain is Z 15) is a group homomorphism with ƒ(7)=6.
a. Determine ƒ(x).
b. Determine ƒ(Z50).
c. Determine ker ƒ.
d. Determine ƒ–1({3}).
I am lost particularly on b, a,c and d I think I will be able to get by myself. Any suggestions to help me for b?
Thanks
First, as $(7,50)=1$ then $<7>=\mathbb{Z}_{50}$ and then $\forall\,x\in\mathbb{Z}_{50}\,\,\exists\,n\in\math bb{Z}\,\,\,s.t.\,\,\,7n=x$ $\Longrightarrow f(x)=f(7n)=nf(7)=6n\!\!\!\pmod{15}$.
Since $ord(6)=5\,\,\,in\,\,\,\mathbb{Z}_{15}$ , you get that $f\left(\mathbb{Z}_{50}\right)=<6>\cong \mathbb{Z}_5$ .
I did two, now you do the other two.
Tonio | 419 | 1,089 | {"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": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-05 | longest | en | 0.794526 |
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Simplifying Square Roots Worksheet Education Com Solving Quadratic Equations Solving Quadratics Square Roots From pinterest.com
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Showing top 8 worksheets in the category - Cube Root. Some of the worksheets for this concept are Math 6 notes name Square roots work Square and square roots grade 8 Powers and roots Finding square root easy Square root work Square roots date period Cubes and cube roots work. 01092020 Solving Square Root Equations Worksheet 8th Grade. Some of the worksheets for this concept are Square roots work Solving quadratic equations by extracting square roots Solving quadratic equations square root law Square root property Quadratic equations square roots Square roots and other radicals Solving quadratic roots Solving quadratic equations by finding square. Worksheet by Lucas Kaufmann.
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If we want to find the square root of 12 then 3 2 9 and 4 2 16. Square Root Worksheets 8th Grade and Square Root Worksheets Grade 8 Worksheets for All. Find the least number which must be added to 2000 to make the sum a perfect square. If we know that the number of children in the village is also a perfect square and is more than. Think of the largest number whose square is equal to or just less than the first pair 0 in our case largest number whose square is equal to or less than 0 is 0.
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Square root worksheets 8th grade Collection Free square root worksheets PDF and html 192258 The Squares and Square Roots A Number Sense Worksheet living. A square park of side 40 cm has a 4 m wide path surrounding it. Class 8 Squares and Square Roots Chapter tests for all important topics covered which can come in your school exams download in PDF. Square Roots 8th Grade - Displaying top 8 worksheets found for this concept. Find the area of the path.
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If we know that the number of children in the village is also a perfect square and is more than. September 1 2020 on Solving Square Root Equations Worksheet 8th Grade. Square Roots 8th Grade. Class 8 Squares and Square Roots Chapter tests for all important topics covered which can come in your school exams download in PDF. Divide the given number by one of the two square roots in step 1.
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Square Roots 8th Grade - Displaying top 8 worksheets found for this concept. Square Roots 8th Grade. By Celestine Aubry on September 1 2020. Showing top 8 worksheets in the category - Cube Root. Discover and save your own Pins on Pinterest.
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Here you find our worksheets with square roots suited for grade levels 6 and 7. Free square root worksheets PDF and html On this page youll find an unlimited supply of printable worksheets for square roots including worksheets for square roots only grade 7 or worksheets with square roots and other operations grades 8-10. Divide the given number by one of the two square roots in step 1. Some of the worksheets for this concept are Math 6 notes name Square roots work Square and square roots grade 8 Powers and roots Finding square root easy Square root work Square roots date period Cubes and cube roots work. If we want to find the square root of 12 then 3 2 9 and 4 2 16.
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Some of the worksheets for this concept are Math 6 notes name Cubes and cube roots work Square roots work Square roots work Square roots date period Math resource studio Powers and roots First published in 2013 by the university of utah in. First we make the pair of the digits of the integral part and decimal part by placing the bar of each pair. If we want to find the square root of 12 then 3 2 9 and 4 2 16. Free square root worksheets pdf and html 402529. Some of the worksheets for this concept are Math 6 notes name Cubes and cube roots work Square roots work Square roots work Square roots date period Math resource studio Powers and roots First published in 2013 by the university of utah in.
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That staying stated many of us give you a assortment of basic however useful content articles. Find the area of the path. Some of the worksheets for this concept are Math 6 notes name Cubes and cube roots work Square roots work Square roots work Square roots date period Math resource studio Powers and roots First published in 2013 by the university of utah in. September 1 2020 on Solving Square Root Equations Worksheet 8th Grade. Find the least number which must be added to 2000 to make the sum a perfect square.
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We have finding square roots worksheets with square roots up to 50 worksheets in which the radical is unknown square root estimation worksheets and worksheets in which the calculator can be used to find the square root. Think of the largest number whose square is equal to or just less than the first pair 0 in our case largest number whose square is equal to or less than 0 is 0. Class Assignments for Grade 8 Squares and Square Roots printable worksheets and practice tests have been prepared as per pattern of worksheets in various schools and topics given in NCERT textbook 2020 2021. A square park of side 40 cm has a 4 m wide path surrounding it. Displaying top 8 worksheets found for - Extracting Square Roots Grade 9.
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If the student is asking a question then you should be able to give them a good answer to it. Find the average of the root and the result in step 2. Simplifying Square Roots Worksheet Education Com Square Roots Solving Quadratic Equations. Showing top 8 worksheets in the category - 8th Grade Square And Cube Roots. By Celestine Aubry on September 1 2020.
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September 21st, 2013, 12:44 AM #1 Newbie Joined: Sep 2013 Posts: 8 Thanks: 0 comparison with modular arithmetic Hi, I am working on operations( addition, subtraction, multiplication) with the residue number system instead of integers, and I have to find minimum, maximum and average of a group of numbers in their residue number representation. Could I find any way to apply comparison with modular arithmetic? I do not if (mod) operation could help or not! Thank you : )
September 21st, 2013, 10:52 AM #2
Senior Member
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Posts: 2,409
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Re: comparison with modular arithmetic
Quote:
Originally Posted by abady-1 Hi, I am working on operations( addition, subtraction, multiplication) with the residue number system instead of integers, and I have to find minimum, maximum and average of a group of numbers in their residue number representation.
Can you clarify the context of your question? There's no order relationship on the integers mod n. That's because they cycle. For example in the integers mod 5, the successor of 4 is zero. There's no way to define a sensible order relation. So you can't define min or max.
You could define the mean as the sum divided by the number of things you're adding ... for example in the integers mod 5, the mean of 1 and 3 is 2. But the mean of 1 and 4 is zero. So I don't know how helpful this is.
September 21st, 2013, 02:33 PM #3 Global Moderator Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: comparison with modular arithmetic these are difficult operations in RNS. You can add easily enough, but division won't typically be possible so I don't even know how you could take the average. Minimum and maximum are really tough, you'd almost be better off converting out of RNS, computing min/max, and converting back.
September 21st, 2013, 05:30 PM #4 Newbie Joined: Sep 2013 Posts: 8 Thanks: 0 Re: comparison with modular arithmetic This is an example of a residue number system(RSN): First: you have to determine the base of modulus that you're working on to determine the range of integers that you can represent in RSN: Let's suppose the base is RNS= { 8,7,5}, all co-primes. So, the range of integers that you can represent uniquely in this RNS is: 8 * 7 * 5 = 280 ( 0 -->279) Now, let we suppose that you want to convert integer (12) to RNS), you will do the following: 12 ---> { 12 mod 8 , 12 mod 7 , 12 mod 5} ---> { 4,5,2} ---> equivalent to 12 in decimal. 17 ---> { 17 mod 8 , 17 mod 7 , 17 mod 5} ---> { 1,3,2} ---> equivalent to 17 in decimal. The question is: Could I compare between 12 and 17 by using there RNS forms {4,5,2} & {1,3,2} with the help of other available information or not? BTW, I can do addition,multiplication easily if these operations help to find min/max and average. Thank you Maschke and CRGreathouse for responding, looking for great solutions Best regards,
September 21st, 2013, 08:40 PM #5 Global Moderator Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: comparison with modular arithmetic The trouble is that, philosophically, working in RNS {a1, a2, ..., ak} is basically just working mod LCM(a1, a2, ..., ak) and so there isn't really any such thing as order. In your example with bases {8, 7, 5} it's easy to see that (with 1 = (1, 1, 1) and (0, 0, 0) = 0 as usual) (7, 6, 4) + 1 = 0 and so (7, 6, 4) = 559 is also -1 -- everything 'wraps around' mod the LCM 560. If you have external information about how to map these to usual integers -- say, to the canonical representation where each number represents the smallest nonnegative number of that residue -- then you can impose an order, but there's no good way to do it from inside the RNS since it doesn't have an order on its own.
September 22nd, 2013, 04:46 PM #6
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Re: comparison with modular arithmetic
Quote:
Originally Posted by CRGreathouse The trouble is that, philosophically, working in RNS {a1, a2, ..., ak} is basically just working mod LCM(a1, a2, ..., ak) and so there isn't really any such thing as order. In your example with bases {8, 7, 5} it's easy to see that (with 1 = (1, 1, 1) and (0, 0, 0) = 0 as usual) (7, 6, 4) + 1 = 0 and so (7, 6, 4) = 559 is also -1 -- everything 'wraps around' mod the LCM 560. If you have external information about how to map these to usual integers -- say, to the canonical representation where each number represents the smallest nonnegative number of that residue -- then you can impose an order, but there's no good way to do it from inside the RNS since it doesn't have an order on its own.
Thank you for this information.
To generalise the problem, I am looking for numerical system that simplified large numbers complexity and overhead when I do operations on them?
So, Is there any suggestion as an alternative of RNS, which has properties that I can do operations such as max/min and so on?
Thanks!
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1. Feb 2, 2017
### franktherabbit
1. The problem statement, all variables and given/known data
This is a problem from some textbook a friend emailed me:
The text goes like this:
When the switch is open the charge on $C1$ equals $Q_o=40uC$ and $C_2$ has no charge on it. The switch closes and some charge $q$ flows through the capacitors (suppose the flow to be from up to down). Find:
a) $q$
b) work of generator $V$
c) the ammount of work transfered into heat during this process
2. Relevant equations
3. The attempt at a solution
Well the voltage when the switch is open exists only on the $C1$ capacitor. When it closes the amount of $q$ charge flows thorugh the capacitors in order to equalize the sum of voltages across the capacitors with V, is that right? That makes $q=-10uC$, right? But why didnt $R$ have any role in here? Usually we have a voltage drop since we have resistors and here we do not?
For the work of the generator we have formula $W=Vq$ right? and that makes is $-100uJ$.
Im not sure about the last part but i guess its just the difference of energies of the system?
$E_1=\frac{1}{2}(Q_o)^2/C_1=400uJ$
$E_2=\frac{1}{2}(Q_o+q)^C_1+\frac{1}{2}(q)^2/C_2+Eq=1/4(900)+1/4(100)-100=150uJ$
so the work transfered into heat is just $E_2-E_1=-250uJ$ right?
2. Feb 2, 2017
### Staff: Mentor
Since you're comparing initial states and final states in order find the amount of charge moved, the resistor doesn't come into play; you're looking at the total charge moved and not at the current. That will come later.
While it's true that there is work associated with the voltage source when the charges move, we can't attribute heat loss to that work since we don't know what the source is doing with the charge it accepts (the charge is flowing into the + terminal of the source). For all we know it behaves like an ideal battery and stores the energy and can return it later.
Similarly, the capacitors move energy in and out of electric fields, and ideal capacitors alone won't lose energy via heat. When energy is lost moving charges around between capacitors, it's always lost in the external circuit through some mechanism (Resistive losses are usually what it's attributed to in basic circuit theory).
So that leaves us with the resistor. That will definitely generate heat when current flows through it. The question then becomes, how might you determine that energy?
3. Feb 3, 2017
### franktherabbit
I can't think of a way to do it.. Is at least a) and b) correct?
4. Feb 3, 2017
### Staff: Mentor
Yes.
You have enough information to determine the energy lost to heat. You know the total change in energy stored on the capacitors and the energy absorbed by the source. If there were no resistance then those values should be the same...
Alternatively, and as a check, you might work out the equation for the current with respect to time. It's an RC circuit and will this have a time constant which you should be able to find using the given component values.
5. Feb 3, 2017
### franktherabbit
Well the difference in the capacitor energies is:
$ΔE_c=(frac{1}{2}(Q_o+q)^C_1+\frac{1}{2}(q)^2/C_2)-\frac{1}{2}(Q_o)^2/C_1=-150uJ$
The work of the generator is $Eq=-100uJ$
Since $-250uJ$ wasnt's correct i guess the equation is
$ΔE=ΔE_c-Eq=-50uJ$ but why is it $-Eq$ we are subtracting the energy of the voltage source from the energy difference of capacitors? I dont understand why thats why i added in the first place.
6. Feb 3, 2017
### Staff: Mentor
The capacitors supplied energy while the source absorbed energy.
7. Feb 3, 2017
### doktorwho
So the $q$ flows from - to + and it then absorbs it and if the voltage source was reversed it would supply it?
So if the q flows into - it supplys and revesed ot absorbs?
For example if the q was -30 and it was shown in orig picture to go into + of V it would mean that it supply's because the q is negative and it actually flows to the -?
8. Feb 3, 2017
### Staff: Mentor
That's the idea. Pictorially: | 1,085 | 4,034 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2017-43 | longest | en | 0.929044 |
https://randomperturbation.wordpress.com/tag/relations-math/ | 1,561,572,194,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000414.26/warc/CC-MAIN-20190626174622-20190626200622-00053.warc.gz | 577,115,727 | 15,105 | # Matrices and Graphs
Recall in the post “Canonical Representation of Finitely Presented Coxeter Groups” we defined the Coxeter matrix of a Coxeter group $G$ by $a_{ij}=m_{ij}$ where $(r_ir_j)^{m_{ij}}=1$ with $r_i,r_j$ as the generators of the group. We can also define the Coxeter graph of a Coxeter group $G$ whose vertices are the generators of the group. Then since $m_{ij}\geq 2$ for $i\neq j,$ we will define edges between two generators iff $m_{i,j}\geq 3,$ and label the edge with $m_{ij}$ when $m_{ij}\geq 4.$
In a generic graph (recall our presentation with $V$ a set and $E$ a symmetric relation on $V^2$), we can define a $|V|\times |E|$ matrix, called the incidence matrix of $E$, denoted $[E],$ where
$a_{i,(j,k)}=\left\{\begin{array}{lcl}1&\mbox{if}&i=j \mbox{~or~} i=k\\ 0&\mbox{otherwise}&\end{array}\right .$
So $a_{i,(j,k)}$ is $1$ iff the vertex $i$ is a vertex on the edge $(j,k).$ This assumes some ordering of $V.$ So if we assume irreflexivity of $E,$ then every column, indexed by $(i,j),$ will have precisely two entries that are $1,$ specifically in the $i$th and $j$th row.
In the case of the incidence matrix of a Coxeter graph, the nonzero rows thus coincide with generators $r_i$ such that $m_{ij}\geq 3$ for some $j.$
We can remove the symmetric requirement of $E$; so $E$ is just a subset of $V^2.$ We can then use “binary relation” synonymously with “digraph“. From here on we will assume a digraph on $V$ is an irreflexive binary relation on $V,$ and that a graph on $V$ is an irreflexive symmetric binary relation on $V.$
Definition 1. Let $E_1$ and $E_2$ be digraphs on $V_1$ and $V_2$ respectively. A map $\varphi:V_1\to V_2$ is an $(E_1,E_2)$-digraph homomorphism if $(x,y)\in E_1\Rightarrow$$(\varphi(x),\varphi(y))\in E_2.$ An $(E_1,E_2)$isomorphism of digraphs is a bijective $(E_1,E_2)$-homomorphism of digraphs.
Definition 2. Let $E$ be a digraph on $V.$ We define the adjacency matrix of $E$, denoted $(E),$ as the $|V|\times |V|$ matrix where $a_{x,y}$ is the number of paths from $x$ to $y.$
So an entry $a_{i,(j,k)}\in [E]$ being $1$ means one of the entries $a_{i,j}$ or $a_{j,i}$ is $1$ in $(E)$ since being a vertex on an edge means there is a path to some other vertex (namely the other vertex on that edge).
Proposition 3. Let $E_1$ and $E_2$ be digraphs on $V.$ There is an $(E_1,E_2)$-automorphism of $V$ iff there is a permutation matrix $P$ such that
$\displaystyle P^\top(E_1)P=(E_2).$
[1] Godsil, Chris and Gordon Royle. Algebraic Graph Theory. Graduate Texts in Mathematics. Vol. 207. Springer Science and Business Media. 2004.
# A Bit on Graphs
I was never a big fan of the presentation of graphs as sets of vertices and edges (say $V$ and $E$) such that elements of $E$ had the form $e_{x,y}$ with $x,y\in V$ and $e_{x,y}=e_{y,x}.$ I prefer to use the language of relations.
Definition 1. A graph on $X$ is a symmetric binary relation $E$ on $X.$ We then call $X$ the set of vertices and $E$ the set of edges.
Recall a binary relation on $X$ is a subset of $X^2.$ They are equivalent notions, but in the latter, one simply has a set with a relation on it–rather than these two separate sets such that the second set backtracks to the first as a relation.
Example 2. A complete graph (one in which any two vertices are adjoined by an edge) is thus simply a graph where every $x\in X$ is almost surjective with respect to $E$, by which I mean if we let
$\displaystyle (x,X-\{x\})=\{(x,y):x\neq y\in X\},$
then $(x,X-\{x\})\subseteq E$ for all $x\in X.$ In particular we hence have
$\displaystyle E=\bigcup_{x\in X}(x,X-\{x\}).$
When I say a vertex is surjective with respect to $E$, I mean it has an edge with all other vertices (where almost-surjective excludes an edge with itself). Note the relation $E$ is always a surjective relation if we assume every vertex has some edge connected to it (i.e. is left-total and hence surjective since domain=codomain). But a surjective relation $E$ of course does not imply every element is surjective with respect to $E.$
Example 3. A connected graph is a graph such that for any distinct two $x,y\in X,$ there is a sequence (or path) $x=x_1,...,x_n=y$ such that $(x_i,x_{i+1})\in E$ for all $1\leq i\leq n.$
A tree is thus a connected graph such that paths between points are unique (note this definition of a tree is different from that of posets: where elements have a well-ordered set of predecessors). And so on and so on….
In fact, recall Abramenko and Brown used this algebraic approach in the construction of Coxeter complexes, although it was slightly more complicated since they were dealing with graphs whose vertices were cosets of the maximal standard subgroups of a Coxeter group $G.$ Lengths of paths between vertices in the sense I have described coincide with lengths of minimal galleries between chambers containing those vertices. Also, be not confused that in that construction of Coxeter complexes, there was another implicit notion of a (di)graph where any simplex was a vertex, and a directed edge connected two simplices if one was the face of another. | 1,548 | 5,135 | {"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": 96, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2019-26 | latest | en | 0.758796 |
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# Tutor profile: Amara E.
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### Subject:Trigonometry
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If the hypotenuse of a right triangle has a value of 5in, and the shortest leg of the triangle has a value of 3in, what is the length of the other leg?
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Amara E.
This can be solved by using the Pythagorean theorem c^2=a^2+b^2. Since we know c=5, and a=3, we can plug and chug to find the answer, by rearranging the equation to c^2-a^2=b^2. This means 5^2-3^2=b^2. So 25-9=16. Since b^2 is 16, take the square root to find b. This leaves the answer to be 4in.
### Subject:Basic Math
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Question:
Using the equation y=3x+14, find y if x=2.
Inactive
Amara E.
Since the question gives you the value for x, all you have to do is plug it in! y=3(2)+14, which then is simplified to y=6+14, and with simple addition, y=20.
### Subject:Biology
TutorMe
Question:
What is a bacteriophage?
Inactive
Amara E.
A virus that infects bacteria. The virus does so by injecting its DNA into the bacteria. The DNA then replicates and matures into more viruses. These viruses then burst out, and essentially kill the bacteria.
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TutorMe homepage | 536 | 2,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2020-45 | latest | en | 0.888998 |
http://agxl.co.in/question.php?classlink=class-viii&topiclink=exponents-and-powers | 1,555,870,950,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578532050.7/warc/CC-MAIN-20190421180010-20190421202010-00176.warc.gz | 6,028,151 | 9,987 | ### Class VIII
Which of the following = (100 - 990) * 100?
1. 10000
2. 100
3. 9900
4. 99000
What is the value of (-1)-1?
1. 0
2. -1
3. 1
4. None of these
Which of the following is the standard form of 0.00001275?
1. 1.275 * 10-5
2. 1.275 * 105
3. 127.5 * 10-7
4. 127.5 * 107
[(1 / 2)-1 + (2 /3 )2 - (3/4)0]-2 is equal to:
1. 81/484
2. . 81/169
3. 169/81
4. 16/81
[(1 / 2)-1 + (2 /3 )2 - (3/4)0]-2 is equal to:
1. 81/484
2. 81/169
3. 169/81
4. 16/81
0
0
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TOP | 276 | 552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2019-18 | latest | en | 0.766731 |
https://codereview.stackexchange.com/questions/44482/jump-game-to-find-minimum-hops-from-source-to-destination | 1,702,103,831,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100800.25/warc/CC-MAIN-20231209040008-20231209070008-00240.warc.gz | 212,717,773 | 43,009 | Jump game to find minimum hops from source to destination
Given an array start from the first element and reach the last by jumping. The jump length can be at most the value at the current position in the array. Optimum result is when u reach the goal in minimum number of jumps.
For example:
Given array A = {2,3,1,1,4}
possible ways to reach the end (index list)
i) 0,2,3,4 (jump 2 to index 2, then jump 1 to index 3 then 1 to index 4)
ii) 0,1,4 (jump 1 to index 1, then jump 3 to index 4)
Since second solution has only 2 jumps it is the optimum result.
A lot of input has been derived from previous review here.
public final class JumpGame {
private JumpGame() {}
/**
* Returns the shortest jump path from the source to destination
*
* @param jump The jump array
* @return Returns one of the shortest paths to the destination.
*/
public static List<Integer> getShortestJumps(int[] jump) {
final List<Integer> list = new ArrayList<Integer>();
for (int i = 0; i < jump.length - 1; ) {
int iSteps = Math.min(jump.length - 1, i + jump[i]); // iSteps is all the consecutive steps reachable by jumping from i.
int maxStep = Integer.MIN_VALUE; // max step is one of the step in iSteps, which has the max potential to take us forward.
/* trying each step of iSteps */
for (int j = i + 1; j <= iSteps; j++) {
/* being greedy and picking up the best step */
if (maxStep < jump[j]) {
maxStep = j;
}
}
}
return list;
}
public static void main(String[] args) {
int[] a1 = {2,3,1,1,4};
List<Integer> expected = new ArrayList<Integer>();
Assert.assertEquals(expected, getShortestJumps (a1));
int[] a2 = {3, 1, 10, 1, 4};
expected = new ArrayList<Integer>();
Assert.assertEquals(expected, getShortestJumps (a2));
}
}
Code-Style
For the most part, this is good:
• return types are good.
• Collections are well typed
• method names are good
• apart from list, a1, and a2, variable names are good
The list is a bad name for a List. The fact that it is a list is obvious... but, what is the list...? I prefer result to list.
a1 and a2 are not bad, but they are not good either. They are just 'OK'.
Algorithm
I have worked the code through, and figure there is a plane-jane O(n) solution to the problem. Your algorithm is something larger than O(n), It is closer to O(n * m) where m is the longest jump.
The alternate algorithm I cam up with sets up a 'range' for the next jump.... It seaches all the values from the current cursor to the current range, and it looks for the jump within that range that would extend the range the furthest.
The method looks like:
public static List<Integer> getShortestJumps(int[] jump) {
final List<Integer> list = new ArrayList<Integer>();
if (jump.length == 0) {
return list;
}
int cursor = 0;
int best = 0;
int range = 0;
int remaining = 1;
while (cursor + 1 < jump.length) {
if (cursor + jump[cursor] > range) {
// jumping from here would extend us further than other alternatives so far
range = cursor + jump[cursor];
best = cursor;
if (range >= (jump.length - 1)) {
// in fact, this jump would take us to the last member, we have a solution
break;
}
}
if (--remaining == 0) {
// got to the end of our previous jump, move ahead by our best.
remaining = range - cursor;
}
cursor++;
}
// always add the last member of the array
Your unit tests could be less verbose, using java.util.Arrays.asList(...).
int[] a1 = {2,3,1,1,4}; | 905 | 3,386 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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} | 3.0625 | 3 | CC-MAIN-2023-50 | longest | en | 0.778757 |
https://www.jamiiforums.com/threads/worlds-toughest-simple-question-know-the-ans-chk-it-out.118800/ | 1,516,447,072,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889567.48/warc/CC-MAIN-20180120102905-20180120122905-00214.warc.gz | 925,489,594 | 20,795 | World's toughest simple question. Know the ans CHK IT OUT. | JamiiForums | The Home of Great Thinkers
# World's toughest simple question. Know the ans CHK IT OUT.
Discussion in 'Tech, Gadgets & Science Forum' started by Good Guy, Mar 17, 2011.
1. ### Good GuyJF-Expert Member
#1
Mar 17, 2011
Joined: Oct 24, 2010
Messages: 4,511
Trophy Points: 145
I can bet you can't answer this question.
Okay so you are driving a bus with 16 people in it. In the first stop 4 people leave the bus while 6 people enter the bus. Again in the other stop 7 leave and 9 enter. In the next two stops a total of 2 people have left the bus. Also in the next 4 stops a total of 9 people have entered the bus and 3 have left the bus.
WHAT IS THE NAME OF THE BUS DRIVER?
Hahaha don't know do ya,
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but i just told you in the begining it's you nut head,hahaha.. Take some time out and chilax u'r brain it's overheating
2. ### BushbabyJF-Expert Member
#2
Mar 18, 2011
Joined: Dec 29, 2010
Messages: 1,577
Trophy Points: 145
3. ### K007Member
#3
Mar 18, 2011
Joined: Jan 30, 2011
Messages: 61
Trophy Points: 0
The name is bushman. Hahahaha
4. ### daughterJF-Expert Member
#4
Mar 18, 2011
Joined: Jun 22, 2009
Messages: 1,274
Trophy Points: 135
Ayaaaa yah!
Na vile niling'ang'ana kucalculate niki-imagine swali ni
'how many people are left in the bus after all the 8 stops?'
5. ### BushbabyJF-Expert Member
#5
Mar 18, 2011
Joined: Dec 29, 2010
Messages: 1,577
Trophy Points: 145
sasa mbona hujaweka jibu?
6. ### CestusJF-Expert Member
#6
Mar 18, 2011
Joined: Jan 23, 2011
Messages: 1,000
Trophy Points: 0
uliniacha jupiter!
7. ### Good GuyJF-Expert Member
#7
Mar 18, 2011
Joined: Oct 24, 2010
Messages: 4,511
Trophy Points: 145
Yeah they always do that.hahaha
8. ### GurtaJF-Expert Member
#8
Mar 19, 2011
Joined: Sep 17, 2010
Messages: 2,246
Trophy Points: 145
I am (Gurta) driving the bus a.k.a Mimi hapa ndo naendesha basi!!
Damn it, it took me two whole days to figure out, really in life there are so many tidbits that we just ignore when they are soo important!!
9. ### g.n.nJF-Expert Member
#9
Mar 21, 2011
Joined: Oct 8, 2010
Messages: 413
Trophy Points: 35
:shock:
10. ### UZEE MVISenior Member
#10
Mar 22, 2011
Joined: Feb 6, 2011
Messages: 192
Trophy Points: 35
Rudi kwa mwalimu wako wa methali nahau na vitendawili.
11. ### MwendabureJF-Expert Member
#11
Mar 22, 2011
Joined: Mar 10, 2011
Messages: 2,054
Trophy Points: 180
Dereva wa basi hilo ni Bw. Okey!
12. ### Good GuyJF-Expert Member
#12
Mar 22, 2011
Joined: Oct 24, 2010
Messages: 4,511
Trophy Points: 145
It's easy,ni kwamba this tries to shows that we tend to ignore the simple things in life and consider complx things which we never need mtu akisoma hilo swali ataanza kucalculate wa2 wanaoshuka na kupanda mwisho swali linauliza dereva ni nani unashidwa kujua mpaka uangalie tena huku dereva ni wewe! Get it.
13. ### BushbabyJF-Expert Member
#13
Mar 22, 2011
Joined: Dec 29, 2010
Messages: 1,577
Trophy Points: 145
The name of Bus driver is: Bushbaby
[/B][/COLOR][/QUOTE]
Kumbe nilipatia!! makofi tafadhali.... tak tak tak
14. ### Good GuyJF-Expert Member
#14
Mar 22, 2011
Joined: Oct 24, 2010
Messages: 4,511
Trophy Points: 145
Kumbe nilipatia!! makofi tafadhali.... tak tak tak[/QUOTE]
Not until you read the question twice!
15. ### BushbabyJF-Expert Member
#15
Mar 22, 2011
Joined: Dec 29, 2010
Messages: 1,577
Trophy Points: 145
Not until you read the question twice![/QUOTE]
any way ni kweli, mwanzo nilianza kupiga hesabu hiyo.... kumbe simple!!
16. ### Good GuyJF-Expert Member
#16
Mar 22, 2011
Joined: Oct 24, 2010
Messages: 4,511 | 1,306 | 3,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.857669 |
https://www.numbersaplenty.com/245574122116004 | 1,611,589,734,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703587074.70/warc/CC-MAIN-20210125154534-20210125184534-00476.warc.gz | 884,827,848 | 3,791 | Search a number
245574122116004 = 221146722157539389
BaseRepresentation
bin110111110101100100101110…
…010100001001001110100100
31012012111122201210201112112022
4313311210232110021032210
5224141441300230203004
62230143105543442312
7102504100450532543
oct6765445624111644
91165448653645468
10245574122116004
117127a444915190
1223661ab0756398
13a7047041bc7b6
151d5ce2e5692be
hexdf592e5093a4
245574122116004 has 48 divisors (see below), whose sum is σ = 469849303909440. Its totient is φ = 111380342520960.
The previous prime is 245574122115991. The next prime is 245574122116021. The reversal of 245574122116004 is 400611221475542.
245574122116004 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a Harshad number since it is a multiple of its sum of digits (44).
It is an unprimeable number.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 455012342 + ... + 455551730.
It is an arithmetic number, because the mean of its divisors is an integer number (9788527164780).
Almost surely, 2245574122116004 is an apocalyptic number.
It is an amenable number.
245574122116004 is a deficient number, since it is larger than the sum of its proper divisors (224275181793436).
245574122116004 is a wasteful number, since it uses less digits than its factorization.
245574122116004 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 562028 (or 562026 counting only the distinct ones).
The product of its (nonzero) digits is 537600, while the sum is 44.
The spelling of 245574122116004 in words is "two hundred forty-five trillion, five hundred seventy-four billion, one hundred twenty-two million, one hundred sixteen thousand, four". | 525 | 1,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.801975 |
https://unapologetic.wordpress.com/2009/08/03/the-determinant-of-a-positive-definite-transformation/?like=1&source=post_flair&_wpnonce=671450efa2 | 1,524,640,079,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947705.94/warc/CC-MAIN-20180425061347-20180425081347-00123.warc.gz | 729,060,101 | 22,860 | # The Unapologetic Mathematician
## The Determinant of a Positive-Definite Transformation
Let’s keep pushing the analogy we’ve got going.
First, we know that the determinant of the adjoint of a transformation is the complex conjugate of the determinant of the original transformation (or just the same, for a real transformation). So what about self-adjoint transformations? We’ve said that these are analogous to real numbers, and indeed their determinants are real numbers. If we have a transformation $H$ satisfying $H^*=H$, then we can take determinants to find
$\displaystyle\det(H)=\det(H^*)=\overline{\det(H)}$
and so the determinant is real.
What if $H$ is not only self-adjoint, but positive-definite? We would like the determinant to actually be a positive real number.
Well, first let’s consider the eigenvalues of $H$. If $\lvert v\rangle$ is an eigenvector we have $H\lvert v\rangle=\lambda\lvert v\rangle$ for some scalar $\lambda$. Then we can calculate
$\displaystyle\langle v\rvert H\lvert v\rangle=\langle v\rvert\lambda\lvert v\rangle=\lambda\langle v\vert v\rangle=\lambda\lVert v\rVert^2$
If $H$ is to be positive-definite, this must be positive, and so $\lambda$ itself must be positive. Thus the eigenvalues of a positive-definite transformation are all positive.
Now if we’re working with a complex transformation we’re done. We can pick a basis so that the matrix for $H$ is upper-triangular, and then its determinant is the product of its eigenvalues. Since the eigenvalues are all positive, so is the determinant.
But what happens over the real numbers? Now we might not be able to put the transformation into an upper-triangular form. But we can put it into an almost upper-triangular form. The determinant is then the product of the determinants of the blocks along the diagonal. The $1\times1$ blocks are just eigenvalues, which still must be positive.
The $2\times2$ blocks, on the other hand, correspond to eigenpairs. They have trace $\tau$ and determinant $\delta$, and these must satisfy $\tau^2<4\delta$, or else we could decompose the block further. But $\tau^2$ is definitely positive, and so the determinant $\delta$ has to be positive as well in any of these blocks. And thus the product of the determinants of the blocks down the diagonal is again positive.
So either way, the determinant of a positive-definite transformation is positive.
August 3, 2009 - Posted by | Algebra, Linear Algebra
## 7 Comments »
1. Just a question: doesn’t the real case follow from the complex case (just treat the real transformation as a complex one in which all numbers happen to be real, the upper-triangularizing basis might not be real but who cares: the determinant is basis-independent)?
Comment by Piotr | August 3, 2009 | Reply
2. The line between “useful technique” and “dirty trick” is fine indeed.
Comment by John Armstrong | August 3, 2009 | Reply
3. @Piotr: since we seem to be talking about self-adjoint and positive definite transformations, in both the real and complex cases it turns out that such a transformation is diagonal with respect to some orthonormal basis. So certainly in the real case we can put the transformation into upper triangular and even diagonal form.
Comment by Todd Trimble | August 4, 2009 | Reply
4. Todd, that’s right, but I haven’t shown it yet, so I can’t use it. The easy way to show it, though, would use even more examination of eigenpairs than I did in the post.
Comment by John Armstrong | August 4, 2009 | Reply
5. Yeah, I figured you hadn’t (I tried the search terms “spectral” and “diagonalization” just to make sure). But I didn’t want Piotr to come away with a wrong impression that “Now we might not be able to put the transformation into an upper-triangular form” means literally what it says — there is an implicit preface “for all we know from our development thus far” in there.
Comment by Todd Trimble | August 4, 2009 | Reply
6. There’s an interesting philosophical point there: the difference between mathematics-as-completed-entity and mathematics-situated-in-context.
I think that the transition is a subtle one, but an important one in terms of the growth of a mathematical student. Through most of school, mathematics can be (and is) taken as a completed thing given down from on-high. It takes a definite shift in perspective to see it as incomplete and in-development.
In a sense, one could say that the mathematics I’m talking about here is different than, say, the mathematics in Axler’s text, because I started at a different place and made different decisions along the way. There’s a sort of “path-dependency” to mathematics. The question, then, is how all these paths converge on the same results? How can the Secret Blogging Seminar doesn’t have to start at the beginning, but can assume a general, common background?
Comment by John Armstrong | August 4, 2009 | Reply
7. […] is a unit complex number, and the determinant of a positive-semidefinite transformation is a nonnegative real number. If is nonsingular, so is actually positive-definite, then will be strictly positive, so the […]
Pingback by Polar Decomposition « The Unapologetic Mathematician | August 19, 2009 | Reply | 1,226 | 5,211 | {"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": 19, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-17 | latest | en | 0.812483 |
https://www.pressreader.com/usa/the-outpost/20230130/281659669177207 | 1,695,310,639,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506028.36/warc/CC-MAIN-20230921141907-20230921171907-00258.warc.gz | 1,059,050,258 | 3,781 | The Outpost
# Looking through the mathematical lens
- Chaplain’s Corner Chaplain Capt. Ryan Pearse
In his book, “The Math of
Life and Death: 7 Mathematical Principles That Shape Our Lives,” author Kit Yates looks at the world and almost everything in it through the lens of mathematics and statistics.
There is an old mathematical joke that says there are only 10 types of people: those who understand binary and those who don’t (10 of course, representing the number two in binary). Even for those who hate math this is a very interesting book with lots of practical applications for everyday life.
The next time you’re at a party with at least 23 people, try The Birthday Paradox party trick. Technically speaking, it’s not a trick and it’s not a paradox either. It’s only a paradox because it’s hard for our brains to understand the compounding power of exponents. If you have 23 random people in a room, what are the chances that two of them share the same birthday? If you have 70 people in a room, what are the chances that two of them share the same birthday? With only 23 people, the chances stand at 50% and with 70 people the odds go up to 99.9%. There is virtually a statistical certainty that there are at least two people who share a birthday in a room of 70 random people.
I have personally tried this out before and I never make it past February before I find two people with matching birthdays. If you’re having a hard time accepting these advanced statistical calculations, you’re not alone. The Birthday Paradox is not intuitive.
When given a choice between a tray with nine jellybeans and a single red one or a tray with 91 white jellybeans and nine red ones, which one would you think has a higher likelihood of selecting a red jellybean?
The second tray is chosen more often because participants think they have a greater chance of selecting red when the greater statistical chance lies with the first tray.
While statistics is rooted in mathematical facts, Yates admits that statistics is as much an art as it is a science and statistics can be cherry-picked. Be very careful to be as judicial as possible when hearing sensational statistics in the media or in a courtroom.
As Pulitzer Prize winner Upton Sinclair put it, “It is difficult to get a man to understand something, when his salary depends upon his not understanding it.” Don’t be a victim of statistical subterfuge, rather use mathematics and statistics honestly and judicially. 1 Thessalonians 5:21 says to, “Test everything, hold fast to what is good”. | 590 | 2,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-40 | latest | en | 0.951568 |
http://www.helpmykidlearn.ie/activities/3-4/detail/octopus-song | 1,579,362,232,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250592636.25/warc/CC-MAIN-20200118135205-20200118163205-00066.warc.gz | 232,689,314 | 6,427 | Video
# Octopus song
### What to do with this activity?
Learn this octopus song together and dance around together. Remember, an octopus has eight legs. Wave your arms around like an octopus, and do the actions that the song suggests.
Here are the words:
Octopus, octopus, swim, swim, octopus.
Octopus, octopus, swim, swim, octopus.
Swim up ...swim down...swim forward...turn around.
Swim up ...swim down...swim forward...turn around.
Octopus, octopus, swim, swim, octopus.
Octopus, octopus, swim, swim, octopus.
Make a simple octopus with an empty toilet roll tube. Just paint the cardboard tube with poster paint and stick on eight strips of coloured paper for the octopus arms. Then with a black marker, draw on two big eyes.
• #### Why am I doing this?
Maths is more than working with numbers. It also consists of shape and space, patterns, measuring – things you do and come across in everyday life. When children begin to learn formal maths at school, they are building on a foundation of early numeracy learning from home. Even though they may not even be aware of it, parents and children engage in numeracy activity as part of their everyday lives.
• #### How can I do more?
Always teach numbers in a natural way through everyday activities and play. Count steps on a stairs, food in your shopping trolley or cows in a field. Compare things when talking big or small, long or short, older or younger and faster or slower: “You carry the small box and I’ll take the big one.” Use the words – up and down, over and under, near or far, more or less when talking to your child. Talk about the shapes of everyday things. Ask your child what shapes they can see around the room they’re in.
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Sign up for more tips | 438 | 1,841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | longest | en | 0.944736 |
https://www.northindiacampus.com/2016/05/download-question-paper-of-digital_36.html | 1,726,772,265,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652055.62/warc/CC-MAIN-20240919162032-20240919192032-00658.warc.gz | 811,455,374 | 55,811 | ## Download Question Paper of "Digital Circuits and Logic Designs 2" , Question Paper of BCA 2nd Semester, Subject Code : BSBC - 205, Paper ID 802091, Paper 8
• Monday, May 30, 2016
Roll No……………..
Total No. of Questions : 07
BCA (Sem.–2)
DIGITAL CIRCUIT & LOGIC DESIGN
Subject Code : BC-205 (2007 to 2010 Batch)
Paper ID : [B0209]
Time : 3 Hrs. Max. Marks : 60
INSTRUCTION TO CANDIDATES :
1. SECTION-A is COMPULSORY consisting of TEN questions carrying
TWO marks each.
2. SECTION-B contains SIX questions carrying TEN marks each and students
has to attempt any FOUR questions.
SECTION-A
1. Write briefly :
i) What are weighted and non weighted codes?
ii) List various applications of logic gates.
iii) What is encoder and decoder? Explain.
iv) List various applications of shift register.
v) What is programmable counter?
vi) What is sequential logic circuit?
vii) Draw logic diagram of 4-input multiplexer.
viii)What are limitations of K-Map?
ix) List various applications of multiplexer and de-multiplexer.
x) Convert binary number (1101.1101) to octal
SECTION-B
2. What is Logic Gate? What are its different types? Explain various applications of logic gates. 10
3. What is binary parallel adder? Draw and explain the working of 4 bit
4. Write short notes on the following :
(a) D Flip Flop
(b) SOP and POS Forms. 10
5. What is Synchronous counter? How it is different from Asynchronous
counter? Draw and explain the working of Mod-6 Asynchronous counter.
10
6. What is SIPO shift register? How it is different from PIPO Shift register?
Draw and explain the working of 4 bit SIPO shift register. 10
7. Explain the following :
(a) Error detection and correction codes
(b) BCD. 10 | 546 | 1,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.468559 |
https://mathhelpboards.com/threads/positive-integer-value-of-n.7962/ | 1,606,290,369,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141181482.18/warc/CC-MAIN-20201125071137-20201125101137-00686.warc.gz | 389,480,759 | 15,891 | # positive integer value of n
#### jacks
##### Well-known member
If $3^{n} +81$ is a perfect square, find a positive integer value of $n$.
My Trail::
When $n\leq 4,$ then easy to know that $3^{n} +81$ is not a perfect square.
Now let $\displaystyle n = k +4 (k\in \mathbb{Z^{+}}),$ then $3^{N} +81 = 81 (3^{k} +1).$
So $3^{N} +81$ is a perfect square, and $81$ is square,
there must be a positive integer $x,$ such that
$3^{k}+1 = x^2\Rightarrow 3^k = (x-1)\cdot (x+1)$
Now How can i solve after that
Help me
Thanks
#### MarkFL
Staff member
I would write:
$$\displaystyle 3^n+81=m^2$$
$$\displaystyle 3^n=(m+9)(m-9)$$
Now, observing that:
$$\displaystyle 18+9=27=3^3$$ and $$\displaystyle 18-9=9=3^2$$
What value do we obtain for $n$?
#### mente oscura
##### Well-known member
If $3^{n} +81$ is a perfect square, find a positive integer value of $n$.
Hello.
$$3^n+81$$, It cannot be a perfect square, For being, both, odd numbers.
Demostration:
$$(2n-1)^2+(2m-1)^2=4(n^2+m^2)-4(n+m)+2=$$
$$=2[2(n^2+m^2)-2(n+m)+1]$$
The square root, of the latter expression, has to be an irrational number, for being divisible, only once for "2".
Regards.
#### mathbalarka
##### Well-known member
MHB Math Helper
mente oscura said:
$3^n+81$, It cannot be a perfect square
False. I can find a precise value $n$ for which the above is a perfect square.
#### mathbalarka
##### Well-known member
MHB Math Helper
Hint : Prove that $18$ is divisible by $m - 9$.
#### mente oscura
##### Well-known member
False. I can find a precise value $n$ for which the above is a perfect square.
Hello.
I am sorry. Really, I have considered "n", even number and, it is not possible, but yes it can be an odd number.
Regards. | 573 | 1,722 | {"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": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2020-50 | longest | en | 0.826204 |
http://www.thefreelibrary.com/Sizing+coil+loops.-a07675013 | 1,474,984,719,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661087.58/warc/CC-MAIN-20160924173741-00289-ip-10-143-35-109.ec2.internal.warc.gz | 777,487,722 | 11,840 | # Sizing coil loops.
Sizing coil loops
It is a common misunderstanding that increasing the distance between coil feed and straightener (a longer loop in the horizontal dimension) will provide a larger amount of slack material in the loop. This is not only untrue, but in fact the result may be less slack in the loop.
Slack is gained only by increasing the loop in the vertical direction. As Figure 2 shows, the amount of slack material available in the loop is the difference between, one, the horizontal distance between feed and straightener, and two, the total length of material in the loop.
You can see from this example that placing the straightener farther from the feed results in substantially less slack material in the loop (3 ft versus 1 ft). The amount of available slack can be further increased by raising the feed-line height, or by installing a loop pit.
Pulls longer loop
In addition to providing less material storage, placing the straightener farther from the feed will also cause it to work harder. The feed will be forced to pull a longer loop (25 ft versus 12 ft) on each cycle. What's more, the reduced amount of slack will impose additional wear on straightener and reel. This is caused by the increased frequency of starting and stopping.
Mild steel can be forced into a radius 360 times its thickness without reintroducing a set into the material. Knowing this, you can calculate the best loop size for any gauge of mild-steel coil stock.
Tougher materials normally can be put into a tighter radius, while softer materials may require a larger radius. In any case, the 360-times-thickness measure is a good starting point.
Guide to length
A useful guide to calculating loop length is given in Figure 3 and its table. Let's look at an example:
The material we will be running is 0.062"-thick mild steel. For this material, our chart recommends a loop length if 96". Remember, this is the horizontal distance from where the material radius begins upon leaving the straightener, and where it ends upon entering the feed. The 96" value is not the total length of material in the loop.
Let's assume our pass line height is 42". At this height, the chart shows we will have 38" of slack material in our loop. (In most cases, the pass line height at the straightener will be different from that at the feed. An average of the two gives a usable figure.)
Now, let's assume the feed length of the job we're running will be 18". A good rule of thumb is that slack accumulation should be at least three times the feed length. In this case, we don't have this much slack (38" actual versus 54" required).
The solution is a loop pit, as shown in Figures 1 and 3. If the original pass line height is beyond 720 times the material thickness (360 x 2 x material thickness), every inch of loop depth will add 2" to slack accumulation.
In our example, an additional 8" of depth would add 16" to the accumulation. This would provide us with the three feed-lengths of slack we require.
PHOTO : 1. In this press installation, a loop pit--at right in the photo--is used to gain the required amount of slack in the loop.
PHOTO : 2. Placing the straightener farther from the feed actually results in less slack material in the loop.
PHOTO : 3. Guide for calculating correct loop length. | 729 | 3,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-40 | latest | en | 0.903371 |
https://cstheory.stackexchange.com/questions/42468/data-strcuture-to-represent-dependencies-amongst-modules | 1,623,851,485,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487623942.48/warc/CC-MAIN-20210616124819-20210616154819-00196.warc.gz | 182,391,584 | 39,669 | # Data Strcuture to represent dependencies amongst modules
Consider several software modules $$m_1, m_2, ... m_n$$. Each module has some inputs and outputs and the inputs to some of the modules are dependent on the outputs of some other modes. For example, in the figure, module $$m_3$$ and $$m_4$$ are dependent on the output of module $$m_1$$. If we are given a set if inputs $$I$$ and a set of outputs $$O$$, the objective is to find a set of modules which produce this output.
For example, if the output set $$O$$ is $$\{o_1, o_2\}$$ then all of the modules would have to be used. However if the output set $$O$$ is $$\{o_2\}$$ then only $$\{m_2, m5, m_7\}$$ needs to be used.
A graph is a natural model for such an objective. Are there any alternative data structures to represent this description?
Consider a modification to the earlier example as shown in the figure below, where all the outputs produced are $$o_1$$ and the output set required is also $$O = \{o_1\}$$. In this scenario we have two possible set of modules which satisfy our goal of $$O$$ : $$\{m_2, m_5, m_7\}$$ and $$\{m_1, m_3, m_4, m_6\}$$. We wish to select the set which minimizes the number of modules used, hence the set $$\{m_2, m_5, m_7\}$$ would be our desired set $$O$$. Would a graph still be the most natural representation of such a scenario or some other data structure would be better off?
• I don't understand what is the input to the algorithm. Are we given the graph of modules? And you want to find a subset of nodes such that some condition holds? If so, what is that condition? You haven't defined what outputs a set of modules produces. For instance, what is the output produced by the set $\{m_1,m_7\}$? Do you require the set to have some property (e.g., if a node is in the set, all its predecessors are as well)? – D.W. Mar 3 '19 at 23:41
• @D.W. The input to the algorithm is the set of modules, where each module has some input and output. In addition, the inputs are sets $I$ and $O$. The outputs of a module say $m_1$ serves as input to modules $m_3$ and $m_4$, this represents the fact that $m_1$ must be executed before and to drive $m_3$ and $m_4$. We are not given this graph but I represent this scenario using a graph. Thus we are required to find a subset of the nodes that take $I$ as input and produce $O$ as output. $I$ are inputs to the modules which are not dependent on the output of any other module. – kauray Mar 4 '19 at 6:32
• Why isn't this just a matter of programming? For each module you're given its inputs, so look at which modules have inputs from $I$. Do you really mean an arbitrary subset of nodes, or do you require that it be closed in some way (e.g., if a node is in the set, then all its predecessors must be in the set too)? The problem doesn't seem well-formulated yet; and before it can be solved, first you must figure out how to formulate it clearly. – D.W. Mar 5 '19 at 7:15
• @D.W. yea if a node is in the set, its predecessors must be in the set too – kauray Mar 5 '19 at 15:14
• Is the problem as follows? Given a directed acyclic graph $G$, and some source vertices $I$ and sink vertices $O$, find a subset $S \subseteq V$ of vertices such that $I \cup O \subseteq S$ and if $v \in S$ then all predecessors of $v$ are in $S$ as well? Is that the problem statement? Do you want any set $S$, or the smallest such $S$? Do you want to know the theoretical complexity of this problem, or are you looking for a practical algorithm? It's important that the problem statement be stated clearly in the question. I'm not finding any of that clearly stated in the post. – D.W. Mar 5 '19 at 17:39
Your problem is the following:
Given a dag $$G$$, with labels $$\ell:V\to L$$ on the vertices, and a set $$O \subseteq L$$, find the smallest subset $$S \subseteq V$$ of vertices such that for each $$o \in O$$, there is a vertex in $$S$$ labelled $$o$$.
This problem is NP-hard, by reduction from set cover.
(The reduction: given sets $$T_1,\dots,T_n$$ and universe $$U$$, define a dag with one source vertex for each $$T_i$$, one sink vertex per element of $$U$$, labelled with that element, and an edge $$T_i \to t$$ for each $$t \in T_i$$; and set $$O=U$$. Then the smallest subset of vertices that meets the requirements is also the minimum set cover.)
Since it is NP-hard, you'll need to use any of the standard methods for dealing with NP-hardness: heuristics, an approximation algorithm, an exponential-time algorithm, etc. I suggest reducing this to integer linear programming. Add a zero-or-one boolean variable $$x_v$$ for each vertex $$v$$; for each edge $$v \to w$$, add the inequality $$x_v \ge x_w$$; for each label $$o \in O$$ in the output set, add the inequality $$\sum_{\ell(v)=o} x_v \ge 1$$, where the sum ranges over all vertices with label $$o$$; and then minimize the objective function $$\sum_{v \in V} x_v$$. Then, apply an off-the-shelf ILP solver. Of course this may be exponential time in the worst case, but I suspect it's about as you're likely to get in practice, within any reasonable amount of programming effort.
The data structure to use is a graph. The problem remains NP-hard no matter what data structure you use, so choice of an appropriate data structure isn't going to make the problem any easier.
• How does the problem statement say that if a vertex is in set $S$ then its predecessors are also in the set? – kauray Mar 7 '19 at 5:03
• @kauray, that's what you wrote in your comment: cstheory.stackexchange.com/questions/42468/…. You tell me -- if that's not the problem statement, then edit your question to make it clearer. I'm finding it very hard to extract what the problem statement is. If we have to ask multiple rounds of questions to try to understand what you're asking, that makes it harder to give you a useful answer. – D.W. Mar 7 '19 at 5:05
• Oh sorry this is what captures the problem statement. I had misinterpreted a part of your statement. Thanks a lot! – kauray Mar 7 '19 at 6:00 | 1,599 | 5,985 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 40, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-25 | latest | en | 0.881789 |
https://stacks.math.columbia.edu/tag/0C6G | 1,718,528,806,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00455.warc.gz | 489,945,328 | 5,975 | Lemma 10.103.5. Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak m$. Let $M$ be a finite $R$-module. Let $x \in \mathfrak m$ be a nonzerodivisor on $M$. Then $M$ is Cohen-Macaulay if and only if $M/xM$ is Cohen-Macaulay.
Proof. By Lemma 10.72.7 we have $\text{depth}(M/xM) = \text{depth}(M)-1$. By Lemma 10.63.10 we have $\dim (\text{Supp}(M/xM)) = \dim (\text{Supp}(M)) - 1$. $\square$
There are also:
• 6 comment(s) on Section 10.103: Cohen-Macaulay modules
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 229 | 675 | {"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": 2, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-26 | latest | en | 0.540698 |
http://www.mathworks.com/matlabcentral/answers/76983-counting-in-a-matrix?requestedDomain=www.mathworks.com&nocookie=true | 1,469,420,596,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824204.27/warc/CC-MAIN-20160723071024-00277-ip-10-185-27-174.ec2.internal.warc.gz | 562,275,186 | 12,530 | Counting in a Matrix
on 26 May 2013
Walter Roberson (view profile)
Hi all,
I have a 3 by 7 matrix as following:
g =
``` 0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0```
if I write
g(9) = 1
I get:
g =
``` 0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 1 0 0 0 0```
when actually want g(9) = 1 to be:
g =
``` 0 0 0 0 0 0 0
0 1 0 0 0 0 0
0 0 0 0 0 0 0```
or in other terms, instead of counting down columns for 1,2,3,4,5...... I want to count across rows. Is there an option in Matlab to change this function.
Matthew.
Products
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Walter Roberson (view profile)
on 26 May 2013
No, there is no option to change the order of the dimensions.
If there is an especially good reason to count across, you could use
```Tg = g.';
Tg(9) = 1;
g = Tg.';
```
There are also calculations that can be done to switch between the two orderings, but my algebra center in my brain is telling me it is the weekend and it wants to sleep in...
Image Analyst (view profile)
on 26 May 2013
You can transpose it, change it, and transpose it. But why not just use row, column indexing? There are also functions subs2ind() and ind2subs() that you might be interested in.
MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi | 491 | 1,487 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | latest | en | 0.924042 |
https://thetextchemistry.org/qa/what-is-the-minimum-distance-between-columns.html | 1,607,169,742,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141747774.97/warc/CC-MAIN-20201205104937-20201205134937-00248.warc.gz | 534,237,572 | 8,301 | # What Is The Minimum Distance Between Columns?
## What should be the distance between two columns?
Thumb rule no.
2 Distance between the columns Try to maintain equal distance between the centres of two columns.
Always plan a column layout on a grid.
The distance between two columns of size 9”x9” should not be more than 4m centre to centre of column..
## What is standard size of beam?
THE STANDARD SIZE OF THE BEAMS In a residential building it is 9 ʺ × 12 ʺ or 225 mm × 300 mm standard size according to (IS codes). The minimum RCC beam size should not be less than 9 ʺ × 9 ʺ or 225mm × 225mm with the addition of a 125mm slab thickness.
## What is the strongest shape for a column?
The strongest column has an equilateral triangle as cross section, and it is tapered along its length, being thickest in the middle and thinnest at its ends. Its buckling load is 61.2% larger than that of a circular cylinder.
## What is the maximum length of a beam?
One can use beams of size 9″ X 12″ (225 MM x 300MM) with a slab thickness of 5″ (125 MM) cast in M20 concrete for spans up to 5m. There are other considerations like secondary and tertiary spans, point loads and wall loads which have to be considered.
## How do you show columns in plan?
Open a column family or start a new column family. Click Create tab Properties panel (Family Category and Parameters). In the Family Category and Parameters dialog, under Family Parameters, select or clear the Show family pre-cut in plan views parameter. If you want the column to display in plan view…
## What is the effective length of a column?
The effective column length can be defined as the length of an equivalent pin-ended column having the same load-carrying capacity as the member under consideration. The smaller the effective length of a particular column,the smaller its danger of lateral buckling and the greater its load carrying capacity.
## How far apart should pillars be?
The distance between pillars should be according to the requirement of construction site and load of RCC slab. Generally, the distance between two pillars may be 10ft. to 20 ft.
## What is minimum eccentricity of column?
For a short column, when the unsupported length l = lex (for the purpose of illustration), we can assume l = 12 D (or 12b when b is considered). Thus, we can write the minimum eccentricity = 12D/500 + D/30 = 0.057D, which has been taken as 0.05D or 0.05b as the maximum amount of eccentricity of a short column.
## What is the maximum span of a beam?
2 grade, there is a method that can be used for estimating beam spans for a preliminary design. When supporting joists that span 12 feet with no overhang beyond the beam, a double ply beam can span in feet a value equal to its depth in inches. A double 2×12 beam can span 12 feet; a (2) 2×10 can span 10 feet and so on.
## What is eccentricity of a column?
The distance between the middle of cross-section of the column and the eccentric load is referred as eccentricity symbolized by e. Increase in the eccentric load increases the axial load and the moment acting on the column. This makes the column to bend additional increasing the bending of column.
## What is the standard size of column?
For this general thumb rule, we will assume a structure of G+1 floors high, using standard 6″ walls. Minimum size of an RCC column should not be less than 9” x 12” (225mm x 300mm) with 4 bars of 12 MM Fe415 Steel. These days the minimum I use in my projects is 9″ x 12″ (225 mm x 300mm) with 6 bars of 12 MM Fe500 steel.
## What is the maximum distance between columns?
The Minimum Spacing or Distance between two column should not be less than 2.5 Meter. It Must be 2.5 Meter minimum distance. Or, Maximum Spacing & Distance of two Columns Should not be less than 7.5 meter.
## How do you determine the size of a column?
Steps to decide the size of the column: After defining the type and size of the slab, the load transferring on the slab to the beam is found out and the size of the beam is decided. Once the loads on beams are calculated then the load transfer from beam to the column is estimated by moment distribution method.
## How do I create a column layout plan?
The column layout plan is very important for a Structure….Numbering the column.Column Shape Choose. … Draw the Column. … Fixed the Column Location. … Set the Grid Line. … Numbering the Grid Line. … Set the Dimension Respect to Grid Line.More items…
## Can a column rest on a beam?
Usually columns rest on the foundation to transfer load from slabs and beams. But the floating column rests on the beam. This means that the beam which supports the column acts as a foundation. That beam is called as a transfer beam.
## What is limit of eccentricity?
Limit of eccentricity (e limit): The maximum distance of load from the centre of column, such that if load acts within this distance there is no tension in the column. The maximum distance is called Limit of eccentricity.
## How do you put columns in a building?
Columns should preferably be located at (or) near the corners of a building, and at the intersection of beams/walls. Select the position of columns so as to reduce bending moments in beams. Avoid larger spans of beams. Avoid larger centre-to-centre distance between columns.
## What are the 3 types of columns?
The three major classical orders are Doric, Ionic, and Corinthian. The orders describe the form and decoration of Greek and later Roman columns, and continue to be widely used in architecture today. | 1,270 | 5,538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-50 | latest | en | 0.906607 |
https://ininet.org/tracking-hurricane-katrina.html | 1,716,142,559,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057819.74/warc/CC-MAIN-20240519162917-20240519192917-00163.warc.gz | 284,371,800 | 6,646 | # Tracking Hurricane Katrina
Date 18.10.2016 Size 25.89 Kb. #1605
Name
Tracking Hurricane Katrina
Hurricane Katrina of the 2005 Atlantic hurricane season was the costliest natural disaster, as well as one of the five deadliest hurricanes in the history of the United States. At least 2,000 people lost their lives in the actual hurricane and in the subsequent floods, making it the deadliest U.S. hurricane since the 1928 Okeechobee hurricane: total property damage was estimated at \$81 billion, nearly triple the damage wrought by Hurricane Andrew in 1992.
Hurricane Katrina formed over the Bahamas on August 23, 2005 and crossed southern Florida as a moderate Category 1 hurricane, causing some deaths and flooding there before strengthening rapidly in the Gulf of Mexico. The storm weakened before making its second landfall as a Category 3 storm on the morning of Monday, August 29 in southeast Louisiana. It caused severe destruction along the Gulf coast from central Florida to Texas, much of it due to the storm surge. The most severe loss of life occurred in New Orleans, Louisiana, which flooded as the levee system catastrophically failed, in many cases hours after the storm had moved inland. Eventually 80% of the city and large tracts of neighboring parishes became flooded, and the floodwaters lingered for weeks. However, the worst property damage occurred in coastal areas, such as all Mississippi beachfront towns, which were flooded over 90% in hours, as boats and casino barges rammed buildings, pushing cars and houses inland, with waters reaching 6–12 miles (10–19 km) from the beach.
Objective: Using the information from the Hurricane Katrina data chart, track the path the hurricane took during the time periods listed using latitude and longitude.
Procedure:
1. Using the Hurricane Katrina data chart plot the positions of the storm from August 23rd-September 1st
2. Label each position
3. Using the Saffir- Simpson Hurricane Scale, determine the storms category for each position. (For example a tropical storm = TS, Hurricane category 3 =3)
4. Connect the positions with a solid line and an arrow to show direction
5. Repeat steps 1-4 for hurricane Igor
Hurricane Katrina
Date Time Location Wind speed in MPH 8/23 P.M 24*N 72*W 36 8/24 A.M P.M 26*N 74*W 26.5*N 76*W 39 44 8/25 A.M P.M 26.5*N 79*W 25*N 80.5*W 62 68 8/26 A.M P.M 25*N 82*W 24*N 85*W 71 84 8/27 A.M P.M 24*N 86*W 25*N 87*W 96 112 8/28 A.M P.M 26*N 89*W 27*N 90*W 149 177 8/29 A.M P.M 29*N 90*W 31*N 89.5*W 154 124 8/30 A.M P.M 32*N 89*W 34*N 87*W 112 91 8/31 A.M P.M 36*N 86*W 37*N 85*W 55 46 9/1 A.M P.M 39*N 84*W 40*N 83*W 37 30
Hurricane Igor
9/12 P.M 20*N 48*W 101 9/14 A.M P.M 22*N 51*W 24*N 54*W 122 121 9/16 A.M P.M 26*N 60*W 27*N 62*W 139 147 9/18 A.M P.M 30*N 66*W 31*N 68*W 121 120 9/20 A.M P.M 35*N 71*W 36*N 70*W 88 79 9/22 A.M P.M 39*N 68*W 40*N 66*W 46 41 9/23 A.M. P.M. 41*N 65*W 42*N 63*W 37 31 9/24 A.M. 42*N 62*W 24
Questions:
1. According to the Saffir-Simpson scale, what is the relationship between air pressure and wind velocity?
1. According to the Hurricane Katrina chart, compare the wind speed and air pressure on August 29th (A.M) to the wind speed and air pressure on August 31st.
1. Considering your answer to the previous question, what might be the source of a hurricanes’ energy?
1. What precautions might someone take a few days before the arrival of a major hurricane like Katrina?
1. Besides wind and rain what other factor is usually most devastating to coastal communities during a major hurricane?
1. Even though hurricane Igor remained over the ocean for its’ entire life, why did it begin to lose intensity around September 17th? | 1,094 | 3,685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-22 | latest | en | 0.951598 |
https://community.roonlabs.com/t/dynamic-range-measurement-is-different-than-http-dr-loudness-war-info/115994 | 1,720,791,812,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514404.71/warc/CC-MAIN-20240712125648-20240712155648-00793.warc.gz | 136,193,054 | 9,619 | # Dynamic range measurement is different than http://dr.loudness-war.info
Roon’s Dynamic Range measurement system is lousy.
For example, Donna Summer’s CD, Bad Girls, which has a dynamic range of +16, according to the web http://dr.loudness-war.info/, is detected by Roon with a Dynamic Range of +7.
So the Roon measurement is very unreliable. Can you do something about it?
Thank you.
It’s not lousy, it uses a different system which generates a different value.
It seems that you would like them to use the system used by DR database. For reasons discussed extensively on this website, they don’t. This is from Roon’s KB explaining why they use R128 instead of the system the DR database uses:
## How is Dynamic Range computed?
As with Volume Leveling Roon’s dynamic range calculation is done based on R128 standards. In technical terms, Roon’s “Dynamic Range” is the same as R128’s “Loudness Range”.
That there are older methods of computing Dynamic range out there–most commonly by measuring the “Crest Factor”.
Crest Factor measurements reflect the difference between the average volume and the peak volume–so they are easily swayed by periods of silence or near-silence (which distorts the average), and by short-duration peaks–which may not represent the volume of the loudest parts of the track accurately.
The R128 method is more resillient. It begins by computing the statistical distribution of loudness values present at different points in the track, ignoring periods of silence. The computed dynamic range represents the difference between the 10th percentile and the 95th percentile of that distribution. In other words, the “top” of the range is the volume level that 95% of the track sits below, and the “bottom” is the volume level that 90% of the track sits above.
Though both methods portray roughly the same information, Crest Factor values aren’t directly comparable to values produced using the R128 method.
12 Likes
Different measurement standards; different master?
I don’t think you’re comparing apples with apples.
Roon’s measurement of dynamic range is different to the six albums (that I can find) measurements of dynamic range in the web dynamic range database. The only ‘16’ I can see is a vinyl rip.
Which master is Roon measuring for you?
Roon uses the R128 standard, the loudness war website uses crest factor.
There’s a feature request to make crest factor a selectable option:
1 Like
This is the disc that I have and the measurement that I do gives me the same data as that published on the web Dynamic Range DB.
The maximum range is +17
http://dr.loudness-war.info/album/view/157209
What result do you get if you analyse that track using the R128 algorithm?
He’s already said in his opening post.
It’s like measuring the temperature if Centigrade or Fahrenheit. Fahrenheit sounds better but 32 degrees is still Bloody warm in centigrade
2 Likes
I’ve just checked three of my cd rips using the Orban loudness meter analysis. Using default settings I get comparable results to those reported by Roon.
What did surprise me was the number or reconstructed samples that were over 0dBFS on two of the files.
Actually… maybe I wasn’t surprised. Black Keys and Amy Winehouse.
No, it’s not like Centigrade or Fahrenheit. Those two scales are measuring the same thing and you can easily convert from one to another. You can’t convert between R128 and the crest factor results from the DR Database.
3 Likes
Moderators have temporarily closed this thread as it was being derailed by continued discussion of moderation. Read this post first and then open a thread in the Roon Community Site section if you wish to discuss moderation.
1 Like
This topic was automatically opened after 34 hours.
I think the bottom line might be that LRA is not a measure of dynamic range, nor should it be used as a proxy for dynamic range.
As an aside, it turns out I have five versions of Brown Eyed Girl, four of which appear on CD ‘samplers’. They are all different.
The version with the highest LRA number turns out to be the mono radio edit; it is by far the poorest ‘sound’. The two channels are not identical, leading me to suspect a transcription from vinyl.
The version with the lowest LRA number is on the ‘Still On Top’ compilation, which I guess is ‘remastered’… and also has many inter-sample peaks above 0dBFS.
What a mess…
1 Like
Thanks, I didn’t know about the Orban loudness meter program. Running it on a smattering of my files reveals a pattern: most music I have that is not expressly mastered in “audiophile” ways has plenty of reconstructed samples over 0dBFS, even if its loudness measures are pretty good.
I suspect that Roon’s volume leveling hides these from my DAC in most cases.
I think that’s my position too. I’ve not bothered adding headroom management. Although I have noticed a couple of albums where volume levelling is increasing level, so I have some nagging doubt…
I sure like using R128 for volume leveling. It is brilliant for that. I wouldn’t want Roon without it!
But I sure wish Roon users had the option to use the DR Database way of measuring DR for displaying DR when comparing album versions.
I haven’t given up hope that we will see this option. They revamped the waveforms for us local file junkies, and comparison of versions utilize both waveforms and DR values frequently.
This is another one of those issues with Roon where they would be better off giving users the choice to select the scale this wish to use and are more familiar with. More options is better than being told to just use an option we did not ask for or want. | 1,225 | 5,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-30 | latest | en | 0.926104 |
https://www.kiasuparents.com/kiasu/question/23890-2/ | 1,611,477,170,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703547475.44/warc/CC-MAIN-20210124075754-20210124105754-00266.warc.gz | 846,881,863 | 25,339 | # Question
Hi, can anyone help with question b please? Thank you.
Source: Nan Chiau Primary
I would solve (b) as follows ;
10 + 20 + 30 + … + 790 = (800 x 39) + 400 = 31600
or
10 + 20 + 30 + … + 790 = (790 x 39) + 790 = 31600
Ans : 316m.
0 Replies 1 Like
You can use any method to find the sum of consecutive numbers from 10cm to 790cm ( in steps of 10 ) | 132 | 365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.878427 |
https://www.articleslash.net/Business/Management/326186__Cost-Volume-Profit-Analysis-What-Is-It-All-About.html | 1,563,287,263,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524568.14/warc/CC-MAIN-20190716135748-20190716161748-00318.warc.gz | 609,983,704 | 12,029 | Cost Volume Profit Analysis - What Is It All About?
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Introduction
Running a successful small business requires adept navigation of the many choices created by an ever changing market place. Cost Volume Profit Analysis (CVPA) is an effective tool that can help its user answer important questions such as “what price should I charge for this product or that service?", “which of my products or services is most profitable?", and “what is the best operating leverage level for my business given current market conditions?"
Understanding Fixed, Semi-Variable, and Variable Costs
Before the CVPA can be used, fixed, semi-variable and variable costs must be determined. Determining these costs is a very useful tool in itself, but that's another white paper.
Fixed costs are those costs that your business incurs regardless of sales volume. These are costs such as rent, insurance, and annual business licensing fees. Sales volume, not exceeding your current capacity, has no effect.
Variable costs are those costs that are directly affected by sales volume. These include items such as cost-of-goods sold, sales commissions, and travel expenses, if you are a service provider that travels as a result of service provision.
Semi-variable costs, as you have determined by now, are those costs that increase with sales volume but not directly as with variable costs. An example of a semi-variable cost for an auto body shop might be equipment maintenance expense. At some point, equipment begins to break down if not maintained at a level consistent with increased use. Therefore, in order to avoid equipment breakdown due to hyper-use, the business owner must spend additional funds on maintaining equipment.
Break-Even
There are several benefits to using CVPA. First, it shows what the break-even point, in units or dollars, for a given product or service is, given a specified sales price. Break-even is the point at which sales revenue covers all fixed costs for the year plus all variable costs up to that sales point. For example, if fixed costs for the year are \$1,000, variable costs per unit total \$1.00, and the product is priced at \$5.00, then 250 units must be sold to cover fixed and variable costs totaling \$1,250.
As you may have noticed, not only does CVPA show break-even, but it can be used for analyzing price sensitivity. For instance, if your competitor is able to price the same product at \$2.50, but you are not able to go below \$3.00, then it may be time to consider several options: discontinue the product, find a way to reduce fixed and variable costs so you can price it at \$2.50, tweak the product in some way that distinguishes it in a positive way from your competitor's-a square hamburger vs. a round hamburger-or use the product as a “loss leader" to get customers in the door.
Contribution Margin
Determining the contribution margin for your business is an additional benefit of CVPA. Contribution margin is simply the amount of each sales dollar left after all variable costs have been covered. It is that portion of the sales dollar that can be devoted to covering fixed costs.
Knowing your overall contribution margin is beneficial because it can be compared to prior periods to determine if it is trending positively or negatively. Additionally, contribution margin analysis can be applied to individual products, product lines, services, or service lines. Knowing the contribution margin of a particular product or service can help determine if carrying that product or performing that service over another is the best decision. Moreover, understanding contribution margin is very helpful in developing the best pricing strategy for your business.
One final benefit to knowing how to determine contribution margin is that it can point out your most profitable products or services, even though sales may indicate something different. For example, if product A has sales of \$100K and product B has sales of \$80K, it would appear, based on total sales alone, that product A is the more attractive product to emphasize. But a quick contribution margin analysis reveals that product B contributes 0.49 cents of every sales dollar to covering fixed costs vs. product A's 0.34 cents. Clearly, product B is a real contributor and should be part of this retailer's product mix.
Operating Leverage
In gaining an understanding of operating leverage, let's reconsider our hypothetical auto body shop owner. She has seen her maintenance and service expense increase because of all the additional use her machinery is getting due to a recent and significant up-trend in sales.
She is faced with a decision: should she invest in additional fixed assets to handle the additional sales volume or just continue with her current fixed asset platform?
Without understanding operating leverage, this business owner doesn't have valuable information that could help her make the best decision. Operating leverage is the degree to which a business uses fixed costs to generate profit. The greater the degree of fixed cost reliance, the greater the increase in profits during a sales up-trend and the greater the loss in a sales down-trend.
As fixed assets usually carry fixed costs, financed payments for the equipment, additional insurance, etc. , investing in additional equipment is something our auto body shop owner will want to seriously consider if the up-ward sales trend she is experiencing is something she believes to a be long-term phenomenon. If she believes the sales up-trend to indeed be long-term, then investing in additional fixed assets may be just the thing for her to do.
CVPA is one tool our auto body shop owner can use to help her determine what to do in this situation. By using her break-even model and considering contribution margins, she can perform sensitivity analyses to help her determine whether or not to increase her operating leverage in an effort to take advantage of a sales up-trend.
Summary
CVPA is a tool that can be used to help answer questions you may have about pricing your products and services, whether or not to invest in additional capital items, and which products and services to emphasize. While there is no one magic bullet, CVPA is a nice tool to have in your business analysis bag to help you make good decisions when answering these types of questions.
Jay Hickman is president of Advantage Business Solutions, LLC. He has over ten years of business consulting experience working with Arthur Andersen, LLC and Protiviti and holds an MBA from the University of Utah's David Eccles School of Business. His firm specializes in business management consulting for small business owners. Please visit http://www.advantagebusinesssolutions.net to access additional articles.
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by: Imogen Brown (September 21, 2008) | 1,606 | 7,815 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | longest | en | 0.947172 |
https://qutip.readthedocs.io/en/latest/guide/guide-steady.html | 1,696,189,871,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510924.74/warc/CC-MAIN-20231001173415-20231001203415-00647.warc.gz | 527,957,665 | 9,576 | ## Introduction¶
For time-independent open quantum systems with decay rates larger than the corresponding excitation rates, the system will tend toward a steady state as $$t\rightarrow\infty$$ that satisfies the equation
$\frac{d\hat{\rho}_{ss}}{dt}=\mathcal{L}\hat{\rho}_{ss}=0.$
Although the requirement for time-independence seems quite resitrictive, one can often employ a transformation to the interaction picture that yields a time-independent Hamiltonian. For many these systems, solving for the asymptotic density matrix $$\hat{\rho}_{ss}$$ can be achieved using direct or iterative solution methods faster than using master equation or Monte Carlo simulations. Although the steady state equation has a simple mathematical form, the properties of the Liouvillian operator are such that the solutions to this equation are anything but straightforward to find.
## Steady State solvers in QuTiP¶
In QuTiP, the steady-state solution for a system Hamiltonian or Liouvillian is given by qutip.steadystate.steadystate. This function implements a number of different methods for finding the steady state, each with their own pros and cons, where the method used can be chosen using the method keyword argument.
Method
Keyword
Description
Direct (default)
‘direct’
Direct solution solving $$Ax=b$$ via sparse LU decomposition.
Eigenvalue
‘eigen’
Iteratively find the zero eigenvalue of $$\mathcal{L}$$.
Inverse-Power
‘power’
Solve using the inverse-power method.
GMRES
‘iterative-gmres’
Solve using the GMRES method and optional preconditioner.
LGMRES
‘iterative-lgmres’
Solve using the LGMRES method and optional preconditioner.
BICGSTAB
‘iterative-bicgstab’
Solve using the BICGSTAB method and optional preconditioner.
SVD
‘svd’
Steady-state solution via the dense SVD of the Liouvillian.
The function qutip.steadystate.steadystate can take either a Hamiltonian and a list of collapse operators as input, generating internally the corresponding Liouvillian super operator in Lindblad form, or alternatively, a Liouvillian passed by the user. When possible, we recommend passing the Hamiltonian and collapse operators to qutip.steadystate.steadystate, and letting the function automatically build the Liouvillian (in Lindblad form) for the system.
As of QuTiP 3.2, the direct and power methods can take advantage of the Intel Pardiso LU solver in the Intel Math Kernel library that comes with the Anacoda (2.5+) and Intel Python distributions. This gives a substantial increase in performance compared with the standard SuperLU method used by SciPy. To verify that QuTiP can find the necessary libraries, one can check for INTEL MKL Ext: True in the QuTiP about box (qutip.about).
Solving for the steady state solution to the Lindblad master equation for a general system with qutip.steadystate.steadystate can be accomplished using:
>>> rho_ss = steadystate(H, c_ops)
where H is a quantum object representing the system Hamiltonian, and c_ops is a list of quantum objects for the system collapse operators. The output, labeled as rho_ss, is the steady-state solution for the systems. If no other keywords are passed to the solver, the default ‘direct’ method is used, generating a solution that is exact to machine precision at the expense of a large memory requirement. The large amount of memory need for the direct LU decomposition method stems from the large bandwidth of the system Liouvillian and the correspondingly large fill-in (extra nonzero elements) generated in the LU factors. This fill-in can be reduced by using bandwidth minimization algorithms such as those discussed in Additional Solver Arguments. However, in most cases, the default fill-in reducing algorithm is nearly optimal. Additional parameters may be used by calling the steady-state solver as:
rho_ss = steadystate(H, c_ops, method='power', use_rcm=True)
where method='power' indicates that we are using the inverse-power solution method, and use_rcm=True turns on a bandwidth minimization routine.
Although it is not obvious, the 'direct', eigen, and 'power' methods all use an LU decomposition internally and thus suffer from a large memory overhead. In contrast, iterative methods such as the 'iterative-gmres', 'iterative-lgmres', and 'iterative-bicgstab' methods do not factor the matrix and thus take less memory than these previous methods and allowing, in principle, for extremely large system sizes. The downside is that these methods can take much longer than the direct method as the condition number of the Liouvillian matrix is large, indicating that these iterative methods require a large number of iterations for convergence. To overcome this, one can use a preconditioner $$M$$ that solves for an approximate inverse for the (modified) Liouvillian, thus better conditioning the problem, leading to faster convergence. The use of a preconditioner can actually make these iterative methods faster than the other solution methods. The problem with precondioning is that it is only well defined for Hermitian matrices. Since the Liouvillian is non-Hermitian, the ability to find a good preconditioner is not guaranteed. And moreover, if a preconditioner is found, it is not guaranteed to have a good condition number. QuTiP can make use of an incomplete LU preconditioner when using the iterative 'gmres', 'lgmres', and 'bicgstab' solvers by setting use_precond=True. The preconditioner optionally makes use of a combination of symmetric and anti-symmetric matrix permutations that attempt to improve the preconditioning process. These features are discussed in the Additional Solver Arguments section. Even with these state-of-the-art permutations, the generation of a successful preconditoner for non-symmetric matrices is currently a trial-and-error process due to the lack of mathematical work done in this area. It is always recommended to begin with the direct solver with no additional arguments before selecting a different method.
Finding the steady-state solution is not limited to the Lindblad form of the master equation. Any time-independent Liouvillian constructed from a Hamiltonian and collapse operators can be used as an input:
>>> rho_ss = steadystate(L)
where L is the Louvillian. All of the additional arguments can also be used in this case.
Keyword
Options (default listed first)
Description
method
‘direct’, ‘eigen’, ‘power’, ‘iterative-gmres’,’iterative-lgmres’, ‘svd’
Method used for solving for the steady-state density matrix.
sparse
True, False
Use sparse version of direct solver.
weight
None
Allows the user to define the weighting factor used in the 'direct', 'GMRES', and 'LGMRES' solvers.
permc_spec
‘COLAMD’, ‘NATURAL’
Column ordering used in the sparse LU decomposition.
use_rcm
False, True
Use a Reverse Cuthill-Mckee reordering to minimize the bandwidth of the modified Liouvillian used in the LU decomposition. If use_rcm=True then the column ordering is set to 'Natural' automatically unless explicitly set.
use_precond
False, True
Attempt to generate a preconditioner when using the 'iterative-gmres' and 'iterative-lgmres' methods.
M
None, sparse_matrix, LinearOperator
A user defined preconditioner, if any.
use_wbm
False, True
Use a Weighted Bipartite Matching algorithm to attempt to make the modified Liouvillian more diagonally dominate, and thus for favorable for preconditioning. Set to True automatically when using a iterative method, unless explicitly set.
tol
1e-9
Tolerance used in finding the solution for all methods expect 'direct' and 'svd'.
maxiter
10000
Maximum number of iterations to perform for all methods expect 'direct' and 'svd'.
fill_factor
10
Upper-bound on the allowed fill-in for the approximate inverse preconditioner. This value may need to be set much higher than this in some cases.
drop_tol
1e-3
Sets the threshold for the relative magnitude of preconditioner elements that should be dropped. A lower number yields a more accurate approximate inverse at the expense of fill-in and increased runtime.
diag_pivot_thresh
None
Sets the threshold between $$[0,1]$$ for which diagonal elements are considered acceptable pivot points when using a preconditioner.
ILU_MILU
‘smilu_2’
Selects the incomplete LU decomposition method algorithm used.
Further information can be found in the qutip.steadystate.steadystate docstrings.
## Example: Harmonic Oscillator in Thermal Bath¶
A simple example of a system that reaches a steady state is a harmonic oscillator coupled to a thermal environment. Below we consider a harmonic oscillator, initially in the $$\left|10\right>$$ number state, and weakly coupled to a thermal environment characterized by an average particle expectation value of $$\left<n\right>=2$$. We calculate the evolution via master equation and Monte Carlo methods, and see that they converge to the steady-state solution. Here we choose to perform only a few Monte Carlo trajectories so we can distinguish this evolution from the master-equation solution.
import numpy as np
import matplotlib.pyplot as plt
import qutip
# Define paramters
N = 20 # number of basis states to consider
a = qutip.destroy(N)
H = a.dag() * a
psi0 = qutip.basis(N, 10) # initial state
kappa = 0.1 # coupling to oscillator
# collapse operators
c_op_list = []
n_th_a = 2 # temperature with average of 2 excitations
rate = kappa * (1 + n_th_a)
if rate > 0.0:
c_op_list.append(np.sqrt(rate) * a) # decay operators
rate = kappa * n_th_a
if rate > 0.0:
c_op_list.append(np.sqrt(rate) * a.dag()) # excitation operators
# find expectation value for particle number in steady state
fexpt = qutip.expect(a.dag() * a, final_state)
tlist = np.linspace(0, 50, 100)
# monte-carlo
mcdata = qutip.mcsolve(H, psi0, tlist, c_op_list, [a.dag() * a], ntraj=100)
# master eq.
medata = qutip.mesolve(H, psi0, tlist, c_op_list, [a.dag() * a])
plt.plot(tlist, mcdata.expect[0], tlist, medata.expect[0], lw=2)
# plot steady-state expt. value as horizontal line (should be = 2)
plt.axhline(y=fexpt, color='r', lw=1.5)
plt.ylim([0, 10])
plt.xlabel('Time', fontsize=14)
plt.ylabel('Number of excitations', fontsize=14)
plt.title(
r'Decay of Fock state $\left|10\rangle\right.$'
r' in a thermal environment with $\langle n\rangle=2$'
)
plt.show() | 2,378 | 10,275 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-40 | longest | en | 0.75472 |
http://mathhelpforum.com/number-theory/191974-prove-two-composite-numbers-same-print.html | 1,516,469,374,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889677.76/warc/CC-MAIN-20180120162254-20180120182254-00243.warc.gz | 235,238,390 | 3,182 | # Prove two composite numbers are the same
• Nov 15th 2011, 11:18 AM
bourbaki87
Prove two composite numbers are the same
Q: Consider two composite numbers A and B. Since A is composite we can write:
A = (p1^v1)(p2^v2)...(pn^vn)
Where p(i) is prime and v(i) is an integer. Suppose B is made up of the same primes, the difference is their exponent:
B = (p1^w1)(p2^w2)...(pn^wn) : w(i) is an integer
Show that A=B if and only if v(i) = w(i) for all i.
-----
A: (Necessary condition) Suppose v(i) = w(i). Clearly A=B.
But how can one deduce v(i) = w(i) given A=B as the starting point. How does one prove the sufficient condition?
My gratitude in advance to those bright sparks that prove this proposition.
• Nov 15th 2011, 11:54 AM
Plato
Re: Prove two composite numbers are the same
Quote:
Originally Posted by bourbaki87
Q: Consider two composite numbers A and B. Since A is composite we can write:
A = (p1^v1)(p2^v2)...(pn^vn)
Where p(i) is prime and v(i) is an integer. Suppose B is made up of the same primes, the difference is their exponent:
B = (p1^w1)(p2^w2)...(pn^wn) : w(i) is an integer
Show that A=B if and only if v(i) = w(i) for all i.
Comment: it usual in composite numbers for $v_i$ to be non-negative integers.
Take a simple case.
GIVEN: $2^x\cdot 3^y=2^{13}\cdot 3^{14}$.
What must $x~\&~y$ equal and why?
• Nov 15th 2011, 12:02 PM
bourbaki87
Re: Prove two composite numbers are the same | 463 | 1,416 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2018-05 | longest | en | 0.894708 |
https://quant.stackexchange.com/questions/34453/detailing-a-proposition-about-option-pricing-model-coherence | 1,713,664,456,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817699.6/warc/CC-MAIN-20240421005612-20240421035612-00268.warc.gz | 434,392,621 | 39,840 | # Detailing a proposition about option pricing model coherence
On page 4 of this paper, the author states the following:
"Looking at the limit case, when the strike tends towards 0, we should have the price of a forward contract and it should not depend on the equity skew around 0."
I cannot interpret mathematically what the author stated. Could you please provide a mathematical explanation of what the author said in words?
Thank you!
## 1 Answer
The value of a call option at expiry is $V=\mathrm{max}(0, S_t-K)$.
If you set $K=0$, then you have $V=\mathrm{max}(0, S_t)$, and since $S\geqslant0$, $\mathrm{max}(0, S_t) = S_t$ - i.e. ie's equivalent to holding the stock, which at expiry you'll expect to be worth the whatever the forward is.
• The price today of a contract that will pay off $S_T$ at maturity is, pretty much by definition, the forward price. May 30, 2017 at 17:02
• yes, exactly. Is it not clear that that's what i meant?
– will
May 30, 2017 at 17:03
• Same idea, different words. May 30, 2017 at 17:31 | 289 | 1,033 | {"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} | 2.953125 | 3 | CC-MAIN-2024-18 | latest | en | 0.906161 |
https://brainly.in/question/43361 | 1,484,923,161,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280835.22/warc/CC-MAIN-20170116095120-00365-ip-10-171-10-70.ec2.internal.warc.gz | 799,025,063 | 11,304 | # The sides of a rectangular piece of paper is 2:3. If the area is 150 sq.cm, find the length of its sides.An alloy consists of 30% copper and 40% zinc, and the remaining is nickel. Find the amount of nickel in 20 kg of alloy.Which is a better investment? - Rs. 5600 for 8 yrs at 6% or 5000 for 10 yrs at the rate of 5%
2
by tanushichopra21
Dont post more than one question in a single post. Working on these three questions for mere 5 points is not worth of spending time.
well said
i.e. correct
2014-09-29T20:04:22+05:30
1. Let length be 2x
Area= l*b=2x*3x=6x^2
Given area= 150 cm^2
According to the problem, equation is
6x^2=150
x^2=150/6
x^2=25
x=√25
x=5
therefore, length =2x=2*5=10 cm
2. in the alloy, copper and zinc together are of 70%
therefore, nickel will be of 30%
so the amount of nickel in 20kg of alloy= 30/100 of 20 = (30 x 20)/100 =6kg
Therefore, the amount of nickel = 6 kg
3. RS.5600 for 8 years at 6% per annum is.
5600*6/100*8=56*6*8=2688 rs.
RS. 5000 for 10 years at 5% per annum is
5000*5/100*10=50*5* 10=2500 Rs.
Therefore, Rs.5600 for 8 yeras at 6% per annum is the best investment.
Comment has been deleted
2014-09-30T11:26:37+05:30
1.Let the length of the sides of the rectangular paper be 2x and 3x cm .
Area=2x*3x=6x² cm²
Given,6x² cm² =150 cm²
=>x²=25
=>x=√25
=>x=5
Therefore, the length of its sides are 2x= 2 cm*5=10 cm and 3x=3 cm*5=15 cm respectively.
2.Percentage of copper=30%
Percentage of zinc=40%
Percentage of nickel=%(100-30-40)=30%
Ratio of copper,zinc and nickel in the alloy=30%:40%:30%
=3:4:3
Let the quantity of each metal in 20 kg alloy be 3x , 4x and 3x respectively.
By the condition,
3x+4x+3x=20 kg
=>10x=20 kg
=>x=2 kg
Therefore,Amount of nickel in 20 kg alloy=3x=3*2 kg=6 kg.
3.SI on the first investment=Rs.P*R*T/100
=Rs.5600*6*8/100
=Rs.56*6*8
=Rs.2688
SI on the second investment=Rs.P*R*T/100
=Rs.5000*10*5/100
=Rs.2,500
Therefore,The first investment is better.
this one is x-based,so you have a reason to mark it as the best. | 769 | 1,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-04 | latest | en | 0.85087 |
https://in.mathworks.com/matlabcentral/cody/problems/42673-longest-collatz-sequence/solutions/769037 | 1,590,630,917,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347396300.22/warc/CC-MAIN-20200527235451-20200528025451-00192.warc.gz | 393,501,326 | 16,017 | Cody
# Problem 42673. Longest Collatz Sequence
Solution 769037
Submitted on 30 Oct 2015 by James
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
%% x = 10; assert(isequal(euler14(x),9))
2 Pass
%% x = 100; assert(isequal(euler14(x),97))
3 Pass
%% x = 96; assert(isequal(euler14(x),73))
4 Pass
%% x = 1000; assert(isequal(euler14(x),871))
5 Pass
%% x = 870; assert(isequal(euler14(x),703))
6 Pass
%% assert(isequal(euler14(871),871))
7 Pass
%% x = 77030; assert(isequal(euler14(x),52527))
8 Pass
%% x = 77031; assert(isequal(euler14(x),77031))
9 Pass
%% assert(isequal(euler14(500000),410011))
10 Pass
%% z = 900000; y_correct=837799; assert(isequal(euler14(z),y_correct))
11 Pass
%% Projet Euler n°14 solution with x=1000000 | 301 | 876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-24 | latest | en | 0.502656 |
https://www.wikibacklink.com/search/standard-form-equation-quadratics | 1,642,984,690,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304345.92/warc/CC-MAIN-20220123232910-20220124022910-00080.warc.gz | 1,071,695,839 | 17,285 | # Keyword Analysis & Research: standard form equation quadratics
## Keyword Research: People who searched standard form equation quadratics also searched
How do you graph an equation in standard form?
The standard form of a linear equation puts the x and y terms on the left hand side of the equation, and makes the coefficient of the x-term positive. While standard form is commonly, we sometimes rewrite a line in slope-intercept form in order to graph it. The standard form of a linear equation looks like ax+by=c.
How do you convert an equation into standard form?
If you have something similar to and you want to convert it to standard form then choose "slope-intercept_to_standard". Input the equation in standard form x+ y=. or. Enter your equation in slope-intercept form ( form) y=x +. note: make sure you choose the correct form from the drop box above.
How do you solve standard form equations?
Standard form is another way to write slope-intercept form (as opposed to y=mx+b). It is written as Ax+By=C. You can also change slope-intercept form to standard form like this: Y=-3/2x+3. Next, you isolate the y-intercept(in this case it is 3) like this: Add 3/2x to each side of the equation to get this: 3/2x+y=3. | 286 | 1,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2022-05 | latest | en | 0.909827 |
https://numberworld.info/210021221320031 | 1,685,752,913,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648911.0/warc/CC-MAIN-20230603000901-20230603030901-00100.warc.gz | 472,804,640 | 4,053 | # Number 210021221320031
### Properties of number 210021221320031
Cross Sum:
Factorization:
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
bf03604ac55f
Base 32:
5v0dg4lhav
sin(210021221320031)
-0.64121648724999
cos(210021221320031)
0.7673600305455
tan(210021221320031)
-0.8356136125492
ln(210021221320031)
32.978229695445
lg(210021221320031)
14.32226317967
sqrt(210021221320031)
14492108.932796
Square(210021221320031)
4.4108913404757E+28
### Number Look Up
210021221320031 (two hundred ten trillion twenty-one billion two hundred twenty-one million three hundred twenty thousand thirty-one) is a great figure. The cross sum of 210021221320031 is 20. If you factorisate the number 210021221320031 you will get these result 11 * 23 * 14797 * 56100791. The number 210021221320031 has 16 divisors ( 1, 11, 23, 253, 14797, 162767, 340331, 3743641, 56100791, 617108701, 1290318193, 14193500123, 830123404427, 9131357448697, 19092838301821, 210021221320031 ) whith a sum of 239091701764608. 210021221320031 is not a prime number. The figure 210021221320031 is not a fibonacci number. The number 210021221320031 is not a Bell Number. The figure 210021221320031 is not a Catalan Number. The convertion of 210021221320031 to base 2 (Binary) is 101111110000001101100000010010101100010101011111. The convertion of 210021221320031 to base 3 (Ternary) is 1000112121211111000112001202202. The convertion of 210021221320031 to base 4 (Quaternary) is 233300031200102230111133. The convertion of 210021221320031 to base 5 (Quintal) is 210011441430124220111. The convertion of 210021221320031 to base 8 (Octal) is 5760154022542537. The convertion of 210021221320031 to base 16 (Hexadecimal) is bf03604ac55f. The convertion of 210021221320031 to base 32 is 5v0dg4lhav. The sine of 210021221320031 is -0.64121648724999. The cosine of the figure 210021221320031 is 0.7673600305455. The tangent of 210021221320031 is -0.8356136125492. The root of 210021221320031 is 14492108.932796.
If you square 210021221320031 you will get the following result 4.4108913404757E+28. The natural logarithm of 210021221320031 is 32.978229695445 and the decimal logarithm is 14.32226317967. You should now know that 210021221320031 is very special figure! | 818 | 2,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-23 | longest | en | 0.629613 |
https://www.clutchprep.com/chemistry/practice-problems/83111/on-a-cool-rainy-day-the-barometric-pressure-is-730-mmhg-calculate-the-barometric | 1,603,541,543,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882581.13/warc/CC-MAIN-20201024110118-20201024140118-00271.warc.gz | 651,912,325 | 31,914 | # Problem: On a cool, rainy day, the barometric pressure is 730 mmHg. Calculate the barometric pressure in centimeters of water (cmH2O) (d of Hg = 13.5 g/mL; d of H2O = 1.00 g/mL).
###### FREE Expert Solution
Relationship between height and densities of liquids
$\frac{{\mathbf{h}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}{{\mathbf{h}}_{\mathbf{Hg}}}\mathbf{=}\frac{{\mathbf{d}}_{\mathbf{Hg}}}{{\mathbf{d}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}$
hH2O9855 mm H2O
###### Problem Details
On a cool, rainy day, the barometric pressure is 730 mmHg. Calculate the barometric pressure in centimeters of water (cmH2O) (d of Hg = 13.5 g/mL; d of H2O = 1.00 g/mL).
Frequently Asked Questions
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Pressure Units concept. You can view video lessons to learn Pressure Units. Or if you need more Pressure Units practice, you can also practice Pressure Units practice problems.
What professor is this problem relevant for?
Based on our data, we think this problem is relevant for Professor Decker's class at LUC.
What textbook is this problem found in?
Our data indicates that this problem or a close variation was asked in Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition. You can also practice Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition practice problems. | 408 | 1,470 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-45 | latest | en | 0.832441 |
https://docs.coronalabs.com/api/library/physics/rayCast.html | 1,708,645,596,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473871.23/warc/CC-MAIN-20240222225655-20240223015655-00284.warc.gz | 200,076,966 | 4,389 | # physics.rayCast()
Type Function Library physics.* Return value Array of tables describing each hit Revision Release 2024.3703 Keywords ray, raycast, casting, physics, collision See also physics.reflectRay()
## Overview
This function is used to find the objects that collide with a line, and the collision points along that line.
The positions returned are in content space while the normals returned are in local space.
## Gotchas
If the starting position is inside an object, no hit will be registered for that object.
## Syntax
physics.rayCast( fromX, fromY, toX, toY, behavior )
##### fromX (required)
Number. The starting x position of the ray.
##### fromY (required)
Number. The starting y position of the ray.
##### toX (required)
Number. The ending x position of the ray.
##### toY (required)
Number. The ending y position of the ray.
##### behavior (optional)
String. The collision test behavior, in increasing order of performance cost:
• "any" — Return one result, but not necessarily the closest one.
• "closest" — Return only the closest hit from the starting point, if any. This is the default behavior.
• "unsorted" — Return all results in no particular order.
• "sorted" — Return all results, sorted from closest to farthest.
## Result Properties
hits will be an array of elements containing these properties:
• object — The DisplayObject colliding with the line.
• position.x — The x collision position of object, in content space.
• position.y — The y collision position of object, in content space.
• normal.x — The x component of the normal of the surface hit in local space.
• normal.y — The y component of the normal of the surface hit in local space.
• fraction — The fraction (0..1) along the ray where the hit is located. 0 is the start point of the ray cast and 1 is the end point.
## Example
local hits = physics.rayCast( 0, 0, 200, 300, "closest" )
if ( hits ) then
-- There's at least one hit
print( "Hit count: " .. tostring( #hits ) )
-- Output the results
for i,v in ipairs( hits ) do
print( "Hit: ", i, v.object, " Position: ", v.position.x, v.position.y, " Surface normal: ", v.normal.x, v.normal.y )
end
print( "The first object hit is: ", hits[1].object, " at position: ", hits[1].position.x, hits[1].position.y, " where the surface normal is: ", hits[1].normal.x, hits[1].normal.y, " and where the fraction along the ray is: ", hits[1].fraction )
else
-- No hits on raycast
end | 605 | 2,443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-10 | latest | en | 0.745562 |
https://algebra-tutoring.com/algebra-tutoring/fractional-exponents/rationalize-denominator.html | 1,713,160,252,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816942.33/warc/CC-MAIN-20240415045222-20240415075222-00814.warc.gz | 74,460,223 | 6,913 | Free Algebra Tutorials!
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# rationalize denominator?
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# ALGORITHMS
Journal of Algorithms (informationswissenschaftl. Veröffentlichungen)
Acronyms von A bis Z. 2013.
### Schlagen Sie auch in anderen Wörterbüchern nach:
• ALGORITHMS — Journal of Algorithms (informationswissenschaftl. Veoeffentlichungen) … Acronyms
• algorithms — al·go·rithm || ælgÉ™rɪðəm n. (Mathematics) step by step procedure used to solve a problem (often includes repetition of steps); step by step problem solving procedure used within software applications (Computers) … English contemporary dictionary
• Algorithms + Data Structures = Programs — is a book written by Niklaus Wirth covering some of the fundamental topics of computer programming. The book was published by Prentice Hall in 1976 [ [http://portal.acm.org/citation.cfm?id=540029 Algorithms + Data Structures = Programs] ] and is… … Wikipedia
• algorithms, analysis of — Basic computer science discipline that aids in the development of effective programs. Analysis of algorithms provides proof of the correctness of algorithms, allows for the accurate prediction of program performance, and can be used as a measure… … Universalium
• Algorithms for calculating variance — play a major role in statistical computing. A key problem in the design of good algorithms for this problem is that formulas for the variance may involve sums of squares, which can lead to numerical instability as well as to arithmetic overflow… … Wikipedia
• Algorithms for Recovery and Isolation Exploiting Semantics — In computer science, Algorithms for Recovery and Isolation Exploiting Semantics, or ARIES is a recovery algorithm designed to work with a no force, steal database approach; it is used by IBM DB2, Microsoft SQL Server and many other database… … Wikipedia
• Convex hull algorithms — Algorithms that construct convex hulls of various objects have a broad range of applications in mathematics and computer science, see Convex hull applications . In computational geometry, numerous algorithms are proposed for computing the convex… … Wikipedia
• Ant colony optimization algorithms — Ant behavior was the inspiration for the metaheuristic optimization technique. In computer science and operations research, the ant colony optimization algorithm (ACO) is a probabilistic technique for solving computational problems which can be… … Wikipedia
• List of algorithms — The following is a list of the algorithms described in Wikipedia. See also the list of data structures, list of algorithm general topics and list of terms relating to algorithms and data structures.If you intend to describe a new algorithm,… … Wikipedia
• Force-based algorithms — Force based or force directed algorithms are a class of algorithms for drawing graphs in an aesthetically pleasing way. Their purpose is to position the nodes of a graph in two dimensional or three dimensional space so that all the edges are of… … Wikipedia | 600 | 2,916 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-40 | latest | en | 0.783576 |
http://busynessgirl.com/on-the-problem-of-beginning-algebra/ | 1,548,086,387,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583795042.29/warc/CC-MAIN-20190121152218-20190121174218-00372.warc.gz | 35,882,191 | 23,968 | Every year I teach at least one Beginning Algebra class. Every year I find myself frustrated by the variety of skill levels in the course, which range from students who do not know what 4 times 6 is, to students who are basically looking for a review of algebra. After two tests, I can make a list of students who will not pass the class and I will be correct for 95% of the students on the list. Why, then, do we continue to attempt to teach these students in the same way, when we know they are missing the fundamental skills that will make them successful? This is a problem I’ve struggled with for years.
If the educational world was an ideal place, free from bureaucracy (Financial Aid) and administrative issues (Transferability and Course Approvals), this is an example of how I’d do it.
Suppose that from 10-11am, four days a week, there are two sections of Beginning Algebra taught by Instructor A and Instructor B.
By the time students have taken the first two exams (Review of Arithmetic and Solving Equations), we know which students can make it and who is already in over their head.
At this point, each instructor splits their class into two groups: Ready to Succeed and Needs Remediation. Since the classes are offered at the same time on the same days, the students can transfer from one instructor to the other without altering their schedules.
Instructor A takes the Ready to Succeed group and continues teaching them Beginning Algebra. Instructor B takes the Needs Remediation group, and spends the rest of the semester on signed numbers, fractions, solving equations, and basics of some of what is to come in Beginning Algebra (basic exponent rules, factoring a GCF, finding solutions to a line, etc.).
What does this gain for us? The group that goes with Instructor A does not have to move at a snail’s pace in an attempt to accommodate students who stand almost no chance of being successful – the instructor can teach with more depth than they would otherwise be able to. The group that goes with Instructor B would likely have failed the class, but would not have filled in any of the gaps that might make them successful in the future. With this system, they would have a chance to work with an instructor and fill in some of those gaps at a pace where they can be successful.
Why can’t we do this? The biggest obstacle that I see is that the second group is transferred to a course with no name, no course number, no transferability, and technically less credit hours – which brings up numerous Financial Aid issues. Also, the Group B students might not be willing to continue to come to a class that “doesn’t count,” even if it is for their own good.
Oh, and how do we decide who gets to be Instructor A and who is Instructor B? For this I propose we arm wrestle.
Possibly Related Posts: | 593 | 2,823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-04 | longest | en | 0.96062 |
https://a.osmarks.net/content/codegolf.stackexchange.com_en_all_2020-04/A/tag/number-theory/1.html | 1,701,624,400,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.42/warc/CC-MAIN-20231203161435-20231203191435-00804.warc.gz | 96,982,843 | 5,802 | ## Tag: number-theory
202 Is this number a prime? 2015-09-11T14:06:29.040
73 Calculate Phi (not Pi) 2015-11-12T23:56:22.273
70 Well that's odd... no wait, that's even! 2015-12-02T10:22:24.003
67 Is this even or odd? 2017-03-21T10:30:34.823
67 Yo boy, must it sum 2017-10-23T15:39:46.720
66 Calculate the number of primes up to n 2016-02-26T19:08:02.277
60 Find the Smoothest Number 2014-08-18T23:30:24.447
51 Coprimes up to N 2016-12-26T18:00:02.387
51 Find the smallest number that doesn't divide N 2017-01-03T08:50:00.077
51 Bringing a pair of integers to equality 2018-05-03T14:15:28.510
46 Count sums of two squares 2015-11-26T00:38:01.640
46 Divisor skyline 2017-05-07T16:31:21.547
44 Theoretically output Graham's number 2016-06-27T07:30:45.190
42 A naturally occurring prime generator 2015-08-25T06:31:23.077
40 Greatest Common Divisor 2016-04-07T05:29:54.567
40 Sharing (characters) is Caring! 2017-01-09T19:14:10.043
40 Am I divisible by double the sum of my digits? 2017-07-03T13:00:47.433
39 Pseudofactorial 2016-06-09T20:19:47.310
38 Is this number evil? 2018-08-01T15:08:21.957
37 Is it a Proth number? 2016-08-10T22:07:56.617
37 Abandon all squares, ye who divide me 2017-05-11T15:56:28.810
36 Catalan Numbers 2015-12-09T17:39:41.920
36 Compute the Carmichael function 2016-09-18T18:07:49.620
36 Incrementing Gray Codes 2016-10-02T21:05:35.577
36 Replace twos with threes 2017-05-04T21:45:59.043
35 The Arithmetic Derivative 2016-04-03T11:23:22.977
35 Is it a Mersenne Prime? 2016-12-25T14:18:05.567
35 Sum the powers that be 2017-02-10T09:00:40.570
34 Can you reach this number by doubling and rearranging? 2015-08-19T08:38:05.383
34 Sum of Modulo Sums 2016-12-09T14:36:54.233
33 Is this number Loeschian? 2016-08-04T15:34:41.337
33 Finding The nth Prime such that the prime - 1 is divisible by n 2016-11-17T03:07:25.513
33 Can square tree rings be generated from primes? 2018-01-30T21:28:35.197
33 (RGS 1/5) Binary multiples 2020-02-24T08:02:19.677
32 Standardise a Phinary Number 2014-09-16T13:00:11.690
32 Array Escape - get out of there 2016-07-25T11:08:57.993
32 Score Tarzan's Olympic Vine-Swinging Routine 2016-08-20T14:38:09.433
32 1, 2, 3, 14... or is it 15? 2018-03-30T22:11:53.260
31 Least Common Multiple 2016-09-30T18:47:53.850
31 Are you lost yet? 2016-12-24T04:18:03.893
31 Find the dot product of Rationals 2018-02-22T22:30:39.610
30 That's a prime... almost 2016-02-22T17:36:42.077
29 Generate Keyboard Friendly Numbers 2015-05-12T04:58:46.287
29 Addition on Elliptic Curves 2016-03-18T20:59:23.433
29 Pascal's Column Sums 2017-03-16T14:27:12.067
29 Primitive Pythagorean Triples 2017-07-25T13:20:56.090
29 Am I not good enough for you? 2019-03-12T02:41:31.763
29 Longest Prime Sums 2019-12-11T23:04:49.113
28 Visualize the greatest common divisor 2016-01-01T19:15:52.260
28 One-zero dividend 2016-02-28T15:11:27.560
28 Is this a Smith number? 2016-09-02T14:27:24.987
28 List all multiplicative partitions of n 2016-12-26T22:38:07.357
28 Smallest unseen, but no sharing digits! 2017-08-10T15:03:15.687
28 Fundamental Solution of the Pell Equation 2019-04-16T13:37:52.573
28 Fermat's Last Theorem, mod n 2019-11-04T16:33:10.583
27 Calculate n % 12 2014-06-20T00:05:32.077
27 Calculate Euler's totient function 2016-06-22T06:50:27.523
27 Numbers of purity 2016-08-12T20:23:51.227
27 Which Day of Christmas is it? 2017-01-27T17:14:49.457
27 Is it a Chen prime? 2017-07-10T11:54:59.060
27 Find prime gaps 2017-08-03T18:31:13.517
27 Is this a three-digit number ending in one? 2017-12-16T09:21:24.847
27 The Add-Multiply-Add Sequence 2018-05-04T12:52:04.347
26 Am I perfect (number)? 2015-09-15T21:39:15.693
26 Count the divisors of a number 2015-11-27T06:19:14.787
26 AGM Series Hole 1: Calculate the Arithmetic–Geometric Mean 2015-12-08T22:57:51.977
26 Residue Number System 2016-02-13T06:16:27.700
26 Swap bits with their neighbours 2016-06-03T22:42:34.207
26 Is it a weak prime? 2017-07-08T16:33:32.613
26 Diluted Integer Sums 2018-01-29T15:04:30.570
26 Who's next to me in the queue? 2019-12-08T20:12:58.900
25 What's a half on the clock? 2015-12-28T23:00:56.100
25 Implement the divisibility-by-7 rule 2016-02-14T19:58:53.913
25 Generate lazy values 2016-02-23T02:49:54.770
25 Approximate Brun's Constant 2017-01-22T00:31:45.340
25 Deranged !Combinatorics: Compute the Subfactorial 2017-03-26T17:00:19.690
25 GET your dubs together 2017-03-28T08:51:40.363
25 Palindromic Residue 2017-03-31T14:57:16.960
24 Print all 3 by 3 sturdy squares 2015-10-21T22:47:52.950
24 Decimal concatenation of squares 2015-11-13T20:59:51.217
24 Write numbers as a difference of Nth powers 2018-01-14T15:06:30.543
24 Pascal's Triangle (Sort of) 2018-02-07T13:51:21.443
24 Bertrand's Primes 2018-02-13T15:27:27.247
24 Find the 10-adic cube root of 3 2018-06-07T10:24:54.223
24 Fermat's polygonal number theorem 2019-10-07T17:13:49.750
24 Next Shared Totient 2019-12-04T20:07:38.677
23 Find the prime factors 2011-04-06T20:01:14.137
23 Co-primality and the number pi 2015-12-26T16:16:32.960
23 The Möbius Function 2016-01-24T13:54:37.050
23 Highly composite numbers 2016-02-29T22:26:04.020
23 Partial factorisations of a positive integer 2016-07-29T06:05:22.490
23 Reuse your code! 2017-04-02T15:53:41.200
23 Divinacci Sequence 2017-06-23T13:51:34.740
23 Factorize a Gaussian integer 2017-07-12T21:37:22.707
23 Modular multiplicative inverse 2017-08-24T15:41:12.173
22 Last non-zero digit of n! 2011-02-08T09:03:33.850
22 Generate Lucky Numbers 2015-02-17T03:57:38.097
22 Factoring a 64-bit integer 2015-04-23T23:22:55.783
22 Last Nonzero Digits of a Factorial in Base 2015-04-30T12:54:34.810 | 2,350 | 5,704 | {"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} | 3.328125 | 3 | CC-MAIN-2023-50 | longest | en | 0.495711 |
https://books.google.co.ve/books?qtid=f97e7842&dq=related:ISBN8474916712&lr=&id=5ncAAAAAMAAJ&output=html_text&sa=N&start=60 | 1,638,493,954,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362571.17/warc/CC-MAIN-20211203000401-20211203030401-00595.warc.gz | 214,573,686 | 6,192 | Search Images Maps Play YouTube News Gmail Drive More »
Libros Libros
If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Elements of Geometry: Containing the Principal Propositions in the First Six ... - Página 164
por Euclid - 1789 - 272 páginas
Vista completa - Acerca de este libro
## The Elements of Euclid, the parts read in the University of Cambridge [book ...
Euclides - 1846
...parallelogram BC. Wherefore, Equal parallelograms *rc. QED PROP. XV. THEOR. Equal triangles, which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional : and triangles, which have one angle...
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## Euclid's Elements of geometry, the first three books (the fourth, fifth, and ...
Euclides - 1846
...therefore ex oequali, BD will be to CG, as I to L (by Prop. 33, B. 5). COR. 2. — Triangles which have one angle of the one equal to one angle of the other, are to each other in a ratio compounded of the ratios of the sides about the equal angles. COR. 3....
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## Elements of Geometry: Containing the First Six Books of Euclid, with a ...
John Playfair - 1846 - 317 páginas
...the parallelogram AB is equal (9. 5.) to the parallelogram BC. PROP. XV. THEOR. Equal triangles which have one angle of the one equal to one angle of the other have their sides about the equal angles reciprocally proportional ; And triangles which have one angle...
Vista completa - Acerca de este libro
## Euclid's Elements: Or, Second Lessons in Geometry,in the Order of Simson's ...
Dennis M'Curdy - 1846 - 138 páginas
...of two triangles, &c. QED Kecite (a) p. 23, 1; Wp.9,5; (4) 32, 1 ; (Op.8,1: p. 4, 6; ax. 1,1. 6 Th. If two triangles have one angle of the one equal to one of the other, and the sides about the equal angles proportionals, the triangles shall be equiangular,...
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## A collection of problems and examples adapted to the 'Elementary course of ...
Harvey Goodwin (bp. of Carlisle.) - 1847 - 80 páginas
...than the sum of the two sides by the diameter of the inscribed circle. 23. If two triangles have an angle of the one equal to one angle of the other, and also another angle of the one equal to the supplement of another angle of the other, the sides about...
Vista completa - Acerca de este libro
## Scholarship examinations of 1846/47 (-1853/54).
Bengal council of educ - 1848
...definition will satisfy one of the tests of the former. 2. Give a proof of the following proposition : " If two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals ; then, if each of the remaining angles be either less,...
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## General Report on Public Instruction, in the Lower Provinces of the Bengal ...
Bengal (India) - 1848
...definition will satisfy one of the tests of the former. 2. Give a proof of the following proposition : " If two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals ; then, if each of the remaining angles be either less,...
Vista completa - Acerca de este libro
## The definitions, postulates, axioms, and enunciations of the propositions of ...
Euclides - 1848
...mean proportional between two given straight lines. PROP. XIV. THEOREM. Equal parallelograms, which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional : and conversely, parallelograms...
Vista completa - Acerca de este libro
## Papers for the Schoolmaster
1852
...to one another. 2. About a given circle, to describe i triangle equiangular to a given triangle. 3. If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals the triangles shall be equiangular, and shall have those...
Vista completa - Acerca de este libro
## A Collection of Problems and Examples Adapted to the "Elementary Course of ...
Harvey Goodwin - 1851 - 173 páginas
...shall have to the second a greater ratio than the fifth has to the sixth. 8. Equal triangles which have one angle of the one equal to one angle of the other have the sides about the equal angles reciprocally proportional ; and conversely. 9. If two planes...
Vista completa - Acerca de este libro | 1,139 | 4,521 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-49 | latest | en | 0.714225 |
https://www.smartick.com/blog/mathematics/geometry/trapezium-trapezoid/ | 1,713,261,766,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817081.52/warc/CC-MAIN-20240416093441-20240416123441-00545.warc.gz | 925,361,003 | 15,579 | Smartick is a fun way to learn math!
Jun24
# Trapezium-Trapezoid: Definition, Types and Exercises
In today’s post, we are going to study a bit of geometry. We will focus on one flat figure: the trapezium (UK – trapezoid).
There is considerable confusion over the definition of ‘trapezoid’ and ‘trapezium’ due to differences in the British and US versions. As you can see from the table above, the meanings of the two words are exactly reversed between the US and British interpretations.
In this post, we will be referring to the USA trapezium, which is the same as a British trapezoid.
Trapeziums are one of the lesser known geometric shapes and raise a lot of questions with elementary aged students. We are going to learn to define, characterize, identify, and construct trapeziums in all forms.
Index
## Definition of a trapezium
It is a flat geometric figure with 4 sides of which none are parallel to another.
Exercise: Which of the following figures is a trapezium?
Hint: Since all of the figures have 4 sides, you should look closely at the parallels that exist, or don’t, between all of the sides in each figure.
Solution: Only the purple figure is a trapezium.
## Types of trapeziums
Let’s go a little further and move on to learn how to differentiate between trapeziums and others. All of them should fulfill the conditions we have defined (flat geometric figure, 4 sides, none of the sides are parallel to another) but they can be grouped into different types.
### Symmetrical or asymmetrical
Depending on whether or not it has an axis of internal symmetry we can differentiate between symmetrical (has internal symmetry axis) and asymmetrical (doesn’t have internal symmetry axis) trapeziums.
Exercise: Identify the symmetrical and asymmetrical trapeziums.
Hint: Look closely at which of the two figures has an internal axis of symmetry drawn.
Solution: Trapezium 1 is symmetrical and trapezium 2 is asymmetrical.
### Concave or convex
Whether their diagonals are internal or external is what makes them considered convex (all of their diagonals are internal) or concave (some diagonals are external).
Exercise: Identify the concave and convex trapeziums.
Hint: Look closely at the diagonal lines drawn on each figure. Try to see which ones are drawn outside of the figure.
Solution: Trapezium 1 is concave and trapezium 2 is convex.
## Trapezium Exercise
Now you know how to identify and classify trapeziums. All that’s left to do is practice!
Exercise: Classify the following flat figures.
I hope that with these explanations you have a much clearer idea about trapeziums. If you want more exercises like this and much more, log on to Smartick, register, and try our method for free. | 639 | 2,730 | {"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} | 4.9375 | 5 | CC-MAIN-2024-18 | latest | en | 0.931309 |
http://www.haskell.org/pipermail/haskell-cafe/2006-June/016349.html | 1,411,513,277,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657140379.3/warc/CC-MAIN-20140914011220-00148-ip-10-234-18-248.ec2.internal.warc.gz | 545,484,396 | 2,053 | Donald Bruce Stewart dons at cse.unsw.edu.au
Tue Jun 27 04:34:20 EDT 2006
```hankgong:
>
> Hi, all
>
> I'm just a newbie for Haskell and functional programming
> world. The idea I currently read is quite different and
> interesting.
>
> I have one general question about the recursively looping
> style. For example:
>
> myMax [ ] = error "empty list"
>
> myMax [x] = x
>
> myMax [x:xs] = if x>= (myMax xs) then x else (myMax xs)
>
>
> I just list out this kind of very simple program. However,
> if the list size if a big number such as 10000000, the
> Does it mean that the functional programming is lacking of
> scalability? I do know that we can manually change the stack
> size for it. But that's not a good solution according to my
> opinion.
>
No, your code is just really inefficient (think about how many times its
traversing the list). Try rewriting it as a simple accumulating pass over the
list, carrying the largest element you've seen so far.
mymax [] = undefined
mymax (x:xs) = f x xs
where
f x [] = x
f x (y:ys) | y > x = f y ys
| otherwise = f x ys
However, 'f' is just a foldl inlined:
import Data.List
mymax [] = undefined
mymax (x:xs) = foldl' (\a b -> if b > a then b else a) x xs
And the lambda is just 'max':
import Data.List
mymax [] = undefined
mymax (x:xs) = foldl' max x xs
Now, we already check for the empty list, so avoid checking again:
import Data.List
mymax [] = undefined
mymax xs = foldl1' max xs
And that'll do.
-- Don
``` | 457 | 1,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2014-41 | latest | en | 0.908383 |
https://www.studypool.com/discuss/1087422/how-do-you-do-this-algebra-problem-7?free | 1,481,304,679,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542714.38/warc/CC-MAIN-20161202170902-00161-ip-10-31-129-80.ec2.internal.warc.gz | 1,013,271,242 | 14,020 | ##### how do you do this algebra problem
Algebra Tutor: None Selected Time limit: 1 Day
6-3x=-2x
Jul 22nd, 2015
6 - 3x = -2x
step 1. Add 3x to both sides
6 - 3x + 3x = -2x + 3x
step 2. simplify both sides
6 = 1x
So x=6
double check by substituting 6 for x
6 - 3 (6) = -2 (6)
6 - 18 = -12
-12 = -12 CHECKS OUT, so x=6 is correct
Jul 22nd, 2015
...
Jul 22nd, 2015
...
Jul 22nd, 2015
Dec 9th, 2016
check_circle | 186 | 424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.766689 |
https://la.mathworks.com/matlabcentral/cody/problems/233-reverse-the-vector/solutions/495787 | 1,610,853,416,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703509104.12/warc/CC-MAIN-20210117020341-20210117050341-00390.warc.gz | 419,643,333 | 17,110 | Cody
Problem 233. Reverse the vector
Solution 495787
Submitted on 4 Sep 2014 by wayne jackson
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
%% x = 1; y_correct = 1; assert(isequal(reverse(x),y_correct))
y = 1
2 Pass
%% x = -10:1; y_correct = 1:-1:-10; assert(isequal(reverse(x),y_correct))
y = 1 y = 1 0 y = 1 0 -1 y = 1 0 -1 -2 y = 1 0 -1 -2 -3 y = 1 0 -1 -2 -3 -4 y = 1 0 -1 -2 -3 -4 -5 y = 1 0 -1 -2 -3 -4 -5 -6 y = 1 0 -1 -2 -3 -4 -5 -6 -7 y = 1 0 -1 -2 -3 -4 -5 -6 -7 -8 y = 1 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 y = 1 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10
3 Pass
%% x = 'able was i ere i saw elba'; y_correct = 'able was i ere i saw elba'; assert(isequal(reverse(x),y_correct))
y = 97 y = 97 98 y = 97 98 108 y = 97 98 108 101 y = 97 98 108 101 32 y = 97 98 108 101 32 119 y = 97 98 108 101 32 119 97 y = 97 98 108 101 32 119 97 115 y = 97 98 108 101 32 119 97 115 32 y = 97 98 108 101 32 119 97 115 32 105 y = 97 98 108 101 32 119 97 115 32 105 32 y = 97 98 108 101 32 119 97 115 32 105 32 101 y = 97 98 108 101 32 119 97 115 32 105 32 101 114 y = 97 98 108 101 32 119 97 115 32 105 32 101 114 101 y = 97 98 108 101 32 119 97 115 32 105 32 101 114 101 32 y = 97 98 108 101 32 119 97 115 32 105 32 101 114 101 32 105 y = Columns 1 through 16 97 98 108 101 32 119 97 115 32 105 32 101 114 101 32 105 Column 17 32 y = Columns 1 through 16 97 98 108 101 32 119 97 115 32 105 32 101 114 101 32 105 Columns 17 through 18 32 115 y = Columns 1 through 16 97 98 108 101 32 119 97 115 32 105 32 101 114 101 32 105 Columns 17 through 19 32 115 97 y = Columns 1 through 16 97 98 108 101 32 119 97 115 32 105 32 101 114 101 32 105 Columns 17 through 20 32 115 97 119 y = Columns 1 through 16 97 98 108 101 32 119 97 115 32 105 32 101 114 101 32 105 Columns 17 through 21 32 115 97 119 32 y = Columns 1 through 16 97 98 108 101 32 119 97 115 32 105 32 101 114 101 32 105 Columns 17 through 22 32 115 97 119 32 101 y = Columns 1 through 16 97 98 108 101 32 119 97 115 32 105 32 101 114 101 32 105 Columns 17 through 23 32 115 97 119 32 101 108 y = Columns 1 through 16 97 98 108 101 32 119 97 115 32 105 32 101 114 101 32 105 Columns 17 through 24 32 115 97 119 32 101 108 98 y = Columns 1 through 16 97 98 108 101 32 119 97 115 32 105 32 101 114 101 32 105 Columns 17 through 25 32 115 97 119 32 101 108 98 97
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 1,176 | 2,544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.354482 |
https://codereview.stackexchange.com/questions/238440/efficient-implementation-of-priority-queue | 1,721,841,887,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518427.68/warc/CC-MAIN-20240724171328-20240724201328-00669.warc.gz | 154,654,240 | 43,508 | # Efficient implementation of priority queue
I am trying to solve a maze problem.
The input data is a matrix, where the individual element represents the height of the mountain in that particular area, I have to start from position (0,0) and go to the position (n-1, n-1); where n is the size of the maze, my goal is to return the minimum number of possible climb round that I have to make in order to achieve this goal.
For example:
b = 010
010
010
In the matrix b,If I start from (0,0) I have to climb to height 1, now I am on the hill and I have to go down, so the total number of climb round is
abs(0(height of starting position) - 1(height of mountain)) + abs(1(height of mountain) - 0(height of the third column)) = 2
If I reach to third column then I do nnot have to climb any further round as my goal (2,2) would be at same level
c = 010
101
010
Similarly for c, the answer would be 4
A typical example would be
75218176
00125954
30062751
01192976
24660156
14932066
44532310
60429682
"""
self.location stores the location of current node
self.cost stores the cost or path length from adjacent node approaching from a particular path, for example in the above example If I have coming along the path 752
then cost of the node whose location is (0,2) would be 3, as the difference of height 5 and height 2 is 3
self.altitude simply stores the number which is at the location
self.totalCost represent the totalCost from the begenning, for example: for the path 752, the total cost of '2' would be 7- 5 + 5 -2 = 5
getTotalcost() simply add the parent cost with the difference in altitude of parent and self
neighbours() returns the possible node which can be reached from the current position.
Open is basically implementation of priority queue, I have not used heapq because in order to maintain the invariancy in structure I am not supposed to delete any element at random, although I can re-heapify after deletion but that seems to an obvious choice for inefficiency
"""
class node:
def __init__(self, loc):
self.location = loc
self.cost = 0
self.altitude = 0
self.parent = None
self.totalCost = 0
def __eq__(self, other):
return self.location == other.location
def __hash__(self):
return hash(self.location)
def getTotalcost(self):
if self.parent:
self.totalCost = self.cost + self.parent.totalCost
return self.totalCost
def __lt__(self, other):
return self.getTotalcost() < other.getTotalcost()
def neighbours(S, Node):
a,b = Node.location
options = []
for i in (a, b + 1), (a, b - 1), (a + 1, b), (a - 1, b):
if min(i) >= 0 and max(i) < S:
options.append(node(i))
return options
class Open:
def __init__(self, point):
self.container = [point]
self.se = set(self.container)
self.l = 1
def push(self, other):
if not(other in self.se):
self.properPlace(other)
else:
a = other
b = self.container.index(other)
k = self.container[b]
if k.getTotalcost() > a.getTotalcost():
del self.container[b]
self.l -= 1
self.se.remove(k)
self.properPlace(other)
def __iter__(self):
return iter(self.container)
def rem(self):
self.l -= 1
return self.container.pop(0)
def properPlace(self, other):
i = 0
while i < self.l and self.container[i].getTotalcost() < other.getTotalcost():
i += 1
self.container.insert(i, other)
self.l += 1
def path_finder(maze):
maze= maze.split("\n")
l = len(maze)
start = node((0,0))
start.altitude = int(maze[0][0])
start.totalCost = 0
r = Open(start)
visited = set()
while True:
i = r.rem()
if i.location == (l - 1, l- 1):
return i.getTotalcost()
for j in neighbours(l, i):
j.altitude = int(maze[j.location[0]][j.location[1]])
if j not in visited:
j.cost = abs(i.altitude - j.altitude)
j.parent = i
r.push(j)
I have some thoughts that my implementation of priority queue is inefficient, am I right? If yes, then how can I make it more efficient? How can I make the program efficient?
• Do you need to implement the priority queue yourself (for homework, just for the fun/challenge, etc.), or would a pre-built solution be just as good? Commented Mar 5, 2020 at 22:53
• It would be best if someone helps me to find the chunk and suggest to improve where the maximum computational time goes in my code, else even the more efficient implementation would be good too. Commented Mar 6, 2020 at 6:38
The advice I think most important:
• stick to the Style Guide for Python Code
• Naming: name things for what they can/will be used for, their raison d'être.
For example, the class featuring push(), rem(), and _properPlace() is not used to be open or open something. It looks a priority queue with increase_priority().
The instance r in path_finder() holds a set of tentative costs, or, if you insist, "open" nodes.
The usual suggestion in the context of path finding is to use a Fibonacci_heap - there is PyPI fibheap.
• Modelling
node is a strange hybrid of node and edge, given its single parent and cost:
cost is not an attribute of a node, but an attribute of an edge, a pair of nodes
I suspect del and insert on Open.container to be the performance culprits.
You could try and adapt heapq:
locate nodes using index() as presented (O(n))
update cost and use _siftdown() to restore heap property | 1,355 | 5,145 | {"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} | 3.328125 | 3 | CC-MAIN-2024-30 | latest | en | 0.918359 |
http://intranet.math.vt.edu/netmaps/hurwitz/m5n5/m5n5hurwitz303_Main.output | 1,579,607,914,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-05/segments/1579250603761.28/warc/CC-MAIN-20200121103642-20200121132642-00289.warc.gz | 81,341,534 | 1,808 | These Thurston maps are NET maps for every choice of translation term. They are primitive and have degree 25. PURE MODULAR GROUP HURWITZ EQUIVALENCE CLASSES FOR TRANSLATIONS {0,lambda1} {lambda2,lambda1+lambda2} Since no Thurston multiplier is 1, this modular group Hurwitz class contains only finitely many Thurston equivalence classes. The number of pure modular group Hurwitz classes in this modular group Hurwitz class is 8. ALL THURSTON MULTIPLIERS c/d IN UNREDUCED FORM 0/5, 1/5, 2/5 Since every Thurston multiplier is less than 1, every NET map in this modular group Hurwitz class is rational. EXCLUDED INTERVALS FOR THE HALF-SPACE COMPUTATION (-129.596443,0.992499 ) ( 1.007501,124.434165) SLOPE FUNCTION INFORMATION There are no slope function fixed points. Number of excluded intervals computed by the fixed point finder: 2314 No nontrivial cycles were found. The slope function maps some slope to the nonslope. The slope function orbit of every slope p/q with |p| <= 50 and |q| <= 50 ends in the nonslope. If the slope function maps slope p/q to slope p'/q', then |p'| <= |p| for every slope p/q with |p| <= 50 and |q| <= 50. FUNDAMENTAL GROUP WREATH RECURSIONS When the translation term of the affine map is 0: NewSphereMachine( "a=<1,d^-1,a*b,d^-1,c^-1,1,d^-1,d,1,d,1,1,d,c^-1,1,1,1,1,1,c,1,c*d,1,c,1>(2,8)(3,22)(4,10)(5,24)(6,21)(7,13)(9,15)(11,16)(12,18)(14,20)(17,23)(19,25)", "b=(1,21)(2,3)(4,5)(6,16)(7,8)(9,10)(12,13)(14,15)(17,18)(19,20)(22,23)(24,25)", "c=(1,2)(3,4)(5,25)(6,7)(8,9)(10,20)(11,12)(13,14)(16,17)(18,19)(21,22)(23,24)", "d=(1,22)(2,6)(3,24)(4,8)(7,11)(9,13)(10,25)(12,16)(14,18)(15,20)(17,21)(19,23)", "a*b*c*d"); When the translation term of the affine map is lambda1: NewSphereMachine( "a=(1,6)(2,8)(3,22)(4,10)(5,24)(7,13)(9,15)(11,21)(12,18)(14,20)(17,23)(19,25)", "b=<1,1,1,1,1,b,1,1,1,1,1,1,1,1,1,1,1,1,c^-1,c,b^-1,1,1,1,1>(2,3)(4,5)(6,21)(7,8)(9,10)(11,16)(12,13)(14,15)(17,18)(19,20)(22,23)(24,25)", "c=(1,2)(3,4)(6,7)(8,9)(10,25)(11,12)(13,14)(15,20)(16,17)(18,19)(21,22)(23,24)", "d=(1,22)(2,6)(3,24)(4,8)(5,10)(7,11)(9,13)(12,16)(14,18)(15,25)(17,21)(19,23)", "a*b*c*d"); When the translation term of the affine map is lambda2: NewSphereMachine( "a=(1,7)(2,21)(3,9)(4,23)(5,20)(6,12)(8,14)(10,15)(11,17)(13,19)(16,22)(18,24)", "b=(1,2)(3,4)(5,25)(6,7)(8,9)(10,20)(11,12)(13,14)(16,17)(18,19)(21,22)(23,24)", "c=(1,21)(2,3)(4,5)(6,16)(7,8)(9,10)(12,13)(14,15)(17,18)(19,20)(22,23)(24,25)", "d=(1,16)(2,23)(3,7)(4,25)(5,9)(6,11)(8,12)(10,14)(13,17)(15,19)(18,22)(20,24)", "a*b*c*d"); When the translation term of the affine map is lambda1+lambda2: NewSphereMachine( "a=(1,7)(2,21)(3,9)(4,23)(5,25)(6,12)(8,14)(10,20)(11,17)(13,19)(16,22)(18,24)", "b=(1,2)(3,4)(6,7)(8,9)(10,25)(11,12)(13,14)(15,20)(16,17)(18,19)(21,22)(23,24)", "c=<1,1,1,1,1,b,1,1,1,1,1,1,1,1,1,1,1,1,c^-1,c,b^-1,1,1,1,1>(2,3)(4,5)(6,21)(7,8)(9,10)(11,16)(12,13)(14,15)(17,18)(19,20)(22,23)(24,25)", "d=(1,21)(2,23)(3,7)(4,25)(5,9)(6,16)(8,12)(10,14)(13,17)(15,19)(18,22)(20,24)", "a*b*c*d"); | 1,384 | 3,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.64475 |
http://blog.csdn.net/Singleton1900/article/details/46815601 | 1,506,296,280,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690228.57/warc/CC-MAIN-20170924224054-20170925004054-00538.warc.gz | 42,803,019 | 18,948 | # 基于无向图且权重单一的最短路径Dijkstra算法——JAVA实现
2760人阅读 评论(0)
http://blog.csdn.net/javaman_chen/article/details/8254309
Coordinate.java
Coordinate中包含相邻坐标的List，以及距离起始点的距离。
package com.harlan.dijkstra;
import java.util.LinkedList;
/**
* 坐标点的数据结构
*
* @author Harlan
*
*/
public class Coordinate {
//x坐标
public int x;
//y坐标
public int y;
//相邻坐标
public LinkedList<Coordinate> adj;
//距离
public int steps;
// 最短路径中的前一个顶点
public Coordinate lastPoint;
;
public Coordinate(){
}
public Coordinate(int newX, int newY) {
x = newX;
y = newY;
adj=new LinkedList<Coordinate>();
reset();
}
public void reset(){
steps=Integer.MAX_VALUE;
lastPoint=null;
}
@Override
public boolean equals(Object obj) {
if (!(obj instanceof Coordinate))
return false;
Coordinate other = (Coordinate) obj;
if (x == other.x && y == other.y) {
return true;
}
return false;
}
@Override
public int hashCode() {
return x*10000+y;
}
/**
* 以JSON格式展示坐标
*/
@Override
public String toString() {
return "{\"x\":" + x + ",\"y\":" + y + "}";
}
}
PathInfo.java
import java.util.List;
/**
* 路径信息
* @author Harlan
*
*/
public class PathInfo {
//目标点的坐标
private Coordinate targetCd;
//去往目标点的最佳路径
private List<Coordinate> cdList;
public Coordinate getTargetCd() {
return targetCd;
}
public void setTargetCd(Coordinate targetCd) {
this.targetCd = targetCd;
}
public List<Coordinate> getCdList() {
return cdList;
}
public void setCdList(List<Coordinate> cdList) {
this.cdList = cdList;
}
}
<span style="white-space:pre"> </span> /**
* 和周围的四个点建立关系
*
* @param node
*/
private void getContactWithF(Coordinate node) {
Coordinate coordinate = getCoordinate(node);
Coordinate EAST = new Coordinate(node.x + 1, node.y);
Coordinate SOUTH = new Coordinate(node.x, node.y + 1);
Coordinate WEST = new Coordinate(node.x - 1, node.y);
Coordinate NORTH = new Coordinate(node.x, node.y - 1);
if (isCellSafe(EAST, mRoads)) {
EAST = getCoordinate(EAST);
coordinate.adj.add(EAST);
}
if (isCellSafe(SOUTH, mRoads)) {
SOUTH = getCoordinate(SOUTH);
coordinate.adj.add(SOUTH);
}
if (isCellSafe(WEST, mRoads)) {
WEST = getCoordinate(WEST);
coordinate.adj.add(WEST);
}
if (isCellSafe(NORTH, mRoads)) {
NORTH = getCoordinate(NORTH);
coordinate.adj.add(NORTH);
}
}
/**
* 判断周围的位子是不是道路
*
* @param head
* @return
*/
public boolean isCellSafe(Coordinate park, Set<Coordinate> roads) {
boolean isSafe = false;
// 在道路集合里面，就是安全的，否则，不安全
for (Coordinate info : roads) {
if (info.equals(park)) {
isSafe = true;
}
}
return isSafe;
}
<span style="white-space:pre"> </span>// 无权最短路径计算
public void unweighted(Coordinate enter) {
if (enter == null)
throw new NoSuchElementException("Start vertex not found!");
LinkedList<Coordinate> q = new LinkedList<Coordinate>();
clearAll();
enter = vertexMap.get(enter.toString());
System.out.println("unweighted Harlan:" + enter.adj.toString());
q.addLast(enter);
enter.steps = 0;
while (!q.isEmpty()) {
Coordinate v = q.removeFirst();
for (Iterator<Coordinate> itr = v.adj.iterator(); itr.hasNext();) {
Coordinate w = itr.next();
if (w.steps == Integer.MAX_VALUE) {
w.steps = v.steps + 1;
w.lastPoint = v;
q.addLast(w);
}
}
}
}
<span style="white-space:pre"> </span>private List<Coordinate> getPath(Coordinate dest, List<Coordinate> cdList) {
if (dest.lastPoint != null) {
cdList = (getPath(dest.lastPoint, cdList));
}
cdList.add(dest);
return cdList;
}
<span style="white-space:pre"> </span>// 显示一条路径
public void printPath(String coodrStr) throws NoSuchElementException {
Coordinate coord = vertexMap.get(coodrStr);
if (coord == null)
throw new Exception(No path found!");
else if (coord.steps == Integer.MAX_VALUE)
System.out.println(coord.toString() + "is unreachable!");
else {
printPath(coord);
System.out.println();
}
}
// 显示实际最短路径
private void printPath(Coordinate dest) {
if (dest.lastPoint != null) {
printPath(dest.lastPoint);
System.out.print(",");
}
System.out.print(dest.toString());
}
GetDijkstraPath.java
package com.harlan.dijkstra;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
public class GetDijkstraPath {
private static final String TAG = GetDijkstraPath.class.getSimpleName();
/**
* 主函数，测试类
* @param args
*/
public static void main(String[] args) {
Coordinate enter = new Coordinate(2, 0);
Set<Coordinate> roads = new HashSet<Coordinate>();
roads.add(new Coordinate(3, 10));
roads.add(new Coordinate(3,11));
roads.add(new Coordinate(3, 8));
roads.add(new Coordinate(3, 9));
roads.add(new Coordinate(3, 6));
roads.add(new Coordinate(3, 7));
roads.add(new Coordinate(3, 4));
roads.add(new Coordinate(3, 5));
roads.add(new Coordinate(3, 2));
roads.add(new Coordinate(3, 3));
roads.add(new Coordinate(3, 1));
roads.add(new Coordinate(6, 1));
roads.add(new Coordinate(1, 9));
roads.add(new Coordinate(1, 8));
roads.add(new Coordinate(1, 11));
roads.add(new Coordinate(1, 10));
roads.add(new Coordinate(1, 5));
roads.add(new Coordinate(1, 4));
roads.add(new Coordinate(1, 7));
roads.add(new Coordinate(1, 6));
roads.add(new Coordinate(1, 1));
roads.add(new Coordinate(1, 3));
roads.add(new Coordinate(1, 2));
roads.add(new Coordinate(4, 1));
roads.add(new Coordinate(4, 11));
roads.add(new Coordinate(7, 5));
roads.add(new Coordinate(7, 4));
roads.add(new Coordinate(2, 11));
roads.add(new Coordinate(7, 7));
roads.add(new Coordinate(7,6));
roads.add(new Coordinate(7, 1));
roads.add(new Coordinate(7, 3));
roads.add(new Coordinate(7,2));
roads.add(new Coordinate(2, 1));
roads.add(new Coordinate(7, 9));
roads.add(new Coordinate(7,8));
roads.add(new Coordinate(7, 11));
roads.add(new Coordinate(7, 10));
roads.add(new Coordinate(5,11));
roads.add(new Coordinate(5, 10));
roads.add(new Coordinate(5, 9));
roads.add(new Coordinate(5,8));
roads.add(new Coordinate(5,7));
roads.add(new Coordinate(5,6));
roads.add(new Coordinate(5,5));
roads.add(new Coordinate(5,4));
roads.add(new Coordinate(5,3));
roads.add(new Coordinate(5,2));
roads.add(new Coordinate(5,1));
System.out.println("nearest roads.size(): "+roads.size());
Set<Coordinate> trags = new HashSet<Coordinate>();
trags.add(new Coordinate(5, 4));
trags.add(new Coordinate(5, 5));
PathInfo nearest = getNearestPathInfo(roads,trags,enter);
System.out.println("nearest : "+nearest.getCdList());
}
/**
* 对外的接口(如果计算多入口的最短路径的时候使用)
* 获取多入口的最佳路径
* @param roads
* @param trags
* @param enters
* @return
*/
public static PathInfo getNearestPathInfoFromDiffEnter(Set<Coordinate> roads,
Set<Coordinate> trags, Set<Coordinate> enters){
List<PathInfo> list = new ArrayList<>();
for(Coordinate enter:enters){
list.add(getNearestPathInfo(roads,trags,enter));
}
//每条路径的步长
int steps = Integer.MAX_VALUE;
PathInfo nearste = new PathInfo();
for(PathInfo pathInfo:list){
if(pathInfo.getCdList().size()<steps){
steps = pathInfo.getCdList().size();
nearste = pathInfo;
}
}
return nearste;
}
/**
* 对外的接口(如果计算单一入口的最短路径时候使用)
* 获取单一入口的最佳路径
*
* @param roads
* @param trags
* @param enter
* @return
*/
public static PathInfo getNearestPathInfo(Set<Coordinate> roads,
Set<Coordinate> trags, Coordinate enter){
List<PathInfo> list = getAllAvailablePathInfo(roads,trags,enter);
// for(PathInfo info:list){
// System.out.println("getNearestPathInfo targ:"+info.getTargetCd());
// System.out.println("getNearestPathInfo route:"+info.getCdList());
// System.out.println("getNearestPathInfo *********************");
// }
//
//每条路径的步长
int steps = Integer.MAX_VALUE;
PathInfo nearste = new PathInfo();
for(PathInfo pathInfo:list){
if(pathInfo.getCdList().size()<steps){
steps = pathInfo.getCdList().size();
nearste = pathInfo;
}
}
return nearste;
}
/**
* 获取到达所有目标点的所有可用路径
*
* @param roads
* @param trags
* @param enter
* @return
*/
private static List<PathInfo> getAllAvailablePathInfo(Set<Coordinate> roads,
Set<Coordinate> trags, Coordinate enter) {
Set<Coordinate> availableRoadTar = getAllRoadNearTarg(roads,trags);
//计算出起始点到各个可达点的距离
HarlanDijkstra test=new HarlanDijkstra(roads,enter);
test.unweighted(enter);
//得出到停车位的可达点的最短路径
List<PathInfo> availableList = new ArrayList<>();
for(Coordinate info:availableRoadTar){
PathInfo pathInfo = test.getPathInfo(info.toString());
availableList.add(pathInfo);
}
return availableList;
}
/**
* 获取通往所有目标的有效道路点(一个目标的临近道路点的集合)
* @param roads
* @param tragSet
* @return
*/
private static Set<Coordinate> getAllRoadNearTarg(Set<Coordinate> roads,
Set<Coordinate> tragSet) {
Set<Coordinate> allOfNearList = new HashSet<>();
for (Coordinate targ : tragSet) {
// System.out.println("getAllRoadNearTarg targ:"+targ);
Set<Coordinate> childSet = getRoadNearTarg(roads,targ);
allOfNearList.addAll(childSet);
}
return allOfNearList;
}
/**
* 获取通往一个目标的临近道路点
*
* @param roads
* @param targ
* @return
*/
private static Set<Coordinate> getRoadNearTarg(Set<Coordinate> roads,
Coordinate targ) {
Set<Coordinate> nearList = new HashSet<>();
Coordinate EAST = new Coordinate(targ.x + 1, targ.y);
Coordinate SOUTH = new Coordinate(targ.x, targ.y + 1);
Coordinate WEST = new Coordinate(targ.x - 1, targ.y);
Coordinate NORTH = new Coordinate(targ.x, targ.y - 1);
for (Coordinate info : roads) {
if (EAST.equals(info)) {
nearList.add(EAST);
}
if (SOUTH.equals(info)) {
nearList.add(SOUTH);
}
if (WEST.equals(info)) {
nearList.add(WEST);
}
if (NORTH.equals(info)) {
nearList.add(NORTH);
}
}
return nearList;
}
}
nearest roads.size(): 49
nearest : [{"x":2,"y":0}, {"x":2,"y":1}, {"x":3,"y":1}, {"x":4,"y":1}, {"x":5,"y":1}, {"x":5,"y":2}, {"x":5,"y":3}]
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最新评论 | 2,823 | 9,508 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-39 | longest | en | 0.172486 |
https://forum.civiljungle.com/how-is-concrete-measured/ | 1,696,070,088,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510671.0/warc/CC-MAIN-20230930082033-20230930112033-00478.warc.gz | 275,380,845 | 73,068 | Please briefly explain why you feel this question should be reported.
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# How Is Concrete Measured?
Calculating the volume for concrete requires using three dimensions: length times width (which is area) times the height (thickness) or l x w x h. If you are using feet as your standard unit of measurement, this formula will give you the volume of concrete in cubic feet.
## How Is Concrete Measured?
Concrete is a versatile material used in construction for a variety of purposes. It is the most widely used construction material in the world and is used to create foundations, walls, driveways, and other structures.
Concrete is a mixture of cement, water, and aggregate, such as sand and gravel. Each of these components is measured when concrete is being mixed to ensure that it has the correct strength and consistency.
When mixing concrete, it is important to know the exact proportions of each component. This is necessary to create a strong and durable concrete.
The proportions of each component depend on the type of concrete being used, such as its intended purpose and the environment in which it will be used.
The proportions will also affect the strength and durability of the concrete. When measuring concrete, the concrete mix is typically measured in cubic yards.
A cubic yard is a unit of volume that measures 3 feet by 3 feet by 3 feet. Each cubic yard contains 27 cubic feet of concrete.
This amount can then be divided into the amount of each component needed for the particular concrete mix. The amount of cement used in a concrete mix is usually measured in sacks.
A sack is a unit of weight that is equal to 94 pounds of material. One sack of cement | 348 | 1,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-40 | longest | en | 0.944694 |
https://blog.tanyakhovanova.com/2009/08/physics-puzzle-by-levy-ulanovsky/ | 1,714,008,072,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296820065.92/warc/CC-MAIN-20240425000826-20240425030826-00190.warc.gz | 118,393,669 | 14,185 | ## Physics Puzzle, by Levy Ulanovsky
My guest blogger is Levy Ulanovsky, a maven of physics puzzles. Here is one of his favorite puzzles:
There are n points in 3-dimensional space. Every point is connected to every other point by a wire of resistance R. What is the resistance between any two of these points?
Share:
I got 2R/n.
2. #### Austin:
Ditto Leo. I only know very basic physics, but I think this can be puzzled out from Ohm’s Law, in the form V/I = R.
Let us fix the power source and let n vary. By analogy to pressure, the voltage everywhere in the circuit is inversely proportional to the total length of wire, which is n(n-1)/2. Meanwhile, the effective current between any pair of nodes is inversely proportional to the average travel time between the two nodes. Since the current forks in n-1 equally attractive directions at each junction, this travel time is proportional to n-1.
So, between any pair of nodes, V/I is in linear proportion to n-1/[n(n-1)/2] = 2/n.
If the resistance in each wire is R, then the resistance between any pair of nodes is 2R/n (since this gives the right constant of proportionality when n=2).
I imagine a real physics maven could explain this more succinctly and persuasively!
3. #### misha:
Here is an approach I would imagine a physicist could use. Let us connect 2 points of our grid to a battery. Then all the rest of the points will have the same potential because of the symmetry, and the current in all the wires connecting them will be zero. So these wires can be removed, and all we are left with is one wire of resistanse R and n-2 wires of resistance 2R connecting the terrminals of the battery. Now use the fact that the resistance of several resistors in parallel connection is equal to the harmonic mean of the resistance of all the resistors to get the formula.
4. #### misha:
Oops! Sorry, it’s not the harmonic mean, it’s m times smaller, where m is the number of resistors in the parallel connection.
5. #### misha:
Here is another one.If you throw a ball upwards, will it take more time to reach the maximum height or to fall down from it? If we disregard the air reistance, the times will be the same. But what if we take it into account? It is an “applied project” in Stewart’s 4-th edition Calculus, pages 602-603 and includes the drag force propotional to the velocity and some dreadful calculations. Can you tell without any calculations? | 568 | 2,421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | longest | en | 0.933782 |
https://www.exam-answer.com/osfp-route-selection | 1,722,676,692,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640365107.3/warc/CC-MAIN-20240803091113-20240803121113-00204.warc.gz | 593,775,109 | 11,016 | OSPF Route Selection:
# Selecting Routes in OSPF
### Question
When OSPF learns multiple paths to a network, how does it select a route?
### Explanations
Click on the arrows to vote for the correct answer
A. B. C. D.
C
OSPF (Open Shortest Path First) is a link-state routing protocol that determines the shortest path to a destination network based on the lowest metric or cost. OSPF learns multiple paths to a network through the exchange of link-state advertisements (LSAs) between neighboring routers.
When OSPF receives multiple paths to a network, it uses a Dijkstra algorithm to calculate the shortest path to the network based on the lowest cost or metric. The cost of a path is determined by the bandwidth of the links along the path.
To calculate the cost of a path, OSPF uses the following formula:
Cost = Reference bandwidth / Interface bandwidth
The reference bandwidth is a value that represents the theoretical maximum bandwidth of a network. By default, OSPF uses a reference bandwidth of 100 Mbps. However, this value can be changed with the auto-cost reference-bandwidth command.
The interface bandwidth is the actual bandwidth of the outgoing interface that connects to the next hop towards the destination network. OSPF uses the bandwidth command to determine the bandwidth of an interface.
Once OSPF calculates the cost of all the paths to the destination network, it selects the path with the lowest cost as the preferred path. If there are multiple paths with the same lowest cost, OSPF can load balance traffic across the paths.
Therefore, the answer that correctly explains how OSPF selects a route when learning multiple paths to a network is C. OSPF divides a reference bandwidth of 100 Mbps by the actual bandwidth of the outgoing interface to calculate the route with the lowest cost. | 368 | 1,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-33 | latest | en | 0.878875 |
http://www.pas.rochester.edu/~blackman/ast104/aristotle8.html | 1,620,966,591,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991737.39/warc/CC-MAIN-20210514025740-20210514055740-00348.warc.gz | 79,687,935 | 3,992 | The Universe of
Aristotle and Ptolemy and the Role of Eratosthenes
## Aristotle and Eratosthenes
Although Aristotle was not a scientist, it is important to note that he did point out that the Earth had to be sphere since its shadow was always circular. This was in fact a key scientific insight. It allowed Eratosthenes around 200 BC to calculate the circumference of the Earth. The method was very clever for the time:
The southern Egyptian city of Syene had a well in which sun-rays fell directly vertical during the summer solstice. The sun was therefore at the Zenith there during this time.
However, during the same time in Alexandria, the sun was 7 degrees south of the Zenith (1/50 of the circumference of the sky). Now since the Earth was spherical (from Aristotle) the 7 degree angle subtended at the Earths surface divided by 360 degrees equaled the distance between Alexandria and Syene divided by the Earth's circumference. That is, by using
7 / 360 = distance / circumference
he could measure the distance (in units of stadia) and then infer the circumference in stadia (about 40,000). Hard to infer the actual distance in standard modern units since stadia are of varying sizes, but the technique was clever at the time and if one uses typical stadium lengths of the time the estimate was only off by a number bewteen 4 and 14 percent.
## Aristotle's Spheres
The celestial sphere that we introduced previously is a convenient "fiction" to locate objects in the sky. However, the Greek philosopher Aristotle (many of Aristotle's works are available at the Internet Classics Archive) proposed that the heavens were literally composed of 55 concentric, crystalline spheres to which the celestial objects were attached and which rotated at different velocities (but the angular velocity was constant for a given sphere), with the Earth at the center. The following figure illustrates the ordering of the spheres to which the Sun, Moon, and visible planets were attached.
(The diagram is not to scale, and the planets are aligned for convenience in illustration; generally they were distributed around the spheres.) There were additional "buffering" spheres that lay between the spheres illustrated. The sphere of the stars lay beyond the ones shown here for the planets; finally, in the Aristotelian conception there was an outermost sphere that was the domain of the "Prime Mover". The Prime Mover caused the outermost sphere to rotate at constant angular velocity, and this motion was imparted from sphere to sphere, thus causing the whole thing to rotate.
By adjusting the velocities of these concentric spheres, many features of planetary motion could be explained. However, the troubling observations of varying planetary brightness and retrograde motion could not be accommodated: the spheres moved with constant angular velocity, and the objects attached to them were always the same distance from the earth because they moved on spheres with the earth at the center.
## Epicycles and Planetary Motion
The "solution" to these problems came in the form of a clever proposal: planets were attached, not to the concentric spheres themselves, but to circles attached to the concentric spheres, as illustrated in the adjacent diagram. These circles were called "Epicycles", and the concentric spheres to which they were attached were termed the "Deferents". Then, the centers of the epicycles executed uniform circular motion as they went around the deferent at uniform angular velocity, and at the same time the epicycles (to which the planets were attached) executed their own uniform circular motion.
The net effect was as illustrated in the following animation. As the center of the epicycle moves around the deferent at constant angular velocity, the planet moves around the epicycle, also at constant angular velocity. The apparent position of the planet on the celestial sphere at each time is indicated by the line drawn from the earth through the planet and projected onto the celestial sphere. The resulting apparent path against the background stars is indicated by the blue line.
Now, in this tortured model one sees that it is possible to have retrograde motion and varying brightness, since at times as viewed from the earth the planet can appear to move "backward" on the celestial sphere. Obviously, the distance of the planet from the Earth also varies with time, which leads to variations in brightness. Thus, the idea of uniform circular motion is saved (at least in some sense) by this scheme, and it allows a description of retrograde motion and varying planetary brightness.
## More Sophisticated Epicycles: The Ptolemaic Universe
In practice, even this was not enough to account for the detailed motion of the planets on the celestial sphere. In more sophisticated epicycle models further "refinements" were introduced:
• In some cases, epicycles were themselves placed on epicycles, as illustrated in the adjacent figure.
• In actual models, the center of the epicycle moved with uniform circular motion, not around the center of the deferent, but around a point that was displaced by some distance from the center of the deferent.
That ancient astronomers could convince themselves that this elaborate scheme still corresponded to "uniform circular motion" is testament to the power of three ideas that we now know to be completely wrong, but that were so ingrained in the astronomers of an earlier age that they were essentially never questioned:
1. All motion in the heavens is uniform circular motion.
2. The objects in the heavens are made from perfect material, and cannot change their intrinsic properties (e.g., their brightness).
3. The Earth is at the center of the Universe.
These ideas concerning uniform circular motion and epicycles were cataloged by Ptolemy in 150 A.D. His book was called the "Almagest" (literally, "The Greatest"), and this picture of the structure of the Solar System has come to be called the "Ptolemaic Universe".
## Medieval Aristotelian Astronomy
By the Middle Ages, such ideas took on a new power as the philosophy of Aristotle (newly rediscovered in Europe) was wedded to Medieval theology in the great synthesis of Christianity and Reason undertaken by philosopher-theologians such as Thomas Aquinas. The Prime Mover of Aristotle's universe became the God of Christian theology, the outermost sphere of the Prime Mover became identified with the Christian Heaven, and the position of the Earth at the center of it all was understood in terms of the concern that the Christian God had for the affairs of mankind.
Thus, the ideas largely originating with pagan Greek philosophers were baptized into the Catholic church and eventually assumed the power of religious dogma: to challenge this view of the Universe was not merely a scientific issue; it became a theological one as well, and subjected dissenters to the considerable and not always benevolent power of the Church. | 1,404 | 6,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-21 | latest | en | 0.97455 |
https://butchixanh.edu.vn/worksheets-with-areas-and-perimeters-of-polygons-and-shapes-made-for-primary-maths-students-wjcluxq4/ | 1,718,482,211,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00402.warc.gz | 123,199,071 | 41,620 | Area and Perimeter of Composite Plane Shapes Part 1
Area and Perimeter of Composite Plane Shapes Part 1
These are our Compound Shapes worksheets. Click on the previews to download the free area and perimeter worksheet.
On this page you find our area and perimeter worksheets for grades 3 to 6. We have basic area of rectangles and square worksheets, drawing areas and perimeter worksheets, areas in grids worksheets, area of compound shapes, find the length of an unknown side given either the area or the perimeter of a shape, area of triangle worksheets, area of triangular shapes worksheets, area of regular polygons worksheets, circle and circumference worksheets, find the radius and diameter worksheets.
We also have some great worksheets in which you need to find the area of circular objects. Very challenging.
Our area and perimeter worksheets for math grades 3 to 6 cover: finding area worksheets, finding perimeter worksheets, drawing areas worksheets, drawing perimeter worksheets, areas of compound shapes worksheets, areas of squares and rectangles, unknowns sides in areas worksheets, areas in triangles, perimeter of triangles, areas of shapes, circumference of circles, circular shapes, area of quart circles worksheets, area of half circles, find the height worksheets, find the radius worksheets, find the diameter, area of parallelograms, area of trapeziums, are of trapezoids, area of rhombus worksheets, area of regular polygons worksheets.
• Concept of Area and Perimeter on a plane
• Knowing the formulas to calculate area and perimeter
• The use of standard units for area, perimeter and volume
• The ability to solve word problems involving area and perimeter and volume
• Finding the length of unkowns sides given othe parameters
• Finding the area and perimeter of compound shapes
• Finding the area and perimeter of triangles
• Finding the area and circumference, diameter and radius of circles
• Being able to calculate area of circles
• Being able to calculate area of circular objects
You are watching: Worksheets with areas and perimeters of polygons and shapes, made for primary maths students.. Info created by Bút Chì Xanh selection and synthesis along with other related topics. | 436 | 2,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-26 | latest | en | 0.876255 |
http://mathhelpforum.com/algebra/214196-math-challenge-print.html | 1,495,724,820,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608084.63/warc/CC-MAIN-20170525140724-20170525160724-00465.warc.gz | 233,142,996 | 2,690 | # Math Challenge
• Mar 4th 2013, 07:09 AM
srirahulan
Math Challenge
If $10(2009)^4+11(2009)^3+12(2009)^2+13(2009)+14$ Divided by 2009 ,Find out the value? In this problem I cannot calculate the value of 2009^n please help meee????
• Mar 4th 2013, 10:11 AM
HallsofIvy
Re: Math Challenge
It's fairly obviously $10(2009)^3+ 11(2009)^2+ 12(2009)+ 13+ \frac{14}{2009}$. What do you mean by "cannot calculate the value of 2009^n"? If you are not allowed to calculate those powers you can just leave it the way it is.
• Mar 4th 2013, 03:27 PM
srirahulan
Re: Math Challenge
They simplify the answers,and it is in M.C.Q--
Here is the choices
A)0 B)12 C)14 D)1985 E)2008 | 247 | 663 | {"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} | 3.140625 | 3 | CC-MAIN-2017-22 | longest | en | 0.817776 |
http://s3-us-west-1.amazonaws.com/nitwitpdfdl/beginners-website-templates-free-50.html | 1,563,673,232,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526799.4/warc/CC-MAIN-20190720235054-20190721021054-00280.warc.gz | 129,923,201 | 8,811 | ## Beginners website templates free 5.0,online number worksheets,export promotion strategy wikipedia,digital marketing courses oxford - How to DIY
Whenever you need to quickly render a three dimensional object geometrically accurate, this is the perspective style to go with. You can opt to place these vanishing points at equal distance from your canvas center, on or outside of your canvas limits for example.
When working digitally, in Photoshop for example, you can create a much larger canvas than needed, and isolate the actual visible area with a layer you use as a "Crop-mask". You can see what I'm referring to in the image below, showing you how I've used this technique for the first image in this article. To ensure that you have a firm grasp of the basic concepts presented here, let's take a look at this quick exercise. Start defining the base of your building, as usual, making use of as many construction lines as needed.
Now that the placement has been clarified, let's start extruding the volume upwards to identify exactly how tall the building will appear to the viewer.
Separate your lateral and front sides of the volume by using two different shades and layers.
Now that we've decided upon the final dimensions of our volume, it's time to go deeper into refining the proportions. Identify the exact center of your lateral wall by drawing out the diagonal lines, and extending a vertical construction line. To define the lateral margins and slope of your roof, you simply need to extend a line from the highest point towards the ground, and pass through a corner of your initial volume, in the direction you're extending to. Having done that for all four corner points, I strongly suggest that you now separate the surfaces both visually, with different shades, as well as on their own layers, for easier later edits (if you'd want to add roof shingles later, this would make perfect sense). To create the porch, we will simply extend the base of our building by about two meters towards the viewer, and around 40 centimeters in height. Firstly, we want the pillars to be located exactly at the corners of our building, but we also want them to be visible, so we're going to have the walls pushed in slightly.
The easiest method would be that of identifying the base of each pillar and extruding it upwards. In this step, it's important to keep in mind the fact that due to the porch, your position as a viewer is slightly lower, and this in turn will mean that your door will seem taller.
As the number of construction lines is constantly growing, I suggest you take the time ever so often to shade the faces that you defined, and to separate them on dedicated layers as often as possible. Similarly to how you managed to push the walls towards the inside of your volume, you will now have to define the thickness of the front wall.
Now that the construction lines have been drawn, once more, clean up the image and make sure to use the same shades, depending on the orientation of the resulting faces. As the porch is a bit higher than the average "comfortable" height, we will have to create a step to ease the ascent onto it. The key is drawing as many construction lines as needed, and always projecting relevant points on already existent faces. Add the same lighter shading on the top faces, the darkest shade to the faces that are not in direct light (our light source is to the right). In order to be able to do this, you will first have to decide upon a good position for the chimney. Finally, draw two construction lines connected to your right vanishing point, across the front face of the roof. Given the rather steep perspective, a great part of your fence may fall outside of the canvas area. This step of course is entirely up to you, and in a real usage scenario, requires a bit more work.
Please keep in mind that drawing in perspective takes far less time if you're doing it on paper, rather than digitally.
With enough practice, you will only need a minimal amount of construction lines, and still be able to achieve convincing perspectives.
Hope you enjoyed reading this, and if you have any questions, don't hesitate to write in the comments below! I was worried that because I did not have great computer skills I would fall behind in class, however with the great support from you and your team, and of course the friendly and comfortable environment, I did manage. I now have my website that I am very happy with and I also have the knowledge and skills to manage it myself, I can now update or fix anything in my webpage whenever I want! I had an existing website that was tired and ordinary looking, and engaged Training Lab to help us build a fresh, new site, with loads of flexibility that we can easily update anytime (even from my iphone!). The Training Lab team were committed to honouring their claim to have a working website at the end of the weekend course.
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The Dynamic Flash template Photo Portfolio presented on this page is a high-quality professional web design solution for professional photographers and beginners that seek effective presentation of their photo works. You can edit this Dynamic Flash template Photo Gallery using HTML editor, Flash and Photoshop software. Two point perspective is by far the best suited perspective drawing styles available for rendering environment concept art. That doesn't mean that the three point perspective wouldn't be equally, if not better suited, but for obvious reasons, the current style is far easier to implement. Please note that the closer one of the vanishing points will be to the center of your canvas, the more biased the image will seem.
This is useful when you decide to place your vanishing points far apart, outside of your main canvas.
I've decided to place one of them right on the left margin of our canvas, and the other a bit further outside. In this step, it's important to get the base of your volume defined, and see how it will lay on the ground.
The height at which you decide to extend this will represent the highest point of your roof.
Of course, as we're not actually measuring in our case, I'm just providing you with some dimensions worth keeping in mind, so that you can identify correct proportions. Aim to repeat the same process used when defining the base shape in the second step of the exercise, only this time, within the initial boundaries of your volume. You do not have to spend a lot of time on this step, but the more time you invest now, the more precise your selections will be further on, when you'll want to start adding details. Don't worry about it, but do not skip the construction process, because the human eye is very receptive to perspective errors, and what might save you a couple of minutes, would have a negative effect on your final image. To enhance the depth perception, I recommend using a gradient for the sky, and maybe something more solid, but much darker for the ground. The scope of this tutorial is that of explaining the work flow, and helping you learn by doing. No matter how trivial my query might seem she has been able to provide guidance and a ready solution.
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The reason for this is the fact that realism and proportions can be achieved fairly quickly, and the focus remains on the composition, rather than on the technical details. As opposed to one point perspective and two point perspective, with the two vanishing points, offers you the ability to focus more on your actual subject, than on the potential negative effects of choosing an inappropriate single vanishing point and perspective direction.
The vanishing point placement however, is far less strict than in the case of the one point perspective. What this means is that the side of your object affected by this vanishing point will seem "squeezed". In hand-drawing you could make use of a bigger piece of paper, but digitally you just need to consider a few extra steps. Be sure to repeat this procedure on both sides of your volume to obtain the correct slope in the next step. Project these on the front face, then connect the resulting points to your left vanishing point. That being said however, I recommend that in actual usage scenarios you just keep these techniques in mind, and rely on them just to build so called "bounding boxes" (less detailed containment volumes) to which you add details as it comes naturally. The best bit being that you then have the ability to make changes as you see fit down the track. I had to contact them to make some admin changes and their customer service was excellent, considering the amount I am paying I have to give them an A+. Dynamic Flash template Portfolio is very easy to edit despite the sophisticated and effective look. This meant, in many ways we felt uncertain about what we needed on our website and how best to represent our product.
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### Comments to «Beginners website templates free 5.0»
1. SEVEN_OGLAN Says:
You, that might not be the very best that.
2. KINQ_BOXINQ Says:
High-priced hammer as you phrased it - that demands what all.
3. Lapuli4ka Says:
Concluded on Tough drive trial of debut??but the video capture is genuinely. | 2,370 | 11,733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-30 | latest | en | 0.939317 |
https://puzzling.stackexchange.com/questions/103967/how-to-know-which-head-belongs-to-which-tail-in-a-bunch-of-wires?noredirect=1 | 1,723,599,986,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641086966.85/warc/CC-MAIN-20240813235205-20240814025205-00200.warc.gz | 391,976,141 | 34,702 | # How to know which head belongs to which tail in a bunch of wires? [duplicate]
I saw a puzzle in a contest in Iran held by an association called "Rahneshan" and the original text of the contest is uploaded HERE and of course both website and text are in Persian.
Between "City A" and "City B", there is a big cable which contains 100 separate wires inside it. Unfortunately, the wires are knotted together and we can't tell which head of a wire in one side (City A), belongs to which tail on the other side (City B).
The only tools that we have are:
• A power outlet
• A test light (which can be used to determine if electricity is flowing through a wire)
We can make a round-trip from City A to City B and back once.
How can we determine which endpoints belong to which wires?
• No I didn't create it myself. There's a kind of contest and I was asked. Commented Oct 18, 2020 at 21:19
• No it's not active. I was only curious about it and I want to see how each individual may solve the problem, if possible. I myself have no idea. Commented Oct 18, 2020 at 21:24
• To whoever closed it, this is not a duplicate. The alleged duplicate question asks "What is the fewest number of trips from one end of the cable to the other required to identify ...each individual wire." This question stipulates only ONE trip can be made. Commented Oct 18, 2020 at 23:12
• @Earlien Yes I think so, too. Thanks for your attention. But the procedure is completely useful in this puzzle, too. Commented Oct 18, 2020 at 23:15
• The answers to the linked question are answers to this question as well. They lay out a strategy, requiring one round-trip, that can solve the problem. Since the answers are the same the focus on a different part of the solution does not matter. Commented Oct 19, 2020 at 3:09 | 445 | 1,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-33 | latest | en | 0.970966 |
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