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https://www.cs.uaf.edu/2009/fall/cs441/lecture/09_22_tkgate.html | 1,586,221,376,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371662966.69/warc/CC-MAIN-20200406231617-20200407022117-00079.warc.gz | 882,885,099 | 4,388 | # Circuit-Level CPU Simulation with TkGate
CS 441 Lecture, Dr. Lawlor
## Circuit Simulation via TkGate
We're really not going to do spend much time designing actual digital circuits (this isn't an EE course!), but it's important to understand the general way digital logic circuits work.
The easiest way to do this is to just play around with circuits for a few hours. I've prepared the examples below using TkGate, a UNIX-ey open source digital logic designer and simulator.
• In Windows, TkGate is only available via the UNIX compatability layer Cygwin. I've prepared one 55MB .zip file with just TkGate and the stuff it requires, but if you want more of Cywgin you can do a Cygwin Raw Install (install from a normal mirror first, then grab TkGate from the www.cs.ubc.ca/local/software/cygwin mirror).
• In Linux, TkGate is often available from your package manager (in Ubuntu, "sudo apt-get install tkgate").
• In MacOS X, supposedly TkGate can be built from source, which you download here.
The TkGate Documentation is pretty good, and the program runs a simple tutorial when you start it.
Many other circuit simulators exist. In EE 341, I used "LogicWorks", which is a fine commercial program for Windows and MacOS. Sadly, translating circuits between two graphical simulator packages is only rarely possible!
## Simple CPUs built from Digital Logic Circuits
Here are a few steps on the evolutionary path to a CPU:
First, an add circuit. We've set up two 8-bit hex input devices (DIP Switches) in TkGate, hooked them to a Make -> ALU -> Adder, and displayed both input and output binary data in 8-LED arrays. Click the above circuit to download the TkGate Verilog-style circuit description.
Here's almost the same circuit, except now we run the input values past a multiply circuit as well as the adder.
Now we've added a "bus". The little three-input triangles below each arithmetic unit are "tri-state buffer/drivers", which enable us to turn the add and multiply outputs on and off, and so combine the outputs of these two circuits. The idea is we turn on the output we want, which at the moment we do manually, by flipping the appropriate output switch. If you turn multiple outputs on at the same time, or turn no outputs on, TkGate shows a yellow indeterminate output state. In real circuits, driving a bus to several different values causes the buffer drivers to heat up and possibly fry themselves!
Now we add a set of "registers", which are just data storage elements. We hook up the output of each register to a tri-state driver, and use the drivers to select which register we want to read as input. We hook up the data input to each register to the arithmetic output bus. Finally, we can manually "clock in" the arithmetic result into any register (via the triangle-shaped register inputs).
Annoyingly, TkGate starts all the registers in indeterminate (yellow) state, rather than zero, so we have to manually clear the registers at simulated startup by flipping the register clear switch off and then on again.
OK, now we're approaching a real CPU! The only thing we're missing is a control unit that will flip all the switches appropriately to execute some instructions!
The big trapezoidal "0 D 3" modules are called decoders (also known as demultiplexors or demuxes). They take a binary input value (the red bus going in the top), and copy the left input to one of the four outputs along the bottom. We've wired these mux outputs to the input lines of our arithmetic bus, register output bus, and register input bus. This means that rather than manually flipping switches, we just need to load a binary number into the "instruction" register!
In this case, we've connected bits 1:0 (the low two bits) of the instruction to the arithmetic operation demux. So if the low bits are 00, the add circuit turns on. If the low bits are 01, the multiply circuit turns on. Bits 5:4 (the low two bits of the high hex digit) connect to the register output. So 00 means output register 0 to the arithmetic input bus, 01 means output register 1, 10 means output register 2, and 11 is an error. Finally, bits 7:6, the high two bits of the instruction, connect to the register input control lines. To write to register 0, the high bits should be zero, and so on. You need to manually flip the "write" switch to make the registers write (this avoids annoying circuit race conditions).
So overall, this CPU's instructions look like:
<destination register: 2 bits> <source register: 2 bits> <unused: 2 bits> <operation: 2 bits>
<constant: 8 bits>
For example, the instruction (hex) "40" (binary 0100 0000) writes to register 1, reads from register 0, and adds.
Here's a philosophically similar CPU, but I'm using multiplexors instead of tristate busses. You can get an 8-bit mux by double-clicking the mux, clicking the "Port" tab, double-clicking on an input (or output) pin, and typing in the bit width you want (possibly after shuffling dialog boxes). The above CPU uses bits 7:6 as the destination register, bit 5 is reserved, bit 4 is the arithmetic operation, and bits 3:2 and 1:0 are both input register numbers.
So this mux-based CPU's instructions look like:
<destination register: 2 bits> <reserved: 1 bit> <operation: 1 bit> <source register A: 2 bits> <source register B: 2 bits>
For example, the instruction (hex) "13" (binary 0001 0011) writes to register 0 the sum of register zero and register 3 (the constant).
Finally, here's the CPU we developed in class.
All instructions read from register 0 (the accumulator) and another register, specified in the low two bits (1:0) of the instruction. The operation is encoded in bit 3. The results are (curiously!) written to any combination of the four registers, specified via the high four bits of the instruction.
<source register: 2 bits> <unused: 1 bit> <operation: 1 bit> <destination registers: 4 bits>
For example, "0xf0" adds the accumulator to itself, and writes the result across all four registers.
## CPU Design Philosophy
A CPU is actually a very simple circuit that continually does exactly two things:
1. Figure out what to do next. (fetch)
2. Do it. (decode and execute)
CPU's usually encode "what to do next" in a chunk of binary data called a machine language instruction, almost always stored in memory somewhere. CPUs usually have a special register, the "program counter" or "instruction pointer", that points to the address in memory with the instruction to execute next.
So "Figure out what to do next" just means "Load the next instruction from the memory address at the instruction pointer." The instruction pointer is usually moved to point to the next instruction as part of this "fetch" phase.
"Do it" just means to decode the machine language instruction and figure out what it's telling you to do, then actually do what it says (execute). We've basically got that under control in the TkGate circuit above. | 1,633 | 6,956 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-16 | longest | en | 0.872996 |
https://solvedlib.com/task-1-research-and-explore-what-is-artificial,440406 | 1,709,621,395,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948223038.94/warc/CC-MAIN-20240305060427-20240305090427-00148.warc.gz | 548,758,752 | 17,106 | Task 1 Research and explore what is Artificial intelligence (AI) and Blockchain. Critically analyse how are...
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Task 1 Research and explore what is Artificial intelligence (AI) and Blockchain. Critically analyse how are they currently used in accounting and what is their potential future in the field. Task 2
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https://playtaptales.com/10-to-the-power-of-2759490/ | 1,600,657,325,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198887.3/warc/CC-MAIN-20200921014923-20200921044923-00173.warc.gz | 572,586,520 | 4,127 | ## 10 to the power of 2,759,490
What is 10 to the power of 2759490 (10^2759490)? The answer is a Novendecinongentillinovemvigintioctingentillion. This is based on the Conway-Wechsler system for naming numbers.
## 10 to the power of 2,759,490 written out
10^2759490 is written out as:
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| 521,394 | 1,042,882 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.592262 |
https://math.stackexchange.com/questions/1560137/determining-the-conjugacy-classes-of-a-non-abelian-group-g-of-order-8 | 1,571,022,278,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649035.4/warc/CC-MAIN-20191014025508-20191014052508-00455.warc.gz | 700,222,227 | 31,745 | Determining the conjugacy classes of a non-abelian group $G$ of order $8$
Let $G$ be a non-abelian group of order $8$. It is easy to see, that G has center $Z(G)\cong\mathbb{Z}/2\mathbb{Z}$. Set $Z(G)=\{e,z\}$. Is is also easy to see that $G/Z(G)\cong V$ is the Klein four group. Hence set $$G/Z(G)=\{Z(G), aZ(G), bZ(G), abZ(G)\}.$$ As $G/Z(G)$ is abelian, each of its element is the only element in its conjugacy class. I want to find the conjugacy classes of $G$. I suppose that they are $$\{\{1\},\{z\}, \{a,az\},\{b,bz\},\{ab,abz\}\}.$$
How can I find in this specific example (assuming that I don't know from the classification the precise structure of the two possible groups $G$) the conjugacy classes of $G$, if I already know the conjugacy classes of $G/Z(G)$?
Can I perform this strategy for a general $G$, if I already know the conjugacy classes of some factor group $G/N$?
From an answer to this question, I already know that the preimage under $G\to G/Z(G)$ of a conjugacy class in $G/Z(G)$ is a union of conjugacy classes of $G$ but how do I know which of the preimages 'decompose' into conjugacy classes and which preimages don't?
1 Answer
Hint If $y$ is in the center $Z(G)$ of a group, then for all $g \in G$ we have $$g y g^{-1} = g g^{-1} y = y,$$ and so the conjugacy class of $y$ is just the singleton $\{y\}$. How can we modify this observation to say something about the case $y \not\in Z(G)$?
Additional hint Conversely, if $y \not \in Z(G)$, there some $g \in G$ such that $g y \neq y g$, and hence such that $g y g^{-1} \neq y$, and in particular the conjugacy class of $y$ is not a singleton.
(This linear of reasoning is sufficient, by the way, to handle the case $|G| = 8$, but not the general case.)
• Thank you, this is really easy. I already made this observation when guessing the conjugacy classes of $G$ as above, but somehow I did not realize that it suffices :). Unfortunately this strategy does not work for classes with $\geq 3$ elements. If I understand you correctly, this is in general difficult, is it? – user8463524 Dec 5 '15 at 8:36
• It's easy in the sense that there's a naive algorithm of complexity $O(|G|^2)$ to identify the conjugacy classes (and probably one can improve on this some), but perhaps it suffices to say that this sort of observation is generally not enough to determine the conjugacy classes a priori. – Travis Willse Dec 5 '15 at 21:25 | 711 | 2,410 | {"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} | 3.359375 | 3 | CC-MAIN-2019-43 | latest | en | 0.850281 |
http://oeis.org/A285102 | 1,596,517,402,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735860.28/warc/CC-MAIN-20200804043709-20200804073709-00253.warc.gz | 70,487,284 | 4,656 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A285102 a(n) = A007913(A285101(n)). 6
2, 6, 210, 72930, 620310, 278995269860970, 12849025509071310, 492608110538467706074890, 1342951001046021018427857601026746070, 37793589449865555275592120894959094883390892772270, 728982633030274864467458719371654181886452163442582606072870, 28339554655955912942523491885490197708224606885407444005070 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 LINKS FORMULA a(0) = 2, for n > 0, a(n) = lcm(a(n-1),A242378(n,a(n-1))) / gcd(a(n-1),A242378(n,a(n-1))). a(n) = A007913(A285101(n)). Other identities. For all n >= 0: A001221(a(n)) = A001222(a(n)) = A285103(n). A048675(a(n)) = A068052(n). PROG (PARI) A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); }; A242378(k, n) = { while(k>0, n = A003961(n); k = k-1); n; }; A285102(n) = { if(0==n, 2, lcm(A285102(n-1), A242378(n, A285102(n-1)))/gcd(A285102(n-1), A242378(n, A285102(n-1)))); }; (Scheme) (definec (A285102 n) (if (zero? n) 2 (/ (lcm (A285102 (- n 1)) (A242378bi n (A285102 (- n 1)))) (gcd (A285102 (- n 1)) (A242378bi n (A285102 (- n 1))))))) (Python) from sympy import factorint, prime, primepi from sympy.ntheory.factor_ import core from operator import mul def a003961(n): f=factorint(n) return 1 if n==1 else reduce(mul, [prime(primepi(i) + 1)**f[i] for i in f]) def a242378(k, n): while k>0: n=a003961(n) k-=1 return n l=[2] for n in range(1, 12): x=l[n - 1] l+=[x*a242378(n, x), ] print map(core, l) # Indranil Ghosh, Jun 27 2017 CROSSREFS Cf. A003961, A007913, A048675, A068052, A242378, A285101, A285103. Sequence in context: A156517 A333944 A091439 * A285101 A176782 A013083 Adjacent sequences: A285099 A285100 A285101 * A285103 A285104 A285105 KEYWORD nonn AUTHOR Antti Karttunen, Apr 15 2017 STATUS approved
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Last modified August 4 00:59 EDT 2020. Contains 336201 sequences. (Running on oeis4.) | 870 | 2,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.363786 |
https://illinoisbay.com/australian-capital-territory/differential-equations-lecture-notes-pdf.php | 1,585,594,810,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497301.29/warc/CC-MAIN-20200330181842-20200330211842-00021.warc.gz | 537,115,527 | 8,320 | # Differential equations lecture notes pdf
## Differential equations lecture notes mathematics 2.
Most differential equations, cannot be solved explicitly. one aim of this course is to develop methods which al- low information on the behaviour of solutions anyway..
Differential equations lecture notes mathematics 2.
Lecture Notes on Differential Equations & Differential
Ma1506 lecture notes chapter 1 differential equations. Home > courses > mathematics > differential equations > lecture notes. lecture notes below are the lecture notes for every lecture session. some lecture sessions also have supplementary files called "muddy card responses." students pick up half pages of scrap paper when they come into the classroom, jot down on them what they found to be the most confusing point in the day's lecture or ␦. Ma1506 lecture notes chapter 1 differential equations 1.1 introduction a di﬐erential equation is an equation that contains one or more derivatives of a di﬐eren-.
Nine lectures on exterior diп¬ђerential systems informal notes for a series of lectures delivered 12вђ“23 july 1999 at the graduate sum-mer workshop on exterior diп¬ђerential systems at the mathematical sciences research in this lecture notes we are concerned with evolution of plane curves satisfying a geometric equation v = оі(k,x,оѕ) where v is the normal velocity of an evolving family of planar closed curves.
Types of partial diп¬ђerential equations. before reading these notes, students should understand how to solve the simplest ordinary diп¬ђerential equations, such as the equation of exponential growth dy/dx = ky and the equation of simple harmonic motion d2y/dx2 + п‰y = 0, and how these equations arise in modeling population growth and the motion of a weight attached to the ceiling by means of applied linear algebra and differential equations lecture notes for math 2350 jeffrey r. chasnov m m k k k x 1 x 2 the hong kong university of science and technology
Differential equations course notes, lecture 2: types of differential equations, solving separable odes.5 read more (there is a wikipedia page on di erential forms, for example, which is a good 15-424:foundations of cyber-physical systems lecture notes on differential equations & differential invariants andre platzerвґ carnegie mellon university
Notes we will provide examples of analysis for each of these types of equations. added to the complexity of the eld of the pdes is the fact that many problems can be of mixed type. l10.2 differential equations & differential invariants todayвђ™s lecture reinvestigates differential equations from a more fundamental per-spective, which will lead to a way of proving properties of differential equations with-
Ma1506 lecture notes chapter 1 differential equations 1.1 introduction a di﬐erential equation is an equation that contains one or more derivatives of a di﬐eren- lecture notes for math 251:introduction to ordinary and partial differential equations 1wen shenspring 20131these notes are provided to students as a supplement to the textbook.
Lecture notes math.ntua.gr. Types of partial diп¬ђerential equations. before reading these notes, students should understand how to solve the simplest ordinary diп¬ђerential equations, such as the equation of exponential growth dy/dx = ky and the equation of simple harmonic motion d2y/dx2 + п‰y = 0, and how these equations arise in modeling population growth and the motion of a weight attached to the ceiling by means of. 15-424:foundations of cyber-physical systems lecture notes on differential equations & domains andre platzerвґ carnegie mellon university lecture 2.
...Lecture note by s.m alay-e-abbas: partial differential equations crank-nicolson implicit method: this is one class of methods called the implicit methods. the basic difference between the simple implicit method and this method is that the special discretization is taken both at the time t and t + ∆t and mean of both is used to solve the pde..Lecture notes on elliptic partial di↵erential equations luigi ambrosio ⇤ contents 1 some basic facts concerning sobolev spaces 3 2 variational formulation of some....
Lecture notes partial differential equations scribd. Nine lectures on exterior di﬐erential systems informal notes for a series of lectures delivered 12␓23 july 1999 at the graduate sum-mer workshop on exterior di﬐erential systems at the mathematical sciences research. 15-424:foundations of cyber-physical systems lecture notes on differential equations & domains andre platzerⴠcarnegie mellon university lecture 2.
Notes on partial di erential equations pomona college. Moreover, the lecture notes undoubtedly contain typos and misrepresentations that will hopefully be corrected in lectures. note: because html is not very good for mathematical typesetting, material will typically be provided as pdf files.. Lecture notes for math 251:introduction to ordinary and partial differential equations 1wen shenspring 20131these notes are provided to students as a supplement to the textbook..
Traditional Islamic Approaches To Public International Law.pdf - Free download Ebook, Handbook, Textbook, User Guide PDF files on the internet quickly and easily. Public international law pdf free download Download public international law or read online here in PDF or EPUB. Please click button to get public international law book now. All books are in clear copy here, and …
In this lecture notes we are concerned with evolution of plane curves satisfying a geometric equation v = оі(k,x,оѕ) where v is the normal velocity of an evolving family of planar closed curves. moreover, the lecture notes undoubtedly contain typos and misrepresentations that will hopefully be corrected in lectures. note: because html is not very good for mathematical typesetting, material will typically be provided as pdf files.
Lecture notes on elliptic partial di↵erential equations luigi ambrosio ⇤ contents 1 some basic facts concerning sobolev spaces 3 2 variational formulation of some ⻠mit opencourseware ⻠mathematics ⻠differential equations, spring 2004 lecture notes these lecture notes are pdf renderings of simple text files that were provided to the students containing nearly verbatim transcripts of professor miller's lectures.
В» mit opencourseware в» mathematics в» differential equations, spring 2004 lecture notes these lecture notes are pdf renderings of simple text files that were provided to the students containing nearly verbatim transcripts of professor miller's lectures. partial differential equationsвђќ by yehuda pinchover and jacod rubinstein. (2) the two references (2) the two references that are added at the end of the notes.
In this lecture notes we are concerned with evolution of plane curves satisfying a geometric equation v = оі(k,x,оѕ) where v is the normal velocity of an evolving family of planar closed curves. home > courses > mathematics > differential equations > lecture notes. lecture notes below are the lecture notes for every lecture session. some lecture sessions also have supplementary files called "muddy card responses." students pick up half pages of scrap paper when they come into the classroom, jot down on them what they found to be the most confusing point in the day's lecture or вђ¦
Lecture note by s.m alay-e-abbas: partial differential equations crank-nicolson implicit method: this is one class of methods called the implicit methods. the basic difference between the simple implicit method and this method is that the special discretization is taken both at the time t and t + ∆t and mean of both is used to solve the pde. applied linear algebra and differential equations lecture notes for math 2350 jeffrey r. chasnov m m k k k x 1 x 2 the hong kong university of science and technology
Differential equations lecture notes these notes are intended to supplement sections 6.1 and 4.1 from nagle, sa , and snider. they provide some background and stronger connections to linear algebra which are missing from the home > courses > mathematics > differential equations > lecture notes. lecture notes below are the lecture notes for every lecture session. some lecture sessions also have supplementary files called "muddy card responses." students pick up half pages of scrap paper when they come into the classroom, jot down on them what they found to be the most confusing point in the day's lecture or вђ¦ | 1,791 | 8,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-16 | latest | en | 0.899177 |
https://alphabetaprep.com/article/depreciation-and-amortization | 1,603,740,527,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107891624.95/warc/CC-MAIN-20201026175019-20201026205019-00491.warc.gz | 207,312,381 | 6,160 | Preparing for CFA?
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Depreciation and amortization of long-lived assets
IFRS allows a company to carry its fixed assets either at cost (under the cost model) or revalued amount (under the revaluation model), but US GAAP requires application of the cost model.
Under the cost model, PPE is carried at historical cost less accumulated depreciation and accumulated impairment losses, and intangible assets with finite useful lives are carried at cost less accumulated amortization and accumulated impairments.
The amount at which a long-lived asset is reported on the balance sheet is called its carrying amount (also called carrying value or book value).
Depreciation methods
Depreciation is the process through which the cost of a tangible fixed asset is charged to the income statement. Popular depreciation methods include the straight-line method, accelerated method, and units of production method.
Straight-line method
Under the straight-line method, equal depreciation expense is charged in each period. Depreciation under the straight-line method can be calculated as follows:
$$Straight‐line\ Depreciation=\frac{Depreciable\ Amount}{Useful Life}=\frac{Cost - Salvage\ Value}{Useful\ Life}$$
Accelerated depreciation methods
Under the accelerated methods, such as the declining balance method, higher depreciation expense is charged in earlier periods. Depreciation expense in a period under an accelerated depreciation is calculated by applying the depreciation rate to the opening carrying amount (not to the depreciable amount), and depreciation ceases when carrying amount equals the salvage value.
Units of production depreciation method
Under the units of production method, depreciation expense in a period equals depreciable amount (which equals cost less salvage value) multiplied by the units produced during the period (U) divided by total productive capacity (TU) of the asset.
$$Units\ of\ Production\ Depreciation=\frac{(Cost - Salvage Value)}×\frac{U}{TU}$$
Tax accounting implication of depreciation methods
When there is a difference between the depreciation method used for financial reporting and depreciation method allowed under the tax code, it results in temporary differences that may require recognition of deferred tax assets or liabilities.
Component depreciation
Charging depreciation requires estimating useful life and residual value. Longer useful life and higher residual value would reduce depreciation expense. IFRS requires companies to depreciate each component of an asset separately, and US GAAP allows (but does not require) it.
Depreciation expense of assets used in production is capitalized as part of inventories and charged to the income statement as cost of sales, and depreciation expense of non-production assets is charged to the income statement directly under operating (or selling, general and admin expenses).
Depreciation vs amortization
Amortization is the process through which the cost of intangible assets with a finite useful life, such as copyrights, patents, etc. is charged to the income statement. It is calculated using acceptable depreciation methods. Intangible assets with indefinite useful life are not amortized. Just like depreciation, amortization involves estimating useful life, residual value, etc. which is driven by legal, contractual, regulatory and economic factors.
by Obaidullah Jan, ACA, CFA on Mon Mar 09 2020 | 665 | 3,440 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-45 | latest | en | 0.901861 |
https://physics.stackexchange.com/questions/195548/in-terms-of-the-ampere-maxwell-law-why-is-vec-e-0-in-a-wire-of-a-capacitor | 1,721,904,309,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00650.warc.gz | 396,013,151 | 44,489 | # In terms of the Ampere-Maxwell law, why is $\vec {E}=0$ in a wire of a capacitor circuit?
I'm currently studying from "Introduction to Electromagnetics" by D.J. Griffiths. In the book the significance of the displacement current term is explained by looking a non-steady capacitor circuit (shown below).
Using the integral equation and balloon-shaped surface, it is said that $I_{enc}=0$ but $\int \partial\vec{E}/\partial t \cdot d\vec{a}=I/\epsilon_0$. This statement makes sense to me - there is no physical current flowing between the plates but there is a changing electric flux through the balloon surface.
My misunderstanding is with the flat Amperian loop surface where it is mentioned that $\vec{E}=0$ and $I_{enc}=I$ in this case. Obviously there is a current flow $I_{enc}$ in the Amperian loop, but why is $\vec{E}=0$? Just looking at the equations directly, I would have thought there would be a combination of both current terms (including displacement current due to changing electric field in the wire) in a capacitor charging/discharging situation.
I've attempted to seek other explanations on this from other texts, this forum and elsewhere in this stackexchange site, but this has made me more confused. Clarification on this would be appreciated.
Differential form of Ampere's-Maxwell's equation: $$\nabla \times \vec{B}=\mu_0\vec{J}+\mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}$$
Integral form of Ampere's-Maxwell's equation: $$\oint\vec{B}\cdot d\vec{l} = \mu_0I_{enc}+\mu_0\epsilon_0 \int \frac{\partial \vec{E}}{\partial t} \cdot d\vec{a}$$
The derivation assumes the wire is a perfect conductor, and also that it is negligibly thin. If it had some resistivity, then you're right, there would be an electric field in the wire, but even in that case the electric flux $\int \vec{E}\cdot\text{d}\vec{a}$ would be negligible, and so would its time derivative.
• What about electric field on the flat surface outside the wire? The capacitor produces some electric field outside the capacitor since there is positive charge one plate and negative charges on the other plate. Commented Dec 7, 2020 at 20:50
• @JánLalinský If the plates are very large and very close together, the electric field anywhere outside the inward-facing surfaces of the plates is negligible (especially if you stay close to the wire and away from the edges of the plates). The field from the negative plate cancels with the field from the positive plate there (at least, to the extent that the approximations are good).
– pwf
Commented Dec 8, 2020 at 0:39
There is never actually an electric field in a conductor in the electrostatic sense. An E field is always generated perpendicular to a charged surface (the wire). For any wire carrying current, the electric field tends to radiate outward from the wire. The magnetic field will be circulating around the wire such that the Poynting vector, $\vec S = \vec E \times \vec H$ , is pointing in the direction of current flow, which is also the direction of power transfer. So for your flat Amperian loop, the E field is parallel to the radius of the loop, so there is no net electric flux.
I believe if you were operating at high enough frequencies so that there would be a non-negligible E field inside the conductor, this circuit model really wouldn't be applicable anyway.
• There would be no radial electric field if the wire were neutral, i.e. if the current were composed of charges of opposite sign moving at different speeds. This is usually the case in circuits; electrons flow, but their charge is balanced by stationary positive ions in the material.
– pwf
Commented Jul 23, 2015 at 18:04
Suppose a capacitor with capacitance $$C$$ and $$d$$ of separation of plates, with the wires being connected by a resistor $$R$$ with length $$L$$ and cross section $$S$$.
Before closing the circuit, there is a voltage $$V$$ in the capacitor.
After the circuit is closed the voltage falls exponentially:$$V = V_0e^{\frac{-t}{RC}} \implies \frac{\partial V}{\partial t} = -\frac{V_0}{RC}e^{\frac{-t}{RC}}$$ As we want the electric field, we divide by the length in each case: When $$t=0$$, we have for the capacitor: $$E = \frac{V_0}{d}\implies \epsilon_0\frac{\partial E}{\partial t} = \epsilon_0\frac{V_0}{dRC}$$
In order to calculate the integral over the plate of the capacitor, it is necessary to multiply by its area. $$A = \frac{Cd}{\epsilon_0} \implies \int_A \epsilon_0\frac{\partial E}{\partial t} = \epsilon_0\frac{V_0}{dRC}*\frac{Cd}{\epsilon_0} = \frac{V_0}{R} = I_0$$ as expected.
Applying the same integral to the resistor, the voltage have to be divided by $$L$$, and the integral is now over its cross section $$S$$. The term that multiplies $$I_0$$ shows up as the 'capacitance' of the resistor. It is expected to be several orders of magnitude below that of the capacitor (comparing the usual lengths and areas), and this term can reasonably be disregarded.
$$\int_S \epsilon_0\frac{\partial E}{\partial t} = \epsilon_0\frac{V_0S}{LRC} = \frac{C'}{C}I_0$$
For the flat Amperian Loop,
The current flowing through the wire that pierces the surface of the loop) is I. However, there is no field piercing the surface of the loop.
Now, why is there is no field piercing the loop:-
1. Of course the field between the plates of the capacitor no way pierces the surface of the loop
2. "Isn't there a field inside the wire, which is piercing the loop surface?" The answer is no. Note that you are already accounting for this field by taking the current I into the calculation and hence there is no need to consider again the field inside the wire.
• Are you saying that the field inside the wire is 0 or that it is nonzero but its contribution is equal to the current I? Commented Feb 15, 2021 at 16:03 | 1,489 | 5,785 | {"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": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-30 | latest | en | 0.919497 |
https://thethoughtnow.com/difference-between-series-and-parallel-circuits/ | 1,716,478,928,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058642.50/warc/CC-MAIN-20240523142446-20240523172446-00552.warc.gz | 506,413,971 | 22,555 | # Difference between Series and Parallel Circuits
When there are two or more electrical devices in a circuit with a power source, there are two main ways we connect them. They can be connected in series or in parallel.
Series circuit A circuit in which two components share a common node and have the same current flowing through them. However, in a parallel circuit, the components share two common nodes. In this article, we will look at other differences between series and parallel connection circuits.
### What is a Series Circuit?
A circuit is said to be connected in series with the same current flowing in all parts of the circuit. In such circuits, the current has only one path. Let’s look at home decorative string lights as an example of a circle series.
It is nothing but a series of dozens of small batteries connected in series. If one bulb flips over, all the bulbs in the row don’t light up.
### What is a Parallel Circuit?
A circuit is said to be parallel if there are multiple paths through which electricity flows. Components that are part of parallel circuits will have a constant voltage at all ends.
### Difference between Series and Parallel Circuits
Series Parallel The same amount of current flows through all the components The current flowing through each component combines to form the current flow through the source. In an electrical circuit, components are arranged in a line In an electrical circuit, components are arranged parallel to each other When resistors are put in a series circuit, the voltage across each resistor is different even though the current flow is the same through all of them. When resistors are put in a parallel circuit, the voltage across each of the resistors is the same. Even the polarities are the same If one component breaks down, the whole circuit will burn out. Other components will function even if one component breaks down, each has its own independent circuit If Vt is the total voltage then it is equal to V1 + V2 +V3 If Vt is the total voltage then it is equal to V1=V2=V3 | 413 | 2,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.951835 |
http://learnersinbloom.blogspot.com/2013/01/homeschool-preschool-week-19-20.html | 1,371,630,239,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708145189/warc/CC-MAIN-20130516124225-00079-ip-10-60-113-184.ec2.internal.warc.gz | 148,111,457 | 24,966 | Sunday, January 13, 2013
Homeschool Preschool - Week 19 & 20
L and H are almost 3.5 years old. After a long holiday break, we're getting back into our Homeschool Preschool routine. Some days are definitely better than others. Even on the more frustrating days where I feel that the kids watched too much TV and didn't learn anything, looking back I still see that there was plenty of learning going on. The girls always see me taking photos of our activities, so they've been asking for their own cameras to take pictures as well. Here's what we've been up to...
Math
After breakfast, I like to bring out various math manipulatives. Sometimes things that are as simple as Q-tips turn out to be a lot of fun.
I like to give the girls little logic puzzles to help their mathematical reasoning skills. For example, I'll ask "If you want to make a square, how many q-tips do you think you'll need? Can you make another square now with only 3? If I give you one more q-tip, can you make you square into 2 rectangles?" I remember playing games like these with toothpicks or matches as a kid, so I have a lot of fun with them too.
One of the puzzles we've been working on is to count the number of triangles/rectangles in shapes like the ones above - it took a while for the kids to learn to see the larger shapes that are composed of the smaller ones..
We also use the various manipulatives to reinforce the concepts we're covering in RightStart Math A (tallying and parallel lines shown above). We're almost half-way through Level A, but we've been going back a lot to review many of the concepts. There are definitely a lot of logic skills that I think the girls have to grow into. For example, they easily recognize that 5+1=6 and 5+2=7 even when they see the problem in different forms, but when I hold up five fingers on one hand and 3 fingers on another they forget that it is 8. That's fine, because I don't think memorization is the key to learning, but when they don't know the answer, instead of trying to figure it out they guess totally illogical answers (call out random numbers). I try to show them some strategies like if they know that 5 and 2 is 7, 5 + 3 must be more, since we only add one finger, it's probably the number that comes after 8. I don't want them to just memorize the process for figuring it out, but rather formulate their own problem-solving strategies, but maybe they're just a bit young for some of those skills.
We're still working a lot on the tens, and engaging in a wide variety of different math games and activities.
Although I was doing fine teaching the girls to read using my own homemade activities, I thought they could benefit from a more structured program to help with their fluency and to make sure they go through all the reading/spelling rules more systematically. We just completed our first week of All About Reading Level 1, which the kids love. The hands-on activities are giving them a lot of extra practice and boosting confidence in their reading abilities. I like the fact that I don't have to do a lot of preparation before the lessons so we can have something to work on whenever the kids ask for it (and yes, they really do ask to 'read flashcards'.. a lot!)
Free educational games on the iPad are always a big hit, as well.
For read-alouds, in the past two weeks we've already finished The Gorilla Who Wanted to Grow Up, and abridged version of The Wizard of Oz (kids and I absolutely loved it and they reference it often in their imaginary play) and all four Catwings books (they were shorter than I expected to we read one a day for 4 days in a row and both girls adored them). I haven't decided what is up next - maybe The Wishing Chair Again since the girls enjoyed the first one so much.
Everything Else
The girls got real ceramic tea sets for Christmas, so we've been enjoying afternoon tea-time. I incorporated tea with a little lesson on England, and they always like to say that they are going to London to have tea with the Queen.
We've been spending a lot of time playing with some of the new games the girls got for Christmas. Here E is playing with her Imaginets set. She's good at both following the patterns on the cards and doing creative play - both are equally important for learning.
Working on concentration and fine motor skills.
Matching instruments (from Toob) with flashcards (\$1 Target)
Coin matching. I meant to start teaching them about money this week, but I didn't get around to it.
The science experiments this past week were totally child-guided. For example, I gave them water, measuring cups, droppers, etc, and they invented their own experiments - seeing which containers hold more water, dropping objects in to see if they sink or float, adding soap, etc...
We've been doing lots of art, music, dramatic play, dance, and gymnastics, too - I just haven't been good about capturing photos of those.
My favorite part of the past two weeks has been how much my girls are taking the initiative to ask for certain learning activities and also creating their own learning experiences. A lot of times I just sit back and watch what they come up with all by themselves and it's so fun and creative. | 1,185 | 5,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | longest | en | 0.972016 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/1850/2/a/bd/ | 1,653,205,935,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545090.44/warc/CC-MAIN-20220522063657-20220522093657-00170.warc.gz | 990,273,457 | 58,545 | # Properties
Label 1850.2.a.bd Level $1850$ Weight $2$ Character orbit 1850.a Self dual yes Analytic conductor $14.772$ Analytic rank $0$ Dimension $5$ CM no Inner twists $1$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$1850 = 2 \cdot 5^{2} \cdot 37$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 1850.a (trivial)
## Newform invariants
Self dual: yes Analytic conductor: $$14.7723243739$$ Analytic rank: $$0$$ Dimension: $$5$$ Coefficient field: 5.5.1791440.1 Defining polynomial: $$x^{5} - 9x^{3} + 13x - 4$$ x^5 - 9*x^3 + 13*x - 4 Coefficient ring: $$\Z[a_1, \ldots, a_{7}]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 370) Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\beta_2,\beta_3,\beta_4$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q - q^{2} + \beta_1 q^{3} + q^{4} - \beta_1 q^{6} + ( - \beta_{4} + \beta_{3} + \beta_{2} + \beta_1) q^{7} - q^{8} + (\beta_{2} + 1) q^{9}+O(q^{10})$$ q - q^2 + b1 * q^3 + q^4 - b1 * q^6 + (-b4 + b3 + b2 + b1) * q^7 - q^8 + (b2 + 1) * q^9 $$q - q^{2} + \beta_1 q^{3} + q^{4} - \beta_1 q^{6} + ( - \beta_{4} + \beta_{3} + \beta_{2} + \beta_1) q^{7} - q^{8} + (\beta_{2} + 1) q^{9} + ( - \beta_{4} + \beta_{3} - \beta_1) q^{11} + \beta_1 q^{12} + (\beta_{4} + \beta_{3} + \beta_1 + 1) q^{13} + (\beta_{4} - \beta_{3} - \beta_{2} - \beta_1) q^{14} + q^{16} + ( - \beta_{4} + \beta_{3} + \beta_{2} - \beta_1 - 2) q^{17} + ( - \beta_{2} - 1) q^{18} + 2 \beta_{3} q^{19} + ( - 2 \beta_{3} + 2 \beta_1 + 4) q^{21} + (\beta_{4} - \beta_{3} + \beta_1) q^{22} + ( - \beta_{4} - \beta_{3} + \beta_1 - 1) q^{23} - \beta_1 q^{24} + ( - \beta_{4} - \beta_{3} - \beta_1 - 1) q^{26} + (\beta_{4} - 2 \beta_{3} + 1) q^{27} + ( - \beta_{4} + \beta_{3} + \beta_{2} + \beta_1) q^{28} + ( - \beta_{4} + \beta_{3} + \beta_{2} + 2 \beta_1 + 2) q^{29} + (2 \beta_{4} + \beta_{3} + \beta_{2} + 5) q^{31} - q^{32} + ( - \beta_{4} - 2 \beta_{2} - 5) q^{33} + (\beta_{4} - \beta_{3} - \beta_{2} + \beta_1 + 2) q^{34} + (\beta_{2} + 1) q^{36} - q^{37} - 2 \beta_{3} q^{38} + (\beta_{4} - 2 \beta_{3} + \beta_1 + 5) q^{39} + ( - 2 \beta_{4} + 2 \beta_{3} - \beta_{2} + 2 \beta_1 - 3) q^{41} + (2 \beta_{3} - 2 \beta_1 - 4) q^{42} + ( - \beta_{4} - \beta_{3} + \beta_{2} + \beta_1 + 4) q^{43} + ( - \beta_{4} + \beta_{3} - \beta_1) q^{44} + (\beta_{4} + \beta_{3} - \beta_1 + 1) q^{46} + (2 \beta_{4} - 2 \beta_{3} - 2 \beta_{2} - 2) q^{47} + \beta_1 q^{48} + (\beta_{4} - 3 \beta_{3} - \beta_{2} + \beta_1 + 7) q^{49} + ( - 2 \beta_{3} - 2 \beta_{2} - 4) q^{51} + (\beta_{4} + \beta_{3} + \beta_1 + 1) q^{52} + (3 \beta_{4} + \beta_{3} + \beta_{2} - \beta_1 + 2) q^{53} + ( - \beta_{4} + 2 \beta_{3} - 1) q^{54} + (\beta_{4} - \beta_{3} - \beta_{2} - \beta_1) q^{56} + ( - 2 \beta_{3} - 2 \beta_{2}) q^{57} + (\beta_{4} - \beta_{3} - \beta_{2} - 2 \beta_1 - 2) q^{58} + ( - 2 \beta_{3} - 2 \beta_{2} + 4) q^{59} + ( - 2 \beta_{4} + \beta_{3} - \beta_{2} + 2 \beta_1 - 3) q^{61} + ( - 2 \beta_{4} - \beta_{3} - \beta_{2} - 5) q^{62} + (3 \beta_{4} - \beta_{3} + \beta_{2} + \beta_1 + 8) q^{63} + q^{64} + (\beta_{4} + 2 \beta_{2} + 5) q^{66} + (\beta_{4} + 2 \beta_{3} + \beta_1 - 3) q^{67} + ( - \beta_{4} + \beta_{3} + \beta_{2} - \beta_1 - 2) q^{68} + ( - \beta_{4} + 2 \beta_{3} + 2 \beta_{2} - \beta_1 + 3) q^{69} + (4 \beta_{4} + 2) q^{71} + ( - \beta_{2} - 1) q^{72} + ( - 3 \beta_{4} + 4 \beta_{3} - \beta_1 - 1) q^{73} + q^{74} + 2 \beta_{3} q^{76} + ( - 3 \beta_{4} + 3 \beta_{3} - \beta_{2} - 3 \beta_1 - 2) q^{77} + ( - \beta_{4} + 2 \beta_{3} - \beta_1 - 5) q^{78} + ( - 2 \beta_{4} + 4 \beta_{3} + \beta_1 + 2) q^{79} + (\beta_{4} + \beta_{3} - \beta_{2} + \beta_1 - 2) q^{81} + (2 \beta_{4} - 2 \beta_{3} + \beta_{2} - 2 \beta_1 + 3) q^{82} + (2 \beta_{4} - 4 \beta_1 - 2) q^{83} + ( - 2 \beta_{3} + 2 \beta_1 + 4) q^{84} + (\beta_{4} + \beta_{3} - \beta_{2} - \beta_1 - 4) q^{86} + ( - 2 \beta_{3} + \beta_{2} + 4 \beta_1 + 8) q^{87} + (\beta_{4} - \beta_{3} + \beta_1) q^{88} + ( - 2 \beta_{3} + 2 \beta_{2} - 2 \beta_1 + 4) q^{89} + ( - 2 \beta_{4} - 6 \beta_{3} + 2 \beta_{2} + 6) q^{91} + ( - \beta_{4} - \beta_{3} + \beta_1 - 1) q^{92} + (3 \beta_{4} - 5 \beta_{3} - \beta_{2} + 7 \beta_1 + 3) q^{93} + ( - 2 \beta_{4} + 2 \beta_{3} + 2 \beta_{2} + 2) q^{94} - \beta_1 q^{96} + (\beta_{4} + \beta_{3} - 3 \beta_{2} - \beta_1 - 2) q^{97} + ( - \beta_{4} + 3 \beta_{3} + \beta_{2} - \beta_1 - 7) q^{98} + (2 \beta_{3} - 6 \beta_1 - 3) q^{99}+O(q^{100})$$ q - q^2 + b1 * q^3 + q^4 - b1 * q^6 + (-b4 + b3 + b2 + b1) * q^7 - q^8 + (b2 + 1) * q^9 + (-b4 + b3 - b1) * q^11 + b1 * q^12 + (b4 + b3 + b1 + 1) * q^13 + (b4 - b3 - b2 - b1) * q^14 + q^16 + (-b4 + b3 + b2 - b1 - 2) * q^17 + (-b2 - 1) * q^18 + 2*b3 * q^19 + (-2*b3 + 2*b1 + 4) * q^21 + (b4 - b3 + b1) * q^22 + (-b4 - b3 + b1 - 1) * q^23 - b1 * q^24 + (-b4 - b3 - b1 - 1) * q^26 + (b4 - 2*b3 + 1) * q^27 + (-b4 + b3 + b2 + b1) * q^28 + (-b4 + b3 + b2 + 2*b1 + 2) * q^29 + (2*b4 + b3 + b2 + 5) * q^31 - q^32 + (-b4 - 2*b2 - 5) * q^33 + (b4 - b3 - b2 + b1 + 2) * q^34 + (b2 + 1) * q^36 - q^37 - 2*b3 * q^38 + (b4 - 2*b3 + b1 + 5) * q^39 + (-2*b4 + 2*b3 - b2 + 2*b1 - 3) * q^41 + (2*b3 - 2*b1 - 4) * q^42 + (-b4 - b3 + b2 + b1 + 4) * q^43 + (-b4 + b3 - b1) * q^44 + (b4 + b3 - b1 + 1) * q^46 + (2*b4 - 2*b3 - 2*b2 - 2) * q^47 + b1 * q^48 + (b4 - 3*b3 - b2 + b1 + 7) * q^49 + (-2*b3 - 2*b2 - 4) * q^51 + (b4 + b3 + b1 + 1) * q^52 + (3*b4 + b3 + b2 - b1 + 2) * q^53 + (-b4 + 2*b3 - 1) * q^54 + (b4 - b3 - b2 - b1) * q^56 + (-2*b3 - 2*b2) * q^57 + (b4 - b3 - b2 - 2*b1 - 2) * q^58 + (-2*b3 - 2*b2 + 4) * q^59 + (-2*b4 + b3 - b2 + 2*b1 - 3) * q^61 + (-2*b4 - b3 - b2 - 5) * q^62 + (3*b4 - b3 + b2 + b1 + 8) * q^63 + q^64 + (b4 + 2*b2 + 5) * q^66 + (b4 + 2*b3 + b1 - 3) * q^67 + (-b4 + b3 + b2 - b1 - 2) * q^68 + (-b4 + 2*b3 + 2*b2 - b1 + 3) * q^69 + (4*b4 + 2) * q^71 + (-b2 - 1) * q^72 + (-3*b4 + 4*b3 - b1 - 1) * q^73 + q^74 + 2*b3 * q^76 + (-3*b4 + 3*b3 - b2 - 3*b1 - 2) * q^77 + (-b4 + 2*b3 - b1 - 5) * q^78 + (-2*b4 + 4*b3 + b1 + 2) * q^79 + (b4 + b3 - b2 + b1 - 2) * q^81 + (2*b4 - 2*b3 + b2 - 2*b1 + 3) * q^82 + (2*b4 - 4*b1 - 2) * q^83 + (-2*b3 + 2*b1 + 4) * q^84 + (b4 + b3 - b2 - b1 - 4) * q^86 + (-2*b3 + b2 + 4*b1 + 8) * q^87 + (b4 - b3 + b1) * q^88 + (-2*b3 + 2*b2 - 2*b1 + 4) * q^89 + (-2*b4 - 6*b3 + 2*b2 + 6) * q^91 + (-b4 - b3 + b1 - 1) * q^92 + (3*b4 - 5*b3 - b2 + 7*b1 + 3) * q^93 + (-2*b4 + 2*b3 + 2*b2 + 2) * q^94 - b1 * q^96 + (b4 + b3 - 3*b2 - b1 - 2) * q^97 + (-b4 + 3*b3 + b2 - b1 - 7) * q^98 + (2*b3 - 6*b1 - 3) * q^99 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$5 q - 5 q^{2} + 5 q^{4} + q^{7} - 5 q^{8} + 3 q^{9}+O(q^{10})$$ 5 * q - 5 * q^2 + 5 * q^4 + q^7 - 5 * q^8 + 3 * q^9 $$5 q - 5 q^{2} + 5 q^{4} + q^{7} - 5 q^{8} + 3 q^{9} + 3 q^{11} + 6 q^{13} - q^{14} + 5 q^{16} - 9 q^{17} - 3 q^{18} + 4 q^{19} + 16 q^{21} - 3 q^{22} - 6 q^{23} - 6 q^{26} + q^{28} + 11 q^{29} + 23 q^{31} - 5 q^{32} - 20 q^{33} + 9 q^{34} + 3 q^{36} - 5 q^{37} - 4 q^{38} + 20 q^{39} - 7 q^{41} - 16 q^{42} + 17 q^{43} + 3 q^{44} + 6 q^{46} - 12 q^{47} + 30 q^{49} - 20 q^{51} + 6 q^{52} + 7 q^{53} - q^{56} - 11 q^{58} + 20 q^{59} - 9 q^{61} - 23 q^{62} + 33 q^{63} + 5 q^{64} + 20 q^{66} - 12 q^{67} - 9 q^{68} + 16 q^{69} + 6 q^{71} - 3 q^{72} + 6 q^{73} + 5 q^{74} + 4 q^{76} + q^{77} - 20 q^{78} + 20 q^{79} - 7 q^{81} + 7 q^{82} - 12 q^{83} + 16 q^{84} - 17 q^{86} + 34 q^{87} - 3 q^{88} + 12 q^{89} + 16 q^{91} - 6 q^{92} + 4 q^{93} + 12 q^{94} - 3 q^{97} - 30 q^{98} - 11 q^{99}+O(q^{100})$$ 5 * q - 5 * q^2 + 5 * q^4 + q^7 - 5 * q^8 + 3 * q^9 + 3 * q^11 + 6 * q^13 - q^14 + 5 * q^16 - 9 * q^17 - 3 * q^18 + 4 * q^19 + 16 * q^21 - 3 * q^22 - 6 * q^23 - 6 * q^26 + q^28 + 11 * q^29 + 23 * q^31 - 5 * q^32 - 20 * q^33 + 9 * q^34 + 3 * q^36 - 5 * q^37 - 4 * q^38 + 20 * q^39 - 7 * q^41 - 16 * q^42 + 17 * q^43 + 3 * q^44 + 6 * q^46 - 12 * q^47 + 30 * q^49 - 20 * q^51 + 6 * q^52 + 7 * q^53 - q^56 - 11 * q^58 + 20 * q^59 - 9 * q^61 - 23 * q^62 + 33 * q^63 + 5 * q^64 + 20 * q^66 - 12 * q^67 - 9 * q^68 + 16 * q^69 + 6 * q^71 - 3 * q^72 + 6 * q^73 + 5 * q^74 + 4 * q^76 + q^77 - 20 * q^78 + 20 * q^79 - 7 * q^81 + 7 * q^82 - 12 * q^83 + 16 * q^84 - 17 * q^86 + 34 * q^87 - 3 * q^88 + 12 * q^89 + 16 * q^91 - 6 * q^92 + 4 * q^93 + 12 * q^94 - 3 * q^97 - 30 * q^98 - 11 * q^99
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{5} - 9x^{3} + 13x - 4$$ :
$$\beta_{1}$$ $$=$$ $$\nu$$ v $$\beta_{2}$$ $$=$$ $$\nu^{2} - 4$$ v^2 - 4 $$\beta_{3}$$ $$=$$ $$( \nu^{4} - \nu^{3} - 8\nu^{2} + 5\nu + 8 ) / 3$$ (v^4 - v^3 - 8*v^2 + 5*v + 8) / 3 $$\beta_{4}$$ $$=$$ $$( 2\nu^{4} + \nu^{3} - 16\nu^{2} - 8\nu + 13 ) / 3$$ (2*v^4 + v^3 - 16*v^2 - 8*v + 13) / 3
$$\nu$$ $$=$$ $$\beta_1$$ b1 $$\nu^{2}$$ $$=$$ $$\beta_{2} + 4$$ b2 + 4 $$\nu^{3}$$ $$=$$ $$\beta_{4} - 2\beta_{3} + 6\beta _1 + 1$$ b4 - 2*b3 + 6*b1 + 1 $$\nu^{4}$$ $$=$$ $$\beta_{4} + \beta_{3} + 8\beta_{2} + \beta _1 + 25$$ b4 + b3 + 8*b2 + b1 + 25
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
−2.62545 −1.53175 0.332924 1.09441 2.72987
−1.00000 −2.62545 1.00000 0 2.62545 1.83227 −1.00000 3.89300 0
1.2 −1.00000 −1.53175 1.00000 0 1.53175 −4.67211 −1.00000 −0.653743 0
1.3 −1.00000 0.332924 1.00000 0 −0.332924 −3.51336 −1.00000 −2.88916 0
1.4 −1.00000 1.09441 1.00000 0 −1.09441 3.20984 −1.00000 −1.80226 0
1.5 −1.00000 2.72987 1.00000 0 −2.72987 4.14336 −1.00000 4.45216 0
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 1.5 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Atkin-Lehner signs
$$p$$ Sign
$$2$$ $$1$$
$$5$$ $$-1$$
$$37$$ $$1$$
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 1850.2.a.bd 5
5.b even 2 1 1850.2.a.be 5
5.c odd 4 2 370.2.b.d 10
15.e even 4 2 3330.2.d.p 10
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
370.2.b.d 10 5.c odd 4 2
1850.2.a.bd 5 1.a even 1 1 trivial
1850.2.a.be 5 5.b even 2 1
3330.2.d.p 10 15.e even 4 2
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(1850))$$:
$$T_{3}^{5} - 9T_{3}^{3} + 13T_{3} - 4$$ T3^5 - 9*T3^3 + 13*T3 - 4 $$T_{7}^{5} - T_{7}^{4} - 32T_{7}^{3} + 44T_{7}^{2} + 240T_{7} - 400$$ T7^5 - T7^4 - 32*T7^3 + 44*T7^2 + 240*T7 - 400
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$(T + 1)^{5}$$
$3$ $$T^{5} - 9 T^{3} + 13 T - 4$$
$5$ $$T^{5}$$
$7$ $$T^{5} - T^{4} - 32 T^{3} + 44 T^{2} + \cdots - 400$$
$11$ $$T^{5} - 3 T^{4} - 23 T^{3} + 71 T^{2} + \cdots - 1$$
$13$ $$T^{5} - 6 T^{4} - 29 T^{3} + 154 T^{2} + \cdots - 8$$
$17$ $$T^{5} + 9 T^{4} - 4 T^{3} - 216 T^{2} + \cdots - 368$$
$19$ $$T^{5} - 4 T^{4} - 40 T^{3} + 176 T^{2} + \cdots - 640$$
$23$ $$T^{5} + 6 T^{4} - 25 T^{3} - 82 T^{2} + \cdots - 288$$
$29$ $$T^{5} - 11 T^{4} - 9 T^{3} + 251 T^{2} + \cdots - 335$$
$31$ $$T^{5} - 23 T^{4} + 107 T^{3} + \cdots - 10511$$
$37$ $$(T + 1)^{5}$$
$41$ $$T^{5} + 7 T^{4} - 119 T^{3} + \cdots + 5821$$
$43$ $$T^{5} - 17 T^{4} + 52 T^{3} + \cdots + 3056$$
$47$ $$T^{5} + 12 T^{4} - 44 T^{3} + \cdots + 8000$$
$53$ $$T^{5} - 7 T^{4} - 164 T^{3} + \cdots - 32816$$
$59$ $$T^{5} - 20 T^{4} + 52 T^{3} + \cdots - 3136$$
$61$ $$T^{5} + 9 T^{4} - 81 T^{3} + \cdots - 2323$$
$67$ $$T^{5} + 12 T^{4} - 29 T^{3} + \cdots - 392$$
$71$ $$T^{5} - 6 T^{4} - 184 T^{3} + \cdots - 13792$$
$73$ $$T^{5} - 6 T^{4} - 197 T^{3} + \cdots - 14726$$
$79$ $$T^{5} - 20 T^{4} - 13 T^{3} + \cdots - 8168$$
$83$ $$T^{5} + 12 T^{4} - 120 T^{3} + \cdots - 2752$$
$89$ $$T^{5} - 12 T^{4} - 144 T^{3} + \cdots - 55552$$
$97$ $$T^{5} + 3 T^{4} - 244 T^{3} + \cdots + 1008$$ | 6,700 | 12,118 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-21 | latest | en | 0.371852 |
https://softwarekeep.com/ar-qa/blogs/how-to/how-to-use-vlookup-examples-and-formula | 1,721,690,229,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517927.60/warc/CC-MAIN-20240722220957-20240723010957-00282.warc.gz | 445,971,859 | 83,667 | # Master Excel VLOOKUP: A Step-by-Step Guide to Excel's Most Useful Function
Learn how to use the VLOOKUP function in Excel with our step-by-step guide, examples, and formula. Improve your data analysis skills today!
If you're an Excel user, chances are you've heard of the Excel VLOOKUP function. This powerful tool allows you to search for specific data within a large table or range and return related information.
This article will walk you through everything you need about Excel VLOOKUP, including the formula, examples, and step-by-step instructions.
Why should you trust us? Our team of Excel experts has years of experience using and teaching VLOOKUP to users of all levels.
Whether you're a beginner or an advanced user, we've got you covered. By the end of this article, you'll be able to use VLOOKUP with confidence and take your data analysis skills to the next level.
Ready to get started? Let's dive into the world of VLOOKUP and see how this powerful function can transform how you work with Excel.
## What is VLOOKUP?
The VLOOKUP function is a tool in Excel that helps you find information in a table or data set.
Imagine you have a big table with a lot of information and want to find something specific, like the price of foods. You can use the VLOOKUP function to do that!
Formula:
VLOOKUP(lookup_value, table_array, col index num, [range_lookup])
### How to get started
To start using VLOOKUP:
1. Start by clicking the cell where you want to see the results and enter an equal (=) sign, followed by VLOOKUP and parentheses.
2. Inside the parentheses, you'll need to enter a set of arguments. An argument is just a piece of data that the function needs to run.
3. Enter the value you know, such as a product name, as the first argument.
4. Type a comma, then enter the range of cells containing the data you want to search. For example, the block of data that contains your product names and prices.
5. Type another comma and then enter the column index number of the data you want to see. For instance, if you want to see the price of a product, and the price data is in the third column from the left, enter the number 3.
6. Add another comma and enter FALSE for an exact match between the data you know and the data you want to see. If you enter TRUE, you'll get an approximate match.
7. Press Enter to see the results.
8. If you don't enter a value in the left-hand column, you'll get an error message.
9. Once you enter a value in the left-hand column, you'll get the corresponding value in the third column to the right.
### VLOOKUP in financial modeling and financial analysis
VLOOKUP is useful in everyday tasks and finance-related activities such as financial modeling and analysis. Financial models help companies plan their finances and make projections for the future.
VLOOKUP formulas are used in financial modeling to make these models more dynamic and include various scenarios.
For example:
A sales manager may want to analyze Cost per Conversion to determine which conversions are selling best. By using VLOOKUP, they can easily match the Conversions with the Cost per Conversion and extract data.
This information can then be used to make informed decisions on how to allocate resources, adjust pricing strategies, or launch new products in specific regions.
VLOOKUP can also be used to identify top-performing salespeople by matching their names with sales data and extracting key metrics such as total sales, number of deals closed, and average deal size.
Overall, VLOOKUP can be a powerful tool for sales analysis, providing valuable insights into customer behavior and sales performance that can help drive business growth.
### Things to remember about the VLOOKUP Function
The VLOOKUP function is a useful tool in Excel that helps you find information in a table or dataset. Here are some things to remember when using this function:
• If you don't specify whether you want an exact match, VLOOKUP will allow for non-exact matches but will use exact matches if they exist.
• VLOOKUP only looks to the right of the first column in the table, so it's limited in that sense.
• If the column you're looking at has duplicate values, VLOOKUP will only match the first one.
• The function doesn't care about capitalization or lowercase letters, so "apple" and "APPLE" are treated the same.
• Inserting or deleting a column in the table may break the VLOOKUP formula because the column index values don't update automatically.
• You can use wildcards like * and ? in the lookup value to help find matches.
• If numbers are entered as text in the table, you may get an error if the lookup value isn't in text form.
• You might see different error messages when using VLOOKUP, including #N/A!, #REF!, and #VALUE!.
Remembering these tips can help you use the VLOOKUP function more effectively and avoid common mistakes.
### What is the most common use for VLOOKUP?
The VLOOKUP function in Excel is often used when you want to find a specific piece of information in a big table or list of data.
This function is great when you need to search for a particular item and then get some information about that item from another column in the same row.
Suppose we want to know how much the customer paid for the product they bought. We could use the VLOOKUP function to find the Price Per Unit for each order.
### How to troubleshoot errors in VLOOKUP function?
Sometimes when using the VLOOKUP function, you might encounter errors that prevent it from returning the desired result. If you are confident that the data you are searching for is present in your spreadsheet, then the issue might be due to hidden spaces or non-printing characters in the referenced cells.
Therefore, it is important to verify that your cells do not have any of these issues and that they follow the correct data type.
To troubleshoot VLOOKUP errors, you can:
1. Check the format of the lookup value to ensure that it matches the format of the value in the first column of the lookup table.
2. Double-check the range_lookup argument to ensure that it is either TRUE or FALSE, and not a different value.
3. Use the IFERROR function to catch errors and return a custom message instead of an error code.
By taking these steps, you can increase the accuracy of your VLOOKUP function and ensure that you are getting the desired results.
### When should you not use VLOOKUP?
Although VLOOKUP is a very useful function, there are certain situations when it should not be used.
For example:
If you plan to add or delete columns in your data, VLOOKUP may return incorrect results as the column number reference would now be incorrect. To make the column number dynamic, additional formulas, and functions must be used, which can become complicated.
In this case, using the INDEX/MATCH combination function may be better.
The INDEX/MATCH function combination can handle dynamic column ranges without the need to modify formulas every time new data is added or removed. It is also generally faster than VLOOKUP when searching large datasets.
Therefore, it's essential to understand the limitations of VLOOKUP and know when to use INDEX/MATCH combination or other Excel functions, depending on the specific needs of the task at hand.
## Final Thoughts
In conclusion, the VLOOKUP function is a powerful tool in Excel that allows users to quickly search for and retrieve data from a table.
Users can streamline their work by following the step-by-step process for using the VLOOKUP function and save time and effort in their data processing tasks.
With a basic understanding of the VLOOKUP function and its capabilities, even beginners can make the most of Excel's powerful data management tools. | 1,630 | 7,734 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.880422 |
https://math.stackexchange.com/questions/3143268/in-how-many-different-ways-can-i-get-from-a-to-b | 1,563,331,066,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525009.36/warc/CC-MAIN-20190717021428-20190717043428-00044.warc.gz | 464,424,447 | 36,057 | # In how many different ways can I get from A to B?
I can only use horizontal and vertical arrows, like in the picture, and I must get from $$A$$ to $$B$$ using only $$4$$ horizontal arrows and $$3$$ vertical arrows. (One arrow counts as the line connecting only two dots)
My answer is $$4! \cdot 3!$$, however I don't know if I am right
One of the key observations with directed graph problems in this fashion is that it will always take the same number of moves in either of the two directions, regardless of the path.
For example:
$$\text{up, up, up, right, right, right, right}$$
gets you there, as does
$$\text{right, right, up, up, up, right, right}$$ $$\text{up, right, right, up, up, right, right}$$ $$\text{right, right, up, right, up, up, right}$$
or, really, any permutation of these sequences.
This actually allows us to reframe this in terms of a different common combinatorics problem that might be more familiar to you. Let $$U$$ represent up movements and $$R$$ movements to the right. Then we know that any permutation of the letters $$UUURRRR$$ defines a valid path if we do the actions in order sequentially.
So we ask: in this "word" $$UUURRRR$$ (three $$U$$'s, four $$R$$'s), how many distinct arrangements are there?
This is a problem in which the multinomial theorem applies. To see it more directly:
• We know we have $$7$$ letters total. So if they were all distinct, then we would have $$7!$$ permutations, right?
• But some are identical! For any permutation of the $$U$$'s in a given word, the word is still the same: that is, for each of our $$7!$$ permutations, we have duds. That number of duds is equal to the number of arrangements of $$U$$'s we can have, which is $$3!$$. We will need to divide by $$3!$$ to handle that.
• Similarly, we will need to divide by $$4!$$ because of the $$R$$'s.
$$\frac{7!}{3!\cdot 4!} \;\;\; \text{sometimes represented by} \;\;\; \binom{7}{3,4}$$
You can convert each allowable path into a "word" of $$H$$s and $$V$$s, for horizontal and vertical steps respectively. Your example corresponds to the word $$VHVHHVH$$. The word must consist of three $$V$$s and four $$H$$s. How many ways are there to choose three letters in a seven-letter word to be $$V$$s? | 611 | 2,234 | {"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": 31, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-30 | latest | en | 0.938127 |
https://www.plati.com/itm/setting-the-course-quot-general-theory-of-statistics-quot-threads-5-7/1478354?lang=en-US | 1,532,229,604,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593004.92/warc/CC-MAIN-20180722022235-20180722042235-00006.warc.gz | 967,055,499 | 12,179 | # setting the course "General Theory of Statistics" Threads 5, 7
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Content: statistika_3.zip (35,01 kB)
# Description
Task 1.
From the party in an amount of 1 million. Pcs. small-caliber ammunition taken by random sampling to determine the range of a 1000 pieces. The test results are shown in the table below:
With a probability of 0.954, determine the average distance of the fight in the sample, sampling error, the possible limits of the medium-range battlefield for the entire party cartridges. How to change the error in determining the medium-range, if the confidence level is equal to 0.99?
Task 2.
Internal auditors checked the company's invoice processing system revenues. They took a random sample of size n1 = 50 complete accounts check them out. Four of them were faulty. The company has taken measures to reduce the number of billing errors. After a while the second was conducted random sample of size n2 = 60 completed accounts and three of them found defective. At the significance level of 0.05 find out whether you can assume that the error began to do less?
Note. In fact, the task should be to test the hypothesis of equality of the two sample means.
Task 3.
What firms need to check tax inspection area to the proportion of firms error, delayed payment of taxes, does not exceed 5%? According to the previous sampling fraction of these firms was 32%. Take confidence probability equal to 0.954.
Task 4.
The following data on the implementation of fruit retailers district. (Table)
Calculate the composite indexes:
a) turnover;
b) prices;
c) the physical volume of sales.
Determine the absolute value of the savings customers by lowering prices.
(Note that the table shows the turnover, not physical volume of sales.)
# Additional information
the format of Microsoft Word, 8 p., packed with WinZip
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In order to counter copyright infringement and property rights, we ask you to immediately inform us at support@plati.market the fact of such violations and to provide us with reliable information confirming your copyrights or rights of ownership. Email must contain your contact information (name, phone number, etc.) | 514 | 2,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2018-30 | latest | en | 0.898322 |
https://www.sanfoundry.com/engineering-hydrology-questions-answers-gumbels-method-set-2/ | 1,713,822,539,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00566.warc.gz | 852,923,998 | 20,537 | Engineering Hydrology Questions and Answers – Gumbel’s Method – Set 2
This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Gumbel’s Method – Set 2”.
1. In Gumbel’s method, the reduced mean and reduced standard deviation are functions of which of the following?
a) Return period
b) Sample size
c) Flood magnitude
d) Return period and sample size
Explanation: The reduced mean ($$\overline{y_n}$$) and reduced standard deviation (Sn) are important terms in Gumbel’s analysis and are part of the frequency factor. Both these terms depend only the sample size and they are standard table that give these values. For infinite sample size, ($$\overline{y_n}$$) = 0.577 and Sn = 1.2825.
2. Given below are the steps involved in Gumbel’s analysis to find the flood magnitude xT corresponding to a return period T, based on an annual flood series. Identify the correct sequence of steps.
A: Determine mean and standard deviation of the data
B: Determine the reduced variate
C: Determine the flood magnitude of return period T
D: Determine sample size, reduced mean and reduced standard deviation
E: Determine frequency factor
a) A→ B→ E→ D→ C
b) A→ D→ B→ E→ C
c) B→ E→ A→ D→ C
d) D→ E→ B→ A→ C
Explanation: Since the time period and sample size is known and the flood data is given, steps A, B and D can be carried out in any order. Calculation of the frequency factor requires reduced variate, reduced mean and reduced standard deviation. Hence, step E should be carried out only after step B and D. Step C is the final step.
3. As per Gumbel’s analysis for a very large sample size, what is the probability of occurrence of a mean annual flood?
a) 33.3%
b) 42.9%
c) 50%
d) 75.2%
Explanation: A flood discharge with a return period of 2.33 years is a mean annual flood as it is obtained as the average of annual series from Gumbel’s analysis. This is valid when the sample size is very large.
∴ Probability of occurrence =$$\frac{1}{T}=\frac{1}{2.33}$$=42.9%
4. What is the shape of Gumbel distribution when plotted on a Gumbel probability paper?
a) Straight line
b) Bell shaped
c) S-shaped
d) Parabolic
Explanation: Gumbel probability paper is a special graph to plot the Gumbel distribution in the form of discharge (ordinate) vs. return period (abscissa). Since a variate varies linearly with the reduced variate as per Gumbel’s equation, a linear plot is obtained on a Gumbel probability paper. This property is used in flood estimation by simple graphical extrapolation.
5. When using a semi-log graph for plotting discharge and return period in Gumbel’s analysis, only 3 sets of values are required.
a) True
b) False
Explanation: Unlike a Gumbel probability paper, a semi-log or a log-log graph will not give a straight line for a flood discharge vs. return period plot. Hence, a large number of values of return period and its corresponding flood value will be required to obtain the specific curve for the given annual series for the particular area.
6. What is the minimum set of values of return period and its corresponding flood discharge to be computed for Gumbel’s analysis by graphical method on a Gumbel probability paper?
a) 1
b) 2
c) 3
d) 4
Explanation: Since it is known that the relationship between flood discharge and return period is linear when plotted on a Gumbel probability paper, only two sets of values (points) are required to obtain a straight line, which can then be linearly extrapolated to obtain the required flood value.
7. Which of the following terms is used to construct the return period axis on a Gumbel probability paper?
a) Mean of variate
b) Standard deviation of variate
c) Reduced mean
d) Reduced variate
Explanation: The abscissa of a Gumbel probability graph is specially marked for return period scale by using values of reduced variate. The required values of return period are selected and their corresponding reduced variates are calculated. These are then plotted on a separate limited arithmetic scale of reduced variate values constructed below the return period axis. The return period scale is then marked using the values from the reduced variate axis.
8. Compute the frequency factor for Gumbel’s analysis for estimating a flood magnitude of return period 25 years. Assume a very large number of data in the annual flood series.
a) 1.56
b) 2.67
c) 2.04
d) 3.32
Explanation: For very large sample size, ($$\overline{y_n}$$)=0.577 and Sn = 1.2825.
Reduced variate, $$y_T=-ln.ln\frac{T}{T-1}=-ln.ln\frac{25}{24}$$=3.1985
∴Frequency factor, $$K=\frac{y_T-\overline{y_n}}{S_n} =\frac{3.1985-0.577}{1.2825}$$=2.04
Sanfoundry Global Education & Learning Series – Engineering Hydrology. | 1,197 | 4,687 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-18 | latest | en | 0.881313 |
40kwarmaster.blogspot.com | 1,519,550,572,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816351.97/warc/CC-MAIN-20180225090753-20180225110753-00738.warc.gz | 4,135,559 | 28,572 | Thursday, December 11, 2014
Math by Request: Screamers vs Landraider
I had a request from reader Jon Glas for the probability related to screamers attacking a landraider. Landraiders can be tough to kill, especially for daemons.
We are going to look at two cases here: attacking normally, and attacking with prescience.
Attacking Normally:
With 9 screamers there is an overall 20.28% chance of destroying a landraider in one round of combat. Within that chance there is a 15.48% chance of exploding it, and a 4.8% chance or glancing it out.
Initially this doesn't look good. A one in five chance is not all that great.
**Warning: here is my work**
I started by calculating the chances of an 'explodes' result.
2/3 * 1/6 * 1/6 = 1/54
Using this we find that there is an 84.52% chance of not getting an 'explodes' result which means there is a 15.48% chance of getting at least one 'explodes' result.
and the chances to cause a hull point without causing causing it to explode.
2/3 * (1/9 + 1/6 * 5/6)= 1/6
Rather than finding the chances of getting 4-9 hull points I instead found the chances of 0-3
0: (5/6)^9 = 0.1938
1: 9(1/6)(5/6)^8 = 0.3489
2: 36(1/6)^2 * (5/6)^7 = 0.2791
3: 84(1/6)^3 * (5/6)^6 = 0.1302
This gives a 95.2% chance of doing 0-3 hull points which means there is a 4.8% chance of doing 4+ hull points. If we combine this with the 15.48% chance to explode this gives an overall chance of 20.28%
**End of math**
Attacking With Prescience:
Now if we look at what happens when the unit has prescience the results are fairly dramatic. All of a sudden we have a 40% chance of destroying the landraider. The chance to explode increases to 20.15% and the chance to glance out the vehicle increases to 19.85%
**More math!**
Chance to explode:
8/9 * 1/6 *1/6 = 2/81 = 0.2015
Chance to glance:
8/9 * (1/9 + 1/6 * 5/6) = 2/9
Same style as last time:
0: (7/9)^9 = 0.1042
1: 9(2/9)(7/9)^8 = 0.2678
2: 36(2/9)^2 * (7/9)^6 = 0.3061
3: 84(2/9)^3 * (7/9)^8 = 0.1234
This adds up to an 80.15% chance of 0-3 hull points and a 19.85% to do 4+ hull points. Combining this with the chance to explode this leads to a 40% overall chance.
**End math**
Wow! I was completely shocked by this result. The increase to the chance to explode was about what I expected, but the chance to cause 4+ glances increased by a dramatic amount! Having a 40% chance to destroy a landraider in one turn doesn't seem incredible, but consider four melta guns in melta range only have a 44.14% chance of destroying that same landraider. All of a sudden our attacks look much better.
If we manage to knock a single hull point off of the landraider (with all that daemon shooting) our chances of destroying it with the screamers jumps up to 52.34%.
Now if you opponent is letting a squad of 9 screamers run around they are doing things wrong. What happens if you get knocked down to only 5 screamers?
Without prescience 5 screamers have a 9.25% of wrecking a landraider with 8.92% of that chance being an explodes result. With prescience our chances increase to 12.24% overall with 9.4% for exploding.
What we see is that when the unit gets smaller prescience becomes less helpful. With a full unit prescience increases our unit's damage to 197% of our normal damage, but when we are at half strength we only do 132% of the normal damage.
So, if you want to wreck a landraider fast make sure your screamers get there in one piece, or be prepared to dedicate more than one squad to the job. This falls in line from most of what I had learned while playing daemons. Half strength squads are not going to get the job done, and it is generally better to make sure you kill your target the first time.
1. Neat. Help me figure out your equations. You posted this:
I started by calculating the chances of an 'explodes' result.
2/3 * 1/6 * 1/6 = 1/54 = 0.1548
I would think there is a 1/2 chance of hitting, a 1/6 chance of exploding assuming you scored a penetrating hit. So the hard part is calcing the chance of penetrating because there are multiple dice combo results that net a penetrating hit. For example, you need a 10+ on the armorbane roll to penetrate. That is a 5,5 or a 5,6 or a 6,5 or a 6, 6. That is 4 of potentially 36 results. So to me the calc would be: 1/2 * 1/9 * 1/6 = 0.009 or 0.9%.
So obviously my math is wrong. So help me understand yours :)
1. I skipped a couple of steps in what I was showing. The 1/54 was correct but that lead to the 15.48% chance to explode. With the 1/54 there was an 84.52% chance of causing no 'explodes' result giving us the 15.48%.
I also updated the main post. Thanks for pointing this out.
2. Oh and with the chances to penetrate: You are correct we need a 10+: (6,6); (5,6); (6,5); (5,5); (4,6); and (6,4), so there are a total of 6/36 possibilities that will cause a penetrating hit.
3. Oh, doh Lamprey's Bite is S5 not S4... that totally changes the little bit of math I did in the last post. This is what I get for not playing a 40k game for like 3 months.
4. Also to John, the chance to hit is not 1/2. Vehicles that move gain WS1 until the end of the next turn. So it's a 3+ to hit not 4+.
5. How did you calculate the overall % chance of 9 screamers exploding the LR? We determined a single attack has a 1/54 chance of doing it, but where I get lost is factoring in multiple die rolls, like in this case where 9 screamers are making a single attack each. Do you multiply the chance of a single attack causing an explode by the number of attacks? If so that is a 16.66% chance (1/54 * 9) or (0.0185 * 9).
This is fun :) Should change the name of this thread to mathhammer is awesomesauce :)
6. The chances of not getting an explodes result is 53/54 so the chances of 9 guys getting 0 explodes is (53/54)^9 = 0.8452 So, I did 1-0.8452 and this gave me a 0.1548 chance of getting at least one 'explodes' result.
2. I really like this blog.
:-)
1. Thanks! It may sounds cheesy, but it means a lot to hear.
By sheer coincidence when I posted this update yesterday my students were taking their test over probabilities.
3. Totally quiz them on screamers attacking land raiders :)
1. Unfortunately we didn't have time to go over combinatorics, but finding the probability of getting an explodes result is certainly within the realm of possibility.
As far as I know I don't have any students who play 40k, but I did catch one girl reading up on the history of the Dark Angels in class.
4. I have a somewhat related question then. When a Screamer charges and makes it's special attack it says that all of it's normal attacks are consumed, but what about the attack you generate by the charge itself?
Does this attack feed into the special attack (making it 2 special attacks per Screamer), since it's an attack gained by a special circumstance (charging) and thus can not be classified as a 'normal' attack surely (i.e., the amount under the Attacks column). -Or-
Does the Screamer gain 1 Screamer special attack and one 'normal' charge attack to be done at the Screamers usual stats. -Or-
Does the screamer no longer have this attack at all for some reason?
It's always been a question that I've considered and I've never really come to a firm decision myself
1. It loses the charge bonus. It's the same thing as a MC with smash which replaces all of its normal attacks. You get one and that's it.
2. You can think of it exactly like stat modifiers. If you have a x 2 multiplier, that goes first. Then any + or -. And finally if you are supposed to set a stat to a fixed level (in this case one) that overrides everything else.
So you get your base attacks + charge bonus when you charge. If you want to, you can replace all of those for a single lamprey's bite.
5. I would assume that the attack gained by charging is part of its normal attacks, so even on the charge it would only have one special attack.
The FAQ doesn't have anything on this either.
6. One math question: It could be interesting to calculate ho many "weak" models would be needed to make reasonable to assault a strong unit. Ex: how many gretchin, or witches, or eldar guardian, or AM troops wold be needed to assault some of the nonsense space wolf units to balance the match....
1. This is a mathematically difficult problem to solve. I may go so far as to say that a general solution of how to solve your proposed problem is unsolved.
The reason a general solution is (presumably) unsolved is because it would involve solving for an unknown number of exponents at the same time as solving for an unknown number of bases.
However, that doesn't mean I can't answer your question. Specific solutions can be found by trying different numbers of troops until a desired percent is achieved.
I encountered a particularly nasty Space Marine commander on a bike, and I have been wondering how many nurglings it would take to stop him. These calculations may be my distraction from grading finals.
7. :-) yes, maybe I will try by myself with some different units... Then we could check the process to achieve the result desired and maybe find a general equation....
1. +50% of success will be my obj)
2. If you get a result I would be happy to check and see if I get the same thing.
I don't thing a general equation is possible. A solution to this type of general equation would have some profound impacts to mathematics in general. More specifically it would mean that the encryption of the internet could easily be broken.
http://en.wikipedia.org/wiki/Discrete_logarithm
3. So we will be famous, bro'! :-D :-P
4. If you define what you mean by "reasonable odds" then you could make three pseudo-general equations that you just have to plug numbers into. The three equations would be for the different cases of good unit having higher I, same I, or bad unit having higher I.
Basically what you're left with is a general formula in which you plug in number of attacks per model, chance to hit, chance to wound, and chance to fail saves.
It's only general in the sense that the computations are the exact same every time but you have to know the exact statlines of the units fighting each other and which ones get the charge.
I would think the easiest way of going about it would be to calculate expected values rather than probabilities. Something like "how many gretchin need to assault 5 space marines in order to have an expected value of killing all 5 of them" would be a pretty simple calculation. Something like "how many gretchin need to assault 5 space marines to have a 75% probability of killing them all" would require a few more steps.
8. To win a fight is more complicated than just put "n" wound in one turn. I'm thinking that a good start is to have an expected value to put one wound more than the opponent, calculate the possibility of him running away and ours to catch and destroy him (if possible). Then re-do the calculation for the 2nd phase, without the bonus of the charge and see where the fight is going. [Maybe we have some possibility to win the fight on the first turn, but maybe the bigger probability is that the fight goes on and we begin loosing it from the second turn....]
then the problem of the terrain we are charging (or are charged) trough...
What else?
The purpose of this calculation is not the exercise in himself, but rather is to have an idea of what to do when we find our self in some specific situation. Like the poker player that when he find with J10 in his hand and looking at that 3 card on the table knows if the best move for him is to raise or to check/fold the hand. :-)
1. Back in 6th edition I actually wrote a formula to calculate the probability of winning combat. It was very complicated, but theoretically worked. Practically speaking there was too much going on for it to be very helpful, so I attempted to write a program. I am not very good at programing, and as a result never got it to work. Also with the ever-changing rules this would be a continuous endeavor. | 3,121 | 12,004 | {"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} | 4.1875 | 4 | CC-MAIN-2018-09 | longest | en | 0.897165 |
http://www.lahey.com/docs/lfpro73help/f95arsinhfn.htm | 1,529,575,494,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864139.22/warc/CC-MAIN-20180621094633-20180621114633-00232.warc.gz | 458,454,035 | 2,237 | SINH Function
LF Fortran 95
# SINH Function
### Description
The SINH function returns the hyperbolic sine of a REAL argument.
```Syntax
SINH (x)
```
### Arguments
x is an INTENT(IN) scalar or array of type REAL.
### Result
The result is of the same type and kind as x. Its value is a REAL representation of the hyperbolic sine of x.
```Example
real :: x..5,y(2)=(/1.,1./)
write(*,*) sinh(x) ! writes .5210953
write(*,*) sinh(y) ! writes 1.175201 1.175201
``` | 146 | 470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.643301 |
http://topcoder.bgcoder.com/print.php?id=2808 | 1,723,687,055,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00791.warc.gz | 32,544,370 | 2,378 | ### Problem Statement
You are creating a simple one-player computer game. The player must first choose and enter a string of length k, where each character is 'A', 'B', 'C' or 'D'. Each possible string of length k maps to a color, and these mappings will be predetermined by you. To win the game, the player must perform a series of transformations on the string until it is different from the original, but maps to the same color. Only the following two transformations are allowed at each step:
2. Replace an occurrence of before[i] in the string with after[i], for some index i.
You want to create an "Impossible Mode", in which the player can never win. Return the minimum number of colors required to make the game impossible.
### Definition
Class: ImpossibleGame Method: getMinimum Parameters: int, String[], String[] Returns: long Method signature: long getMinimum(int k, String[] before, String[] after) (be sure your method is public)
### Constraints
-k will be between 1 and 30, inclusive.
-before will contain between 1 and 50 elements, inclusive.
-before and after will contain the same number of elements.
-Each element of before will contain between 1 and k characters, inclusive.
-For each index i, before[i] and after[i] will contain the same number of characters.
-Each character in before and after will be 'A', 'B', 'C' or 'D'.
### Examples
0)
`1` ```{ "A" } ``` ```{ "B" } ```
`Returns: 2`
If "A" and "B" are assigned the same color, the player can win by starting with "A" and replacing it with "B". So at least two colors are needed. Two colors are enough to make the game impossible. For example, you can assign red to "A" and "C", and blue to "B" and "D".
1)
`2` ```{ "A", "A", "D" } ``` ```{ "B", "C", "D" } ```
`Returns: 5`
2)
`2` ```{ "B", "C", "D" } ``` ```{ "C", "D", "B" } ```
`Returns: 9`
3)
`6` ```{ "AABBC", "AAAADA", "AAACA", "CABAA", "AAAAAA", "BAAAA" } ``` ```{ "AACCB", "DAAABC", "AAAAD", "ABCBA", "AABAAA", "AACAA" } ```
`Returns: 499`
#### Problem url:
http://www.topcoder.com/stat?c=problem_statement&pm=11326
#### Problem stats url:
http://www.topcoder.com/tc?module=ProblemDetail&rd=14428&pm=11326
lyrically
#### Testers:
PabloGilberto , ivan_metelsky , nika
Graph Theory | 626 | 2,238 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.87586 |
https://chorasimilarity.wordpress.com/2014/08/13/a-symplectic-brezis-ekeland-nayroles-principle/ | 1,529,561,418,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864039.24/warc/CC-MAIN-20180621055646-20180621075646-00557.warc.gz | 587,534,891 | 23,110 | # A symplectic Brezis-Ekeland-Nayroles principle
We submitted to the arXiv the article
Marius Buliga, Gery de Saxce, A symplectic Brezis-Ekeland-Nayroles principle
You can find here the slides of two talks given in Lille and Paris a while ago, where the article has been announced.
UPDATE: The article appeared, as arXiv:1408.3102
This is, we hope, an important article! Here is why.
The Brezis-Ekeland-Nayroles principle appeared in two articles from 1976, the first by Brezis-Ekeland, the second by Nayroles. These articles appeared too early, compared to the computation power of the time!
We call the principle by the initials of the names of the inventors: the BEN principle.
The BEN principle asserts that the curve of evolution of a elasto-plastic body minimizes a certain functional, among all possible evolution curves which are compatible with the initial and boundary conditions.
This opens the possibility to find, at once the evolution curve, instead of constructing it incrementally with respect to time.
In 1976 this was SF for the computers of the moment. Now it’s the right time!
Pay attention to the fact that a continuous mechanics system has states belonging to an infinite dimensional space (i.e. has an infinite number of degrees of freedom), therefore we almost never hope to find, nor need the exact solution of the evolution problem. We are happy for all practical purposes with approximate solutions.
We are not after the exact evolution curve, instead we are looking for an approximate evolution curve which has the right quantitative approximate properties, and all the right qualitative exact properties.
In elasto-plasticity (a hugely important class of materials for engineering applications) the evolution equations are moreover not smooth. Differential calculus is conveniently and beautifully replaced by convex analysis.
Another aspect is that elasto-plastic materials are dissipative, therefore there is no obvious hope to treat them with the tools of hamiltonian mechanics.
Our symplectic BEN principle does this: one principle covers the dynamical, dissipative evolution of a body, in a way which can be reasonably easy amenable to numerical applications.
_______________________________________ | 466 | 2,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-26 | latest | en | 0.911646 |
http://www.talkstats.com/tags/mathematical-statistics/ | 1,659,953,255,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570793.14/warc/CC-MAIN-20220808092125-20220808122125-00076.warc.gz | 106,102,378 | 10,285 | # mathematical statistics
1. ### A basic question about chi-square test
There is a test to examine the usage of different English words by male and female students. So there are two columns of data (male and female students), 30 rows(each row is the frequency of different English words used by male and female students). I find an online chi-square calculator, which...
2. ### Help Determing What Distribution
I believe I have answered a), as L~Gamma(N,1/20), therefore mgf = (20/(20-t))^N (if wrong please feel free to correct me). It has been defined that L=-ln(U), and I am finding difficulty in determing what distribution U would be classified as. Once classfieid, I believe that finding the mean...
3. ### Evaluate performance
i have 5 variables of game's data which is evaluated on some logic . example : level = 100% objective = 100% retry = 100% successful moves = 80% score = 25000 now i have such data of every user, above one is just example , now i want to measure performance of such user and using this 5...
4. ### Book recommendation for Mathematical Stats
So I've read Tanis and Hogg, and Ross's Mathematical Statistics books and understood them fairly well. Still when I try to read some other Stats books I always encounter something I've never seen before that keeps me from making progress. To give some examples, in Gelman's Bayesian Data...
5. ### mathematical model to build a ranking/ scoring system
I want to rank a set of sellers. Each seller is defined by parameters var1,var2,var3,var4...var20. I want to score each of the sellers. Currently I am calculating score by assigning weights on these parameters(Say 10% to var1, 20 % to var2 and so on), and these weights are determined based on...
6. ### Calculating a one-way anova with just means, stdev, and sample sizes
Hi How can you determine the sum of squares total, sum of squares within and between, only knowing the mean, standard deviation and sample size of two groups e.g., group 1: mean 5.67 SD 3.055 sample size 3 group 2: mean 14.33 SD 2.081 sample size 3 Many thanks for any help and...
7. ### Making sense of the parameter delta in this equation.
Hello, I was reading this paper about pollen dispersal directionality and I am trying to make sense of the so-called directionality parameter \delta The Statistical Model On pages 4 and 5 they explain their analysis under the section statistical procedure. More specifically, in the first...
8. ### Maximum Likelihood Estimator - What did I do wrong?
For a X_1,…,X_n random sample from a uniform distribution from [0, 2*theta+1], I was asked to find the MLE. The pdf: f(x|theta) = 1/(2*theta+1) for x in [0, 2*theta+1] and 0 otherwise. I determined the likelihood(theta)=L(theta)=(2*theta+1)^-n Then the log-likelihood to be -n*(2*theta+1)...
9. ### standard deviation for an equation
Hi, I am trying to figure out how to get the standard deviation for two equations. The first one is quite simple; it's just how to find the SD for a part of an average value. I have the SD for an average amount of element in 4 values e.g. average = 118 mg/kg and SD = 48 mg/kg I am...
10. ### I Need Help with Poisson Distribution Problem!!
I don't know where to even start with this Poisson Dist. problem, can anyone help me solve this??? The Hillel Sandwich Shop specializes in delivery of sandwiches in New York's financial district. Customers phone or fax Hillel their orders, which are then delivered directly to the customer's...
11. ### Proof the sum of two CDF's is a CDF.
This is a question my instructor asked in the last midterm exam but nobody was able to solve and he's suggesting it may come out again in the finals: If F(x) and G(X) are two CDF's, prove that H(X) is also a CDF if H(X) = F(X)+G(X)-F(x)G(X) He said something about right...
12. ### Sum of independent beta-Bernoulli random variables
I'm looking for the distribution of the sum of independent beta-Bernoulli random variables, that is Bernoulli random variables with beta-distributed parameters, x_i ~ Bern(p_i) with p_i ~ beta(alpha,beta) (same alpha and beta, of course). I guess one can work with convolutions or characteristic... | 1,018 | 4,151 | {"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-2022-33 | latest | en | 0.922204 |
http://voltageelectricity.tk/electricity-and-magnetism-in-physics/ | 1,550,884,095,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249414450.79/warc/CC-MAIN-20190223001001-20190223023001-00060.warc.gz | 299,356,952 | 10,780 | Electricity And Magnetism In Physics
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5 / 5 ( 1votes )
Home – exploring physics, Work with cool hands-on materials, take hold of your own learning – and do it in a digital environment using the exploring physics curriculum app..
Physics – electricity magnetism, 1.6.1: field of a point charge: 1.6.2: spherical charge distributions: 1.6.3: a long, charged rod: 1.6.4: field on the axis of a ring charge: 1.6.5: field on the axis.
Physics courses – university california san diego, Phys 4c. physics for physics majors—electricity and magnetism (4) continuation of physics 4b covering charge and coulomb’s law, electric field, gauss’s law, electric potential, capacitors and dielectrics, current and resistance, magnetic field, ampere’s law, faraday’s law, inductance, magnetic properties of matter, lrc circuits, maxwell’s equations..
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Magnetism – wikipedia, Magnetism class physical phenomena mediated magnetic fields. electric currents magnetic moments elementary particles give rise magnetic field, acts currents magnetic moments. familiar effects occur ferromagnetic materials, strongly attracted magnetic fields magnetized permanent magnets.
https://en.wikipedia.org/wiki/Magnetism
Electricity & magnetism – physics – university , Electricity & magnetism. electricity & magnetism problems categories. addition definition problems (.. electric force field due point charges), electric force dynamics problems electric energy conservation energy problems..
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Physics4kids.: electricity & magnetism: magnets, Physics4kids.! tutorial introduces magnets physics. sections include motion, heat, electricity, light, modern physics..
http://www.physics4kids.com/files/elec_magnets.html
Author: | 496 | 2,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.694155 |
https://www.knowledgeuniverseonline.com/ntse/Mathematics/relationship-zeros-coefficients.php | 1,656,331,935,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103331729.20/warc/CC-MAIN-20220627103810-20220627133810-00572.warc.gz | 907,883,219 | 9,844 | # Mathematics
### Chapter : Polynomials
#### Relation between the zero and the coefficients of a polynomials
Consider quadratic polynomial P(x) = 2x2 – 16x + 30.
Now, 2x2 – 16x + 30
= (2x – 6) (x – 3)
= 2(x – 3) (x – 5)
The zeroes of P(x) are 3 and 5.
Sum of the zeroes are = 3 + 5 = 8 =
Product of the zeroes are = 3 × 5 = 15 =
So if ax2 + bx + c, a ≠ 0 is a quadratic polynomial and a, b are two zeroes of polynomial then
Example: Find the zeroes of the quadratic polynomial 6x2 – 13x + 6 and verify the relation between the zeroes and its coefficients.
Solution: We have,
6x2 – 13x + 6
= 6x2 – 4x – 9x + 6
= 2x (3x – 2) –3 (3x – 2)
= (3x – 2) (2x – 3)
So, the value of quadratic equation 6x2 – 13x + 6 is 0, when
(3x – 2) = 0 or (2x – 3) = 0 i.e.,
When x = or
Therefore, the zeroes of 6x2 – 13x + 6 are and .
Sum of the zeroes :
+ = =
Product of the zeroes
= × = =
Example: Find the zeroes of the quadratic polynomial 4x2 – 9 and verify the relation between the zeroes and its coefficients.
Solution: We have,
4x2 – 9
= (2x)2 – 32
= (2x – 3) (2x + 3)
So, the value ofa quadratic equation 4x2 – 9 is 0, when
2x – 3 = 0 or 2x + 3 = 0
i.e., when x = or x = .
Therefore, the zeroes of 4x2 – 9 are & .
Sum of the zeroes
= + = 0 = =
Product of the zeroes
= =
Example: Find the zeroes of the quadratic polynomial 9x2 – 5 and verify the relation between the zeroes and its coefficients.
Solution: We have,
9x2 – 5
= (3x)2 – (√5)2
= (3x – √5) (3x +√5)
So, the value of 9x2 – 5 is 0, when
3x –√5 = 0 or 3x +√5 = 0
i.e., when x = or x = .
Sum of the zeroes
== 0 = =
Product of the zeroes
= ×= | 647 | 1,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2022-27 | latest | en | 0.776534 |
https://socratic.org/questions/how-do-you-prove-2-sec-2-theta-sec-2-theta-sec-2-theta-1-2sin-2-theta | 1,582,333,659,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145621.28/warc/CC-MAIN-20200221233354-20200222023354-00278.warc.gz | 565,017,474 | 5,957 | How do you prove (2-sec^2[theta])/(sec^2[theta]) =1-2sin^2[theta]?
(2-sec^2 theta)/(sec^2 theta ) =1-2sin^2 theta
$\frac{2 - {\sec}^{2} \theta}{{\sec}^{2} \theta} = \frac{2 {\cos}^{2} \theta - 1}{\cos} ^ 2 \theta {\cos}^{2} \theta = 2 {\cos}^{2} \theta - 1 = 2 \left(1 - {\sin}^{2} \theta\right) - 1 = 1 - 2 {\sin}^{2} \theta$ | 162 | 327 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2020-10 | latest | en | 0.275883 |
http://mathhelpforum.com/calculus/index1545.html | 1,518,987,740,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812259.18/warc/CC-MAIN-20180218192636-20180218212636-00786.warc.gz | 220,667,043 | 15,671 | # Calculus Forum
Calculus Help Forum
1. ### Calculus Notes
• Replies: 95
• Views: 15,441
Jan 17th 2017, 05:06 AM
2. ### Introductory Tutorial to Multiple Integration
• Replies: 6
• Views: 7,756
Dec 4th 2012, 11:17 AM
3. ### Integral [example thread of why you should post all given info]
• Replies: 7
• Views: 8,004
Jun 12th 2009, 08:31 AM
1. ### Integral
• Replies: 2
• Views: 309
Mar 12th 2008, 09:30 PM
2. ### Related Rates (help)
• Replies: 1
• Views: 1,476
Mar 12th 2008, 09:25 PM
3. ### thought problem involving derivatives
• Replies: 2
• Views: 357
Mar 12th 2008, 09:24 PM
4. ### Parametric Arc Length and Tangent Equations
• Replies: 3
• Views: 516
Mar 12th 2008, 09:10 PM
5. ### rate of change
• Replies: 5
• Views: 435
Mar 12th 2008, 08:59 PM
6. ### integration
• Replies: 5
• Views: 423
Mar 12th 2008, 06:53 PM
7. ### [SOLVED] A D.N.E limit proofs
• Replies: 4
• Views: 626
Mar 12th 2008, 05:36 PM
8. ### Calulating the minimum value of f(x) in a given range of x?
• Replies: 12
• Views: 1,044
Mar 12th 2008, 05:16 PM
9. ### Can anyone help me finish this
• Replies: 4
• Views: 692
Mar 12th 2008, 04:47 PM
10. ### Limit of sequence
• Replies: 4
• Views: 581
Mar 12th 2008, 04:16 PM
11. ### Improper Integrals - need help
• Replies: 1
• Views: 415
Mar 12th 2008, 04:01 PM
12. ### Question about the midpoint rule
• Replies: 2
• Views: 470
Mar 12th 2008, 01:20 PM
13. ### finding the angle
• Replies: 3
• Views: 473
Mar 12th 2008, 01:17 PM
14. ### Local Linearization!
• Replies: 1
• Views: 1,186
Mar 12th 2008, 01:03 PM
15. ### Integrating e^t cos t
• Replies: 8
• Views: 647
Mar 12th 2008, 12:20 PM
16. ### Calc 2: Converge/Diverge
• Replies: 8
• Views: 2,348
Mar 12th 2008, 11:27 AM
17. ### Tangent and parallel lines
• Replies: 1
• Views: 543
Mar 12th 2008, 09:42 AM
18. ### Limits and Improper Integrals
• Replies: 5
• Views: 655
Mar 12th 2008, 09:02 AM
19. ### [SOLVED] Problem
• Replies: 1
• Views: 530
Mar 12th 2008, 08:39 AM
20. ### Pls check my work – bivariate optimization
• Replies: 5
• Views: 1,284
Mar 12th 2008, 08:17 AM
21. ### Ok then, here's a tricky one.
• Replies: 1
• Views: 1,059
Mar 12th 2008, 07:51 AM
22. ### Related Rates word problem
• Replies: 3
• Views: 709
Mar 12th 2008, 06:41 AM
23. ### Multivariable maximums with a spherical limit
• Replies: 0
• Views: 287
Mar 12th 2008, 06:28 AM
24. ### converge or diverge
• Replies: 2
• Views: 354
Mar 12th 2008, 01:13 AM
25. ### Equilibria and Oscillations question
• Replies: 1
• Views: 474
Mar 12th 2008, 01:05 AM
26. ### approximation of sum of Alternating series
• Replies: 1
• Views: 1,058
Mar 12th 2008, 12:32 AM
27. ### Super Easy quick question!
• Replies: 2
• Views: 451
Mar 12th 2008, 12:24 AM
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,
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# calculus forums
Click on a term to search for related topics.
Use this control to limit the display of threads to those newer than the specified time frame.
Allows you to choose the data by which the thread list will be sorted. | 1,171 | 2,980 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | longest | en | 0.606682 |
https://vdare.com/posts/math-geeks-benford-s-law-and-crooked-voting-numbers-in-milwaukee-and-chicago | 1,610,944,958,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514121.8/warc/CC-MAIN-20210118030549-20210118060549-00707.warc.gz | 638,645,406 | 20,489 | Math Geeks, Benford's Law, And Crooked Voting Numbers In Milwaukee And Chicago.
11/06/2020
A+
|
a-
Yes, the voting numbers in this election are more crooked than usual. Is there a way to quantify how crooked they are?
Math geeks have been having fun with Benford's Law. You can read up Benford's Law on Wikipedia. Here's a very short account.
Consider a big list of numbers identifying something in the real world. Wikipedia offers, quote, "electricity bills, street addresses, stock prices, house prices, population numbers, death rates, lengths of rivers, and physical and mathematical constants."
OK, you're looking at your list of house prices (say). How often is the lead-off digit in a price—the leftmost digit—how often is it a 1? Well, since it can be any one of nine digits, and there seems no reason why any digit should be favored over any other, that lead digit will be a 1 one-ninth of the time, which is to say, 11.1 percent of the time. Right?
Wrong! Benford's Law tells us that digit will be a 1 more than thirty percent of the time. At the other end, it'll be a 9 less than five percent of the time.
The distribution of first digits, according to Benford's law. Each bar represents a digit, and the height of the bar is the percentage of numbers that start with that digit.
I know: It's weirdly counterintuitive. Math can be like that. It's a real law, though, with solid theoretical explanations supporting it.
Most to the point here, it offers a way to look at a big list of numbers and see if it is, so to speak, normal. Does it obey Benford's Law? If it doesn't, there might be a good reason; but you'd want to know the reason. You'd particularly want to know if the reason was, that someone just made up the numbers, or cranked them out of a computer random-number generator.
The computer-programming website Github.com has been running the numbers for vote counts by precinct in select cities and counties. You can view results over there: Go to Github.com and search for "2020 Benford's."
The ones I've seen so far are suggestive but not, I'd say, conclusive. The biggest departures from Benford's Law show up in the Joe Biden vote counts for … uh-oh, Milwaukee and … double-uh-oh, Chicago.
Dispositive? No. Worth investigating? Definitely, if anyone's keen to do so. | 549 | 2,302 | {"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} | 3.890625 | 4 | CC-MAIN-2021-04 | longest | en | 0.948523 |
https://www.education.com/worksheets/measuring-length/CCSS-Math-Content/ | 1,621,037,636,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991829.45/warc/CC-MAIN-20210514214157-20210515004157-00150.warc.gz | 758,797,094 | 25,009 | # Search Our Content Library
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Worksheet | 547 | 2,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-21 | latest | en | 0.77784 |
http://www.extremeoptimization.com/QuickStart/FSharp/SparseMatrices.aspx | 1,653,428,589,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662577259.70/warc/CC-MAIN-20220524203438-20220524233438-00690.warc.gz | 80,075,660 | 6,791 | Data Analysis Mathematics Linear Algebra Statistics
New Version 8.1!
Supports .NET 6.0. Try it for free with our fully functional 60-day trial version.
QuickStart Samples
# Sparse Matrices QuickStart Sample (F#)
Illustrates using sparse vectors and matrices using the classes in the Extreme.Mathematics.LinearAlgebra.Sparse namespace in F#.
```// Illustrates using sparse vectors and matrices using the classes
// in the Extreme.Mathematics.LinearAlgebra namespace
// of the Extreme Optimization Numerical Libraries for .NET.
#light
open System
// The sparse vector and matrix classes reside in the
// Extreme.Mathematics.LinearAlgebra namespace.
open Extreme.Mathematics
open Extreme.Mathematics.LinearAlgebra
//
// Sparse vectors
//
// The SparseVector class has three constructors. The
// first simply takes the length of the vector. All elements
// are initially zero.
let v1 = Vector.CreateSparse<float>(1000000)
// The second constructor lets you specify how many elements
// are expected to be nonzero. This 'fill factor' is a number
// between 0 and 1.
let v2 = Vector.CreateSparse<float>(1000000, 1e-4)
// The second constructor lets you specify how many elements
// are expected to be nonzero. This 'fill factor' is a number
// between 0 and 1.
let v3 = Vector.CreateSparse<float>(1000000, 100)
// The fourth constructor lets you specify the indexes of the nonzero
// elements and their values as arrays:
let indexes = [| 1; 10; 100; 1000; 10000 |]
let values = [| 1.0; 10.0; 100.0; 1000.0; 10000.0 |]
let v4 = Vector.CreateSparse(1000000, indexes, values)
// Elements can be accessed individually:
v1.[1000] <- 2.0
printfn "v1.[1000] = %A" v1.[1000]
// The NonzeroCount returns how many elements are non zero:
printfn "v1 has %d nonzeros" v1.NonzeroCount
printfn "v4 has %d nonzeros" v4.NonzeroCount
// The NonzeroElements property returns a collection of
// IndexValuePair structures that you can use to iterate
// over the elements of the vector:
printfn "Nonzero elements of v4:"
for pair in v4.NonzeroElements do
printfn "Element %d = %A" pair.Index pair.Value
// All other vector methods and properties are also available,
// Their implementations take advantage of sparsity.
printfn "Norm(v4) = %A" (v4.Norm())
printfn "Sum(v4) = %A" (v4.Sum())
// Note that some operations convert a sparse vector to a
// DenseVector, causing memory to be allocated for all
// elements.
//
// Sparse Matrices
//
// All sparse matrix classes inherit from SparseMatrix. This is an abstract class.
// There currently is only one implementation class:
// SparseCompressedColumnMatrix.
// Sparse matrices are created by calling the CreateSparse factory
// method on the Matrix class. It has 4 overloads:
// The first overload takes the number of rows and columns as arguments:
let m1 = Matrix.CreateSparse<float>(100000, 100000)
let m2 = Matrix.CreateSparse<float>(100000, 100000, 1e-5)
// The third overload uses the actual number of nonzero elements rather than
// the fraction:
let m3 = Matrix.CreateSparse<float>(10000, 10000, 20000)
// The fourth overload lets you specify the locations and values of the
// nonzero elements:
let rows = [| 1; 11; 111; 1111 |]
let columns = [| 2; 22; 222; 2222 |]
let matrixValues = [| 3.0; 33.0; 333.0; 3333.0 |]
let m4 = Matrix.CreateSparse(10000, 10000, rows, columns, matrixValues)
// You can access elements as before...
printfn "m4.[111, 222] = %A" m4.[111, 222]
m4.[99, 22] <- 99.0
// A series of Insert methods lets you build a sparse matrix from scratch:
// A single value:
m1.InsertEntry(25.0, 200, 500)
// Multiple values:
m1.InsertEntries(matrixValues, rows, columns)
// Multiple values all in the same column:
m1.InsertColumn(33, values, indexes)
// Multiple values all in the same row:
m1.InsertRow(55, values, indexes)
// A clique is a 2-dimensional submatrix with indexed rows and columns.
let clique = Matrix.Create(2, 2, [| 11.0; 12.0; 21.0; 22.0|], MatrixElementOrder.ColumnMajor)
let cliqueIndexes = [| 5; 8 |]
m1.InsertClique(clique, cliqueIndexes, cliqueIndexes)
// You can use the NonzeroElements collection to iterate
// over the nonzero elements of the matrix. The items
// are of type RowColumnValueTriplet:
printfn "Nonzero elements of m1:"
for triplet in m1.NonzeroElements do
printfn "m1.[%d,%d] = %A" triplet.Row triplet.Column triplet.Value
// ... including rows and columns.
let column = m4.GetColumn(22)
printfn "Nonzero elements in column 22 of m4:"
for pair in column.NonzeroElements do
printfn "Element %d = %A" pair.Index pair.Value
// Many matrix methods have been optimized to take advantage of sparsity:
printfn "F-norm(m1) = %A" (m1.FrobeniusNorm())
// But beware: some revert to a dense algorithm and will fail on huge matrices:
try
let inverse = m1.GetInverse()
printfn "%O" inverse
with
| :? OutOfMemoryException as e -> printfn "%s" e.Message
printf "Press Enter key to exit..." | 1,330 | 4,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-21 | latest | en | 0.501478 |
http://forums.wolfram.com/mathgroup/archive/2008/Nov/msg00354.html | 1,585,858,255,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370507738.45/warc/CC-MAIN-20200402173940-20200402203940-00197.warc.gz | 71,788,734 | 7,858 | Re: LatticeReduce
• To: mathgroup at smc.vnet.net
• Subject: [mg93580] Re: LatticeReduce
• From: Artur <grafix at csl.pl>
• Date: Sun, 16 Nov 2008 07:02:56 -0500 (EST)
• References: <200811151103.GAA16591@smc.vnet.net> <491F2366.3040401@csl.pl>
```P.S.
And for the case:
{a1, a2, a3} = {1, 3^4,
5^4}; a = {{1, 0, 0, -a1}, {0, 1, 0, -a2}, {0, 0, 1, -a3}}; b =
LatticeReduce[a]
Out1:*{{1, 0, 0, -1}, {12, -8, 1, 11}, {-6, -23, 3, -6}}
Mathematica should be return 2 solutions:
23 - 8*3^4 + 5^4=0
12 + 23*3^4 - 3*5^4=0
**After good LatticeReduce should begin from : {{23,-8,1,0}, {12,
23,-3,0}, {c, c1,c2, c3}}
What is wrong in LatticeReduce or what I'm doing wrong ?
Artur
*
Artur pisze:
> Dear Mathematica Gurus,
> I want to find such inetegrs x,y,z that
> x + y*2^15 + z*3^8 = 0
> I'm reading in manual that I can use LatticeReduce
>
> {a0, a1, a2} = {1,
> 2^15, -3^8}; a = {{1, 0, 0, -a0}, {0, 1, 0, -a1}, {0, 0,
> 1, -a2}}; b = LatticeReduce[a]
>
> Out1:{{1, 0, 0, -1}, {18, 1, 5, 19}, {117, -171, -854, 117}}
>
> That mean that computer don't find such solution (solution is finded
> when last number in one of rows should be 0)
>
> If I run again:
> {a0, a1, a2} = {37,
> 2^15, -3^8}; a = {{1, 0, 0, -a0}, {0, 1, 0, -a1}, {0, 0,
> 1, -a2}}; b = LatticeReduce[a]
>
> Out2:{{1, 1, 5, 0}, {1, 0, 0, -37}, {170, -7, -34, 12}}
>
> Now solution is finded by Mathematica OK!
> k = Transpose[{{37, 2^15, -3^8, anything}}]; b[[1]].k
>
> Out3: {0}
>
> Is OK!
>
> Is another method finding coefficients x, y, z as LatticeReduce and
> why Mathematica don't reduced
> {a0, a1, a2} = {1, 2^15, -3^8}; a = {{1, 0, 0, -a0}, {0, 1, 0, -a1},
> {0, 0, 1, -a2}}; b = LatticeReduce[a]
>
>
> Best wishes
> Artur
>
```
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• Next by thread: Re: Re: LatticeReduce | 866 | 1,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-16 | latest | en | 0.707353 |
https://testbook.com/question-answer/due-to-rise-of-20-in-the-price-of-apples-a-dealer--5fe9baf2e41f1f40d9ef322a | 1,642,576,241,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301264.36/warc/CC-MAIN-20220119064554-20220119094554-00597.warc.gz | 578,143,436 | 29,594 | # Due to rise of 20% in the price of apples a dealer get 20 kg less for Rs.18,000, find new price per kg of apple.
1. 220 Rs./kg
2. 150 Rs./kg
3. 180 Rs./kg
4. 240 Rs./kg
5. None of these
Option 3 : 180 Rs./kg
## Detailed Solution
Given:
Due to rise of 20% in the price of apples a dealer get 20 kg less for Rs.18,000
Calculation:
Let, Price of per kg apple = Rs.100x
Let, Total purchase capacity initially = 100y kg
As question says,
(100x) × (100y) = 18000
⇒ 100xy = 180 ----(I)
New price after reduction = {(100 + 20)/100} × 100x = Rs.120x
Final purchase capacity after reduction = (100y – 20) kg
As question says,
(120x) × (100y – 20) = 18000
⇒ 100xy – 20x = 150 ----(II)
From equation (I) and (II)
180 – 20x = 150
⇒ x = 1.5
New price of per kg apple = 120 × 1.5 = 180 Rs./kg
180 Rs./kg is reduction price of apple. | 309 | 850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2022-05 | latest | en | 0.816005 |
https://gradehunters.net/define-the-term-standard-deviation/ | 1,656,630,297,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103915196.47/warc/CC-MAIN-20220630213820-20220701003820-00407.warc.gz | 329,201,005 | 15,051 | # Define the term standard deviation
General Requirements:
Use the following information to ensure successful completion of the assignment:
• Read each segment of this assignment carefully. There is information in the segment that will guide your completion.
• Instructors will be using a grading rubric to grade the assignments. It is recommended that learners review the rubric prior to beginning the assignment in order to become familiar with the assignment criteria and expectations for successful completion of the assignment.
• Doctoral learners are required to use APA style for their writing assignments. The APA Style Guide is located in the Student Success Center.
• This assignment requires that at least two additional scholarly research sources related to this topic, and at least one in-text citation from each source be included.
Directions:
In an essay of 250-500 words, thoroughly address the following items and respond to the related questions:
1. Define the term standard deviation. Why is it important to know the standard deviation for a given sample? What do researchers learn about a normal distribution from knowledge of the standard deviation? A sample of n=20 has a mean of M = 40. If the standard deviation is s=5, would a score of X= 55 be considered an extreme value? Why or why not?
2. Hypothesis testing allows researchers to use sample data, taken from a larger population, to draw inferences (i.e., conclusions) about the population from which the sample came. Hypothesis testing is one of the most commonly used inferential procedures. Define and thoroughly explain the terms null hypothesis and alternative hypothesis. How are they used in hypothesis testing?
3. Define the term standard error. Why is the standard error important in research using sample distributions? Consider the following scenario: A random sample obtained from a population has a mean of µ=100 and a standard deviation of σ = 20. The error between the sample mean and the population mean for a sample of n = 16 is 5 points and the error between a sample men and population mean for a sample of n =100 is 2 points. Explain the difference in the standard error for the two samples.
must have two empirical references and no plagiarizm | 442 | 2,249 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-27 | latest | en | 0.916514 |
https://www.spiedigitallibrary.org/ebooks/TT/Aberration-Theory-Made-Simple/Chapter8/Optical-Systems-with-Circular-Pupils/10.1117/3.43000.ch8 | 1,526,924,090,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864461.53/warc/CC-MAIN-20180521161639-20180521181639-00532.warc.gz | 865,218,341 | 25,313 | Home > eBooks > Aberration Theory Made Simple > Optical Systems with Circular Pupils
Access to SPIE eBooks is limited to subscribing institutions. Access is not available as part of an individual subscription. However, books can be purchased on SPIE.Org
Chapter 8:
Optical Systems with Circular Pupils
Abstract
8.1 Introduction In this chapter, we consider optical systems with circular exit pupils and discuss imaging by them based on the diffraction of object radiation at the exit pupil. Our starting point is an equation for the distribution of light in the image of a point object called the diffraction point-spread function (PSF) of the system. This equation is equally suitable for calculating the diffraction pattern of a circular aperture. Since, under certain conditions, the diffraction image of an incoherent object is given by the convolution of its Gaussian image (which is a scaled replica of the object) and the system PSF, the PSF calculations are fundamental to the theory of optical imaging. To understand the effect of aberrations on images, it is essential that we first understand the aberration-free PSF. Accordingly, we give briefly the characteristics of the aberration-free image of a point object. Our discussion on aberrated images is built slowly. First, we discuss defocused images and irradiance along the axis of the pupil. Next, approximate relationships between the ratio of the PSF values at its center with and without aberration, called the Strehl ratio, and the variance of the aberration across the pupil are developed. The approximate results for primary aberrations are compared with the corresponding exact results to determine the range of validity of the simple Strehl ratio formulas. The concept of aberration balancing is introduced in which an aberration of a certain order in pupil coordinates is mixed or balanced with one or more aberrations of lower order to minimize its variance, and thereby maximize the Strehl ratio of the system. Aberration tolerances based on a Strehl ratio of 0.8 are given for primary and balanced primary aberrations. Rayleigh's quarter-wave rule is briefly discussed and balanced primary aberrations are identified with Zernike circle polynomials. Finally, aberrated PSFs for various amounts of primary aberrations are given and their symmetry properties in and about the Gaussian image plane are discussed. Since the diffraction image of an incoherent isoplanatic object is given by the convolution of its Gaussian image and the PSF of the system forming the image, the spatial frequency spectrum of the diffraction image is given by the product of the spectrum of the Gaussian image and the optical transfer function (OTF) of the system. Thus, the OTF of the system, which is equal to the Fourier transform of its PSF, is equally fundamental to the theory of optical imaging. The OTF of an aberration-free system with a circular pupil is given, and how it is affected by an aberration is discussed. The concept of a Hopkins ratio, representing the ratio of the magnitudes of the OTFs at a certain spatial frequency, with and without aberration is introduced. It is related to the variance of an aberration difference function across the region of overlap between two displaced pupils, the displacement depending on the spatial frequency under consideration. Aberration tolerances based on a Hopkins ratio of 0.8 are given for primary aberrations. Finally, the OTFs for various amounts of primary aberrations are discussed. | 704 | 3,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-22 | longest | en | 0.926981 |
https://www.airmilescalculator.com/distance/ejn-to-krl/ | 1,624,545,631,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556133.92/warc/CC-MAIN-20210624141035-20210624171035-00029.warc.gz | 554,591,978 | 20,429 | # Distance between Ejin Banner (EJN) and Korla (KRL)
Flight distance from Ejin Banner to Korla (Ejin Banner Taolai Airport – Korla Airport) is 767 miles / 1234 kilometers / 666 nautical miles. Estimated flight time is 1 hour 57 minutes.
Driving distance from Ejin Banner (EJN) to Korla (KRL) is 916 miles / 1474 kilometers and travel time by car is about 16 hours 8 minutes.
## Map of flight path and driving directions from Ejin Banner to Korla.
Shortest flight path between Ejin Banner Taolai Airport (EJN) and Korla Airport (KRL).
## How far is Korla from Ejin Banner?
There are several ways to calculate distances between Ejin Banner and Korla. Here are two common methods:
Vincenty's formula (applied above)
• 766.670 miles
• 1233.836 kilometers
• 666.218 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 764.673 miles
• 1230.622 kilometers
• 664.483 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Ejin Banner Taolai Airport
City: Ejin Banner
Country: China
IATA Code: EJN
ICAO Code: ZBEN
Coordinates: 42°0′55″N, 101°0′1″E
B Korla Airport
City: Korla
Country: China
IATA Code: KRL
ICAO Code: ZWKL
Coordinates: 41°41′52″N, 86°7′44″E
## Time difference and current local times
The time difference between Ejin Banner and Korla is 2 hours. Korla is 2 hours behind Ejin Banner.
CST
+06
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 132 kg (290 pounds).
## Frequent Flyer Miles Calculator
Ejin Banner (EJN) → Korla (KRL).
Distance:
767
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
767
Round trip? | 519 | 1,880 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-25 | latest | en | 0.815295 |
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# Twenty-five adults reported the amount of time each spent st
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08 May 2013, 07:02
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Refer to the attached file.
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Re: Twenty-five adults reported the amount of time each spent st [#permalink]
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16 May 2013, 20:37
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1. To find the least possible mean the sum of all hours exercised must be the minimum.
For 0-1 hours lets assume all 5 exercised the minimum possible of just over 0 hours and same for all bars.
so SUM of hours exercised = 0*5 + 1*3+2*2+3*3+4*4+5*5+0+0+8*1+0+10*0 = 78
# respondents = 25
average = 78/25 = 3.12
2. # respondents who exercised on average <0.5 hours per day
If Avg hours exercised = 0.5 hours per day the Total hours exercised during the week will be 0.5*7 OR 3.5 hours/week
We need to count respondents who exercise below 3.5 hours per week.
That will include all in 0-1 hour, 1-2 hr, 2-3 hour and since 3.5 lies in between 3-4 band, we can take 2 cases:
CAse 1: where we assume all 4 in 3-4 years exercise MORE than 3.5 hours
Total # respondednts in that case = 5 + 3 + 2 + 4 = 14 (this is the maximum)
CAse 2: where we assume all 4 in 3-4 years exercise LESS than 3.5 hours
Total # respondednts in that case = 5 + 3 + 2 = 10 (this is the minimum)
Range od 10-14 respondents
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Joined: 09 Nov 2013
Posts: 10
Re: Twenty-five adults reported the amount of time each spent st [#permalink]
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03 May 2014, 08:03
2
srcc25anu wrote:
1. To find the least possible mean the sum of all hours exercised must be the minimum.
For 0-1 hours lets assume all 5 exercised the minimum possible of just over 0 hours and same for all bars.
so SUM of hours exercised = 0*5 + 1*3+2*2+3*3+4*4+5*5+0+0+8*1+0+10*0 = 78
# respondents = 25
average = 78/25 = 3.12
2. # respondents who exercised on average <0.5 hours per day
If Avg hours exercised = 0.5 hours per day the Total hours exercised during the week will be 0.5*7 OR 3.5 hours/week
We need to count respondents who exercise below 3.5 hours per week.
That will include all in 0-1 hour, 1-2 hr, 2-3 hour and since 3.5 lies in between 3-4 band, we can take 2 cases:
CAse 1: where we assume all 4 in 3-4 years exercise MORE than 3.5 hours
Total # respondednts in that case = 5 + 3 + 2 + 4 = 14 (this is the maximum)
CAse 2: where we assume all 4 in 3-4 years exercise LESS than 3.5 hours
Total # respondednts in that case = 5 + 3 + 2 = 10 (this is the minimum)
Range od 10-14 respondents
Thanks for the explanation! I fully understand the solution, but it took me almost 4 minutes to do it. Anyone know of a faster way to solve this question?
Thanks!
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Posts: 35
Location: India
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GMAT 1: 700 Q51 V32
Re: Twenty-five adults reported the amount of time each spent st [#permalink]
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23 May 2014, 05:16
1
Hi Ashsim,
Part 1 Solution:
Sum of [Number of respondents * Least value of time spent in the range]
= 5*0 + 3*1 + 2*2 + 4*3 + 4*4 + 5*5 + 1*8 + 1*10
= 78
Average = $$\frac{78}{25} = 3.12$$
Part 2 Solution:
Less than 0.5 hr per day = less than 3.5 hr per week
Minimum number = 5 + 3 + 2 + 0 = 10 (When respondents spending time between 3 and 4 have values greater than 3.5)
Maximum Number = 5 + 3 + 2 + 4 = 14 (When respondents spending time between 3 and 4 have values less than 3.5)
I hope it could be solved in 2 mins
Rgds,
Rajat
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Joined: 28 Dec 2013
Posts: 68
Re: Twenty-five adults reported the amount of time each spent st [#permalink]
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06 Aug 2014, 12:01
rajatjain14 wrote:
Hi Ashsim,
Part 1 Solution:
Sum of [Number of respondents * Least value of time spent in the range]
= 5*0 + 3*1 + 2*2 + 4*3 + 4*4 + 5*5 + 1*8 + 1*10
= 78
Average = $$\frac{78}{25} = 3.12$$
Part 2 Solution:
Less than 0.5 hr per day = less than 3.5 hr per week
Minimum number = 5 + 3 + 2 + 0 = 10 (When respondents spending time between 3 and 4 have values greater than 3.5)
Maximum Number = 5 + 3 + 2 + 4 = 14 (When respondents spending time between 3 and 4 have values less than 3.5)
I hope it could be solved in 2 mins
Rgds,
Rajat
Question : How do we know that less than 0.5 hr per day = less than 3.5 hr per week?
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Joined: 12 Nov 2016
Posts: 136
Concentration: Entrepreneurship, Finance
GMAT 1: 620 Q36 V39
GMAT 2: 650 Q47 V33
Re: Twenty-five adults reported the amount of time each spent st [#permalink]
### Show Tags
15 Nov 2017, 01:49
sagnik242 wrote:
rajatjain14 wrote:
Hi Ashsim,
Part 1 Solution:
Sum of [Number of respondents * Least value of time spent in the range]
= 5*0 + 3*1 + 2*2 + 4*3 + 4*4 + 5*5 + 1*8 + 1*10
= 78
Average = $$\frac{78}{25} = 3.12$$
Part 2 Solution:
Less than 0.5 hr per day = less than 3.5 hr per week
Minimum number = 5 + 3 + 2 + 0 = 10 (When respondents spending time between 3 and 4 have values greater than 3.5)
Maximum Number = 5 + 3 + 2 + 4 = 14 (When respondents spending time between 3 and 4 have values less than 3.5)
I hope it could be solved in 2 mins
Rgds,
Rajat
Question : How do we know that less than 0.5 hr per day = less than 3.5 hr per week?
assuming 7 day week (not weekend break ) 7*0.5 = 3.5
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Joined: 17 Mar 2014
Posts: 426
Re: Twenty-five adults reported the amount of time each spent st [#permalink]
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01 Jan 2018, 19:04
Erjan_S wrote:
sagnik242 wrote:
rajatjain14 wrote:
Hi Ashsim,
Part 1 Solution:
Sum of [Number of respondents * Least value of time spent in the range]
= 5*0 + 3*1 + 2*2 + 4*3 + 4*4 + 5*5 + 1*8 + 1*10
= 78
Average = $$\frac{78}{25} = 3.12$$
Part 2 Solution:
Less than 0.5 hr per day = less than 3.5 hr per week
Minimum number = 5 + 3 + 2 + 0 = 10 (When respondents spending time between 3 and 4 have values greater than 3.5)
Maximum Number = 5 + 3 + 2 + 4 = 14 (When respondents spending time between 3 and 4 have values less than 3.5)
I hope it could be solved in 2 mins
Rgds,
Rajat
Question : How do we know that less than 0.5 hr per day = less than 3.5 hr per week?
assuming 7 day week (not weekend break ) 7*0.5 = 3.5
yeah I assumed weekend break and got this question wrong.
Re: Twenty-five adults reported the amount of time each spent st [#permalink] 01 Jan 2018, 19:04
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# Twenty-five adults reported the amount of time each spent st
Moderators: chetan2u, Bunuel | 2,360 | 7,080 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2019-26 | latest | en | 0.883096 |
https://questions.examside.com/past-years/gate/question/pthe-transition-diagram-of-a-discrete-memoryless-channel-w-gate-ece-network-theory-network-elements-ak558phfeathtda6 | 1,718,894,169,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00008.warc.gz | 431,826,750 | 32,603 | 1
GATE ECE 2022
Numerical
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-0.33
The transition diagram of a discrete memoryless channel with three input symbols and three output symbols is shown in the figure. The transition probabilities are as marked. The parameter $$\alpha$$ lies in the interval [0.25, 1]. The value of .. for which the capacity of this channel is maximized, is __________ (rounded off to two decimal places).
Your input ____
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GATE ECE 2022
Numerical
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Consider communication over a memoryless binary symmetric channel using a (7, 4) Hamming code. Each transmitted bit is received correctly with probability (1 $$-$$ $$\in$$), and flipped with probability $$\in$$. For each codeword transmission, the receiver performs minimum Hamming distance decoding, and correctly decodes the message bits if and only if the channel introduces at most one bit error. For $$\in$$ = 0.1, the probability that a transmitted codeword is decoded correctly is __________ (rounded off to two decimal places).
Your input ____
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GATE ECE 2017 Set 2
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Which one of the following graphs shows the Shannon capacity (channel capacity) in bits of a memory less binary symmetric channel with crossover probability P?
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D
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GATE ECE 2017 Set 1
Numerical
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Let $$\left( {{X_1},\,{X_2}} \right)$$ be independent random variables, $${X_1}$$ has mean 0 and variance 1, while $${X_2}$$ has mean 1 and variance 4. The mutual information I $$\left( {{X_1},\,{X_2}} \right)$$ between $${{X_1}}$$ and $${{X_2}}$$ in bits is ________________.
Your input ____
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© ExamGOAL 2024 | 505 | 1,737 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-26 | latest | en | 0.799635 |
https://life-improver.com/rpg/rpg-how-to-roll-for-damage-with-a-critical-hit/ | 1,670,309,599,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711074.68/warc/CC-MAIN-20221206060908-20221206090908-00075.warc.gz | 392,142,864 | 6,532 | # [RPG] How to roll for damage with a critical hit
critical-hitpathfinder-1e
From the Pathfinder Game Master’s Guide:
Critical Hit
Whenever a creature’s attack roll is a 20 on the d20, the attack might be a critical hit. Make another attack roll for the creature, just like the first one. If this second attack roll would hit the target, the original attack is a critical hit and does double damage—roll the damage for the weapon twice (including modifiers) and add the rolls together to find out how much damage you do. […]
To me this explanation sounds ambiguous, so please help me to understand this right. I'll give you an example fight. Let's assume a Goblin is fighting against Valeros the Fighter. Valeros hits first. The stats are as follows:
Goblin
Armor Class: 16
Valeros the Fighter
Weapon: Longsword
Attack Bonus: +4
Weapon Damage: 1d8+4
What I understand is how a critical hit may fail.
Failing Critical Hit
Valeros rolls a 20 on a d20, so his attack might be critical. He rolls again, but only a 11 on the d20. Even with his attack bonus of +4 his roll is lower than the Goblin's armor. The critical hit fails, but since the first roll was highter than the Goblin's armor, Valeros will deal 1d8+4 damage.
How much damage does Valeros do on a critical hit?
Valeros rolls a 20 on a d20, so his attack might be critical. He rolls again, this time 15 on the d20. With his attack bonus of +4 his roll is higher than the Goblin's armor.
The manual says that "the original attack is a critical hit and does double damage". At this point I haven't rolled for any damage. I would assume that the damage roll I do next will double. So 1d8+4 times two could be 2x(8+4)=24.
In the same sentence I read that I have to "roll the damage for the weapon twice". So I could also roll (8+4)(1+4)=17.
So what is true? How do I roll for damage?
The thing I think you're missing here is that the description of "double damage" is all one sentence.
[...the attack] is a critical hit and does double damage—roll the damage for the weapon twice (including modifiers) and add the rolls together to find out how much damage you do.
Let's break it down.
the attack is a critical hit
Cool, now we've determined what it is, what does that mean?
and does double damage
That's a bit ambiguous. There's more than one way to think about "double damage" (as your post demonstrates.) But here comes the clarification:
roll the damage for the weapon twice (including modifiers) and add the rolls together to find out how much damage you do.
That's not a separate thing, that's the explanation of what they mean by "double damage" immediately before it. In your example, you'd roll (1d8+4)+(1d8+4), which might well be 1+4+8+4=17. Really, you're still doubling the damage you do, just before you roll instead of after. | 697 | 2,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-49 | longest | en | 0.945439 |
http://docs.go-mono.com/monodoc.ashx?link=T%3ASystem.Decimal | 1,513,332,319,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948568283.66/warc/CC-MAIN-20171215095015-20171215115015-00006.warc.gz | 73,727,917 | 7,573 | System.Decimal Structure
Represents a decimal number.
## Syntax
[System.Runtime.InteropServices.ComVisible(true)]
public struct Decimal : IComparable, IComparable<decimal>, IConvertible, IEquatable<decimal>, IFormattable, System.Runtime.Serialization.IDeserializationCallback
## Remarks
The decimal value type represents decimal numbers ranging from positive 79,228,162,514,264,337,593,543,950,335 to negative 79,228,162,514,264,337,593,543,950,335. The decimal value type is appropriate for financial calculations that require large numbers of significant integral and fractional digits and no round-off errors. The decimal type does not eliminate the need for rounding. Rather, it minimizes errors due to rounding. For example, the following code produces a result of 0.9999999999999999999999999999 instead of 1.
code reference: System.Decimal.Class#1
When the result of the division and multiplication is passed to the Math.Round(decimal, int) method, the result suffers no loss of precision, as the following code shows.
code reference: System.Decimal.Class#2
A decimal number is a floating-point value that consists of a sign, a numeric value where each digit in the value ranges from 0 to 9, and a scaling factor that indicates the position of a floating decimal point that separates the integral and fractional parts of the numeric value.
The binary representation of a decimal value consists of a 1-bit sign, a 96-bit integer number, and a scaling factor used to divide the 96-bit integer and specify what portion of it is a decimal fraction. The scaling factor is implicitly the number 10, raised to an exponent ranging from 0 to 28. Therefore, the binary representation of a decimal value the form, ((-2 to 2) / 10), where -(2-1) is equal to decimal.MinValue, and 2-1 is equal to decimal.MaxValue. For more information about the binary representation of decimal values and an example, see the decimal.#ctor(Int32[]) constructor and the decimal.GetBits(decimal) method.
The scaling factor also preserves any trailing zeros in a decimal number. Trailing zeros do not affect the value of a decimal number in arithmetic or comparison operations. However, trailing zeros might be revealed by the erload:System.Decimal.ToString method if an appropriate format string is applied.
## Conversion Considerations
This type provides methods that convert decimal values to and from sbyte, short, int, long, byte, ushort, uint, and ulong values. Conversions from these integral types to decimal are widening conversions that never lose information or throw exceptions.
Conversions from decimal to any of the integral types are narrowing conversions that round the decimal value to the nearest integer value toward zero. Some languages, such as C#, also support the conversion of decimal values to char values. If the result of these conversions cannot be represented in the destination type, an OverflowException exception is thrown.
The decimal type also provides methods that convert decimal values to and from float and double values. Conversions from decimal to float or double are narrowing conversions that might lose precision but not information about the magnitude of the converted value. The conversion does not throw an exception.
Conversions from float or double to decimal throw an OverflowException exception if the result of the conversion cannot be represented as a decimal.
## Performing Operations on Decimal Values
The decimal type supports standard mathematical operations such as addition, subtraction, division, multiplication, and unary negation. You can also work directly with the binary representation of a decimal value by calling the decimal.GetBits(decimal) method.
To compare two decimal values, you can use the standard numeric comparison operators, or you can call the decimal.CompareTo(decimal) or decimal.Equals(decimal) method.
You can also call the members of the Math class to perform a wide range of numeric operations, including getting the absolute value of a number, determining the maximum or minimum value of two decimal values, getting the sign of a number, and rounding a number. | 821 | 4,140 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-51 | longest | en | 0.739583 |
https://braingenie.ck12.org/skills/102350/learn | 1,582,579,886,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145981.35/warc/CC-MAIN-20200224193815-20200224223815-00133.warc.gz | 298,729,933 | 1,775 | Sample Problem
Look at the Train schedule and answer the question.
Jim is at the main street station and it is 7:15 AM When is the next train?
Solution
Look at the schedule. We first find main street and see that there is a train at 7:30, so the answer is the first choice. | 68 | 277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.966588 |
https://www.enotes.com/homework-help/add-following-polynomials-3x-2-5x-7-9x-3-4x-2-9-343927 | 1,484,588,334,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279224.13/warc/CC-MAIN-20170116095119-00175-ip-10-171-10-70.ec2.internal.warc.gz | 916,410,092 | 12,584 | # Add the following polynomials: (3x^2 - 5x - 7) and (-9x^3 + 4x^2 - 9)
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
The sum of the polynomials (3x^2 - 5x - 7) and (-9x^3 + 4x^2 - 9) has to be determined.
3x^2 - 5x - 7 + (-9x^3) + 4x^2 - 9
take the terms with the same power of x together
=> -9x^3 + 3x^2 + 4x^2 - 5x - 7 - 9
=> -9x^3 + 7x^2 - 5x - 16
The sum of (3x^2 - 5x - 7) and (-9x^3 + 4x^2 - 9) is -9x^3 + 7x^2 - 5x - 16
dylee | High School Teacher | (Level 1) eNoter
Posted on
In the summation of polynomial, the coefficient of the term with the same variable should be added together.
For example, x^2 with the coefficient 3 should be added to 4x^2, resulting in (4+3)x^2 = 7x^2
In this way, the calculation can be done:
(3x^2-5x-7)+(-9x^3+4x^2-9) = -9x^3+(3+4)x^2-5x-(7+9)
= -9x^3+7x^2-5x-16 | 388 | 844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | longest | en | 0.782091 |
https://universe-review.ca/R15-17-relativity12.htm | 1,718,485,027,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00866.warc.gz | 527,182,401 | 3,125 | ## Relativity, Cosmology, and Time
### Angular-Size Redshift Relation
The angular-size redshift relation describes the relationship between the angular size observed on the sky of an object of given physical size, and the object's redshift. In elementary Euclidean geometry the relation between angular size , linear size l and distance d (from the Earth) would simply be given by the equation:
= l / d
In an expanding universe the distance d is a function of the redshift z:
d = (c/Ho) {qoz + (qo - 1)[(1 + 2qoz)½ - 1]} / [qo2(1 + z)2]
#### Figure 10m Angular-Size / Redshift [view large image]
where qo = (4G / 3) / H02 is the deceleration parameter with the substitution by
Eq.(20e). In standard cosmology qo = 0.5 for flat space, qo > 0.5 for closed space, and qo < 0.5 for open space (see Figure 10m, angular size in unit of mas = milliarcseconds).
As z 0, d (c/Ho) z , and 1/z
while for z , d (c/Ho) / qoz, and z .
These limiting cases clearly demonstrate the curious effect that the angular size of an object becomes larger as it is further away from Earth. It appears to decrease with distance only for nearby objects. Figure 10n shows the infrared blobs produced by the first stars (high z objects). It is suggested that the appearance of the puffy blobs with large angular size is caused by the expansion of the universe with z > 1.6 as shown in Figures 10m and 10n.
#### Figure 10n Infrared Blobs [view large image]
In principle, the angular-size redshift relation can be used to select the type of space for our universe. However, it is notoriously difficult to collect reliable data in practice because the astronomical yardstick can vary in size and in luminosity over time (the evolutionary and selection effects). In addition, we can only measure the projections on the celestial surface according to the orientation of the objects. All past attempts using data from galaxies, the separation of the lobes of radio sources, quasars, and radio galaxies produced inconclusive results. The observational data in Figure 10oa are based on selective compact radio sources. The best fitting regression analysis gives a value of qo 0.21. More recent study in 2004
#### Figure 10oa Model with Cosmological Constant [view large image]
with ultra-compact radio sources find close match for the model of an universe with cosmological constant, m = 0.24, = 0.76, and spatially flat. The observational data range between 0.6 < z < 2.7 with a population mean for the linear size l ~ 6.2 pc. It indicates a "switch over" from deceleration to acceleration at z = 0.85 (see Figure 10oa).
A novel method to circumvent the small size of astronomical objects is to observe the orientations of pair of galaxies. It is suggested that by correcting for redshift and angular size with a correct geometry of the universe, the orientations of distant pairs of galaxies should be completely random, as shown in Diagram a, Figure 10ob. Otherwise, there would be a preferred direction as shown in Diagram b, Figure 10ob. A 2010 study on distant galaxies with z ~ 0.5 indicates that the geometry of the universe is consistent with the standard cosmological model, with its flat curvature. It also shows that the dark energy
#### Figure 10ob Geometry of the Universe [view large image]
is in agreement with being a vacuum energy, which can be represented by Einstein's famous cosmological constant.
. | 816 | 3,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.888875 |
https://easyelimu.com/high-school-notes/maths/form-3-topical/item/1162-further-logarithms | 1,713,039,535,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816832.57/warc/CC-MAIN-20240413180040-20240413210040-00264.warc.gz | 211,920,003 | 23,810 | # Further Logarithms Questions and Answers - Form 3 Topical Mathematics
## Questions
1. In this question, show all the steps in your calculations, giving the answer at each stage. Use logarithms correct to 4 decimal places, to evaluate;
2. The table below shows monthly income tax rates
Monthly taxable pay in KE Rate of the tax (Kshs/E) 1-342 2 343-684 3 685-1026 4 1027-1368 5 1369-1710 6 1710 and above 7
Mr. Kamau who is a civil servant earns a Monthly salary of Kshs.20000 and is provided with a house at a nominal rent of Kshs.700 per month
1. Taxable pay is the employee’s salary plus 15% less nominal rent. Calculate Mr.Kamau’s taxable pay
2. Calculate the total tax Mr. Kamau pays
3. Mr. Kamau is entitled to a personal relief of Kshs.600 per month. What was his net tax.
4. Mr. Kamau has the following deductions made on his pay;
Loan repayment of Kshs.2100 per month
NSSF Kshs.200 per month
WCPS calculated at 2 % of monthly salary
Calculate Mr. Kipchokes net pay
3. A man bought a matatu at Kshs.400,000 in January 1999. It depreciated at a rate of 16% per annum. If he valued it six months, calculate its value in January 2003
4. The table shows corresponding values of x and y for a certain curve;
Using 3 strips and mid-ordinate rule estimate the area between the curve, x-axis, the lines x =1 and x =2.2
5. Evaluate without using a calculator or mathematical tables.
Log 32 + log 128 − log 729
Log 32 + log 2 − log 27
6. Find the value of x that satisfies the equation:-
log (x+5) = log 4 − log(x+2)
7. Find the least number of terms for which the sum of the GP 100 + 200 + 400 + ……………….. exceeds 3100.
8. A two digit number is formed from the first four prime numbers.
1. Draw the table to show the possible outcomes, if each number can be used only once.
2. Calculate the probability that a number chosen from the digit numbers is an even number
9. Find the gradient of a line joining the centre of a circle whose equation is x² + y² − 6² = 3 – 4y and a point P(6,7) outside the circle.
10. A lady invests shs.10,000 in an account which pays 16% interest p.a. The interest is compounded quarterly. Find the amount in the account after 1½ hrs
11. Use logarithm tables to evaluate
12. Without using logarithms or calculator evaluate:
2log10 5 − 3log10 2 + log10 32
13. Evaluate without using tables or calculators.
Log (3x + 8) − 3 log2 = log x − 4
1.
2.
1. monthly taxable pay;
15% of monthly salary = 15/100 x 20000
= kshs.3000
Monthly pay = Kshs.(20000 + 3000 − 700)
= Kshs. 22300
In Kenya pounds = 22300/20
= KE 1115
2. Total tax payable (Gross tax)
1 − 342_________ 342x2 = Kshs.684
343 − 684 _______ 342 x3=Kshs.1026
685 − 1026 ______ 342 x4= Kshs.1368
1027 − 1368 _____ 89x5 = Kshs.445
Total tax = Kshs.3523
3. Net tax
= Gross tax – relief
= Kshs.(3523 – 600) = Kshs.2923
4. Net pay;
= Kshs.20000 – (2923 + 2100 + 200 + 2/100 x 20000)
= Kshs. (20000 – 5623) = Kshs.14377
3. 6 month depreciation rate = 8%
Number of periods = 8
400,000 (1 – 0.08)8 = 205288
4. Mid ordinate
Area = 1.2 (6.2 + 4.3 + 2.6)
= 15.72
5.
6. Log (x+5) = log(4)/(x+2)
x + 5 = 4/(x + 2)
(x+5) (x+2) = 4
x2+ 2x + 5X + 10 = 4
x2+ 7x + 6 = 0
x2+ 6x + x + 6 =0
x(x+6) + 1(x+6) = 0
(x +1) (x +6) = 0
x = -1 x = -6
7. a = 100
r = 200 = 2
100
a (rn -1) Sn
r - 1
100 (2n – 1) 3,100
2 – 1
2n – 1 > 31
2n >32
2n >25
n >5
n = 6
1. 2 3 5 7
2 32 52 72
3 23 53 73
5 25 35 75
7 27 37 57
2. P(E) = 4/16
= ¼
8. x2 + y2 – 6x = 3 – 4y
x2– 6x + (-6/2)2 + y2 + 4y + (4/2)2 = 3 + (-6/2)2 + (4/2)2
(x – 3)2 (y + 2)2 = 3 + 9 = 4
(x – 3)2 (y + 2)2 = 16
C (3, -2)
Gradient y = 7 - -2 = 3
∆x 6 – 3
9. A = P(1 + r/100 )n
= 10000 (1 + 4/100)6
= 10000(1.04)6
= 12653.19 (12,653)
10.
11.
12.
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http://www.cham.co.uk/phoenics/d_earth/d_core/inplib/402.htm | 1,500,928,654,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424910.80/warc/CC-MAIN-20170724202315-20170724222315-00207.warc.gz | 396,856,711 | 2,652 | ```
PHOTON USE
p
parphi
up z
msg Oil concentration (LOG10(oil)
con logo x 1 fil;0.01
mirror y
msg Press to continue
pause
cl
msg Velocity vectors
vec x 1 sh
msg Press to continue
pause
cl
msg Effective viscosity
con enut x 1 fil;0.01
pause
msg non-dimensional velocity
cl
con wdim x 1 fil;0.01
msg Press e to END
ENDUSE
GROUP 1. Run title and other preliminaries
TEXT(Oil tanker rupture - Turbulent plume
TITLE
DISPLAY
This case simulates an oil tanker rupture. The calculation is stead
with a uniform oil release rate which can be varied by changing
the parameter, "rate". The parabolic mode with expanding
grid is employed for both economy and accuracy. The calculation
includes a residence-time equation.
In-Form is used to set the buoyance force and to calculate the
non-dimensional velocity and oil concentration.
The result shows that the flow is a kind of turbulent plume.
ENDDIS
xulast=0.1
GROUP 4. Y-direction grid specification
CARTES=F
** The oil release rate can be varied by changing the
parameter, rate below
rate=20
rate1=rate/(3.6*24*20)
rate
YFRAC(1)=-40.;YFRAC(2)=1.0/40.
*** Linear grid expansion with slope DYGDZ
AZYV=1.0 ! this dictates that yvlast is linear in z
REAL(DYGDZ); DYGDZ=0.33 ! dygdz is d(yvlast)/dz
ZWADD=YVLAST/DYGDZ ! so YVLAST = YVLAST at the inlet
+ DYGDZ * (ZWLAST + ZWADD)
NZ=200
ZWLAST=3500
GROUP 5. Z-direction grid specification
PARAB=T
AZDZ=PROPY ! propy = grnd2; means dz is proportional to y
DZW1 = 0.077 ! the proportionality factor; so dz=dzw1 * yvlast
GRDPWR(Z,NZ,ZWLAST,1.)
GROUP 7. Variables stored, solved & named
STORE(ENUT,LEN1)
SOLVE(P1,V1,W1,OIL)
** solve an equation for oil residence time
SOLVE(REST)
TERMS(REST,N,Y,N,P,P,P)
PATCH(ELAPSE,PHASEM,1,NX,1,NY,1,NZ,1,1)
COVAL(ELAPSE,REST,FIXFLU,1.0)
GROUP 8. Terms (in differential equations) & devices
DIFCUT=0.0
GROUP 9. Properties of the medium (or media)
** water density =1000 and oil density = 999.97
RHO1=1000.
TURMOD(KEMODL);KELIN=3
ENUL=1.e-6
FIINIT(KE)=0.1**2;FIINIT(EP)=0.1643*FIINIT(KE)**1.5/0.1
** dendif is the density difference between water and oil
REAL(DENDIF)
DENDIF=0.03
GROUP 13. Boundary conditions and special sources
** buoyancy force
PATCH(WHOLE,PHASEM,1,NX,1,NY,1,NZ,1,1)
(SOURCE OF W1 AT WHOLE IS 9.81*OIL*DENDIF WITH FIXFLU)
** free stream
PATCH(NORTH,NORTH,1,1,NY,NY,1,NZ,1,1)
COVAL(NORTH,P1,FIXVAL,0.0)
coval(NORTH,REST,1.0,SAME)
** oil release point
PATCH(LEAK,CELL,1,1,1,NY/2,1,1,1,1)
COVAL(LEAK,KE,FIXFLU,FIINIT(KE))
COVAL(LEAK,EP,FIXFLU,FIINIT(EP))
COVAL(LEAK,OIL,FIXFLU,:RATE1:)
** bottom of the sea
PATCH(OUTSIDE,cell,1,1,NY/2+1,NY,1,1,1,1)
COVAL(OUTSIDE,P1,1.E3,0.0)
COVAL(OUTSIDE,W1,ONLYMS,0.0)
GROUP 14. Downstream pressure for PARAB=T
IPARAB=1
GROUP 16. Termination of iterations
LITHYD=50
RELAX(V1,FALSDT,1.)
RELAX(W1,FALSDT,1.)
RELAX(REST,FALSDT,2.0)
GROUP 19. Data communicated by SATELLITE to GROUND
** Select strain-rate for use in Mixing-Length model
DWDY=T
GROUP 21. Print-out of variables
OUTPUT(P1,Y,Y,Y,Y,Y,Y);OUTPUT(V1,Y,Y,Y,Y,Y,Y)
OUTPUT(W1,Y,Y,Y,Y,Y,Y)
GROUP 22. Monitor print-out
IZMON=10;IYMON=1;ITABL=1;NPLT=1;IPLTL=LITHYD;TSTSWP=-3
** parabolic file dumping
TSTSWP=-5;IDISPA=10;IDISPB=1;IDISPC=NZ
GROUP 23. Field print-out and plot control
libref=402
(stored var logo is log10(oil))
(store var wdim is w1*(zzw/:zwlast:)^0.3333)
(store var odim is oil*(zzw/:zwlast:)^1.6667)
patch(axis,profil,1,1,1,1,1,nz,1,1)
coval(axis,w1,0.0,0.0)
coval(axis,oil,0.0,0.0)
coval(axis,logo,0.0,0.0)
coval(axis,wdim,0.0,0.0)
coval(axis,odim,0.0,0.0)
coval(axis,rest,0.0,0.0)
``` | 1,373 | 3,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.728096 |
http://onebuilding.org/archive/bldg-rate-onebuilding.org/2011-June/000136.html | 1,597,273,480,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738950.31/warc/CC-MAIN-20200812225607-20200813015607-00102.warc.gz | 81,564,072 | 2,379 | # [Bldg-rate] Fan Power
Jason Wendel jwendel at heatheng.com
Fri Jun 17 07:00:22 PDT 2011
```So, which is it, I have received LEED Review comments back stating two different ways to calculate fan power and no matter what I do, the next reviewer for the next project sees it differently and has me revise it.
1. Calculate the baseline fan power from the sum of all the supply, return & exhaust airflows combined to get a total fan power. Then proportion out the total fan power between each of the fans (supply, return, exhaust, etc).
2. Calculate the baseline fan power for the supply fan only, and then proportion out the power from that calculation into each of the fans (supply, return, exhaust, etc.)
Method #1 takes all the airflow in all the fans in the building added together, calculates the fan power to move that much air, and then proportions it out to each of the fans. For example, 100,000 cfm supply air, 90,000 return air and 10,000 exhaust air. Totals to 200,000 cfm airflow in the building. Calculate the airflow for the 200,000 cfm and lets say you get 200 kW. Then you proportion out the fan power between the fans - but the sum is 200 kW of power.
So, method #2 doesn't make sense to me. You take calculate the amount of power required to blow the supply air (lets say 100,000 cfm, for example) throughout the building (say 100 kW for simplicity), and then say that the 100 kW is then divided out between all the fans in the building, resulting in the supply fan using ½ of the power that you calculated it should, and the rest of the power is divided out between each of the other fans in the building. So you calculate that 100 kW of energy is needed to move 100,000 cfm of air, but then you change your mind and say that the baseline system is actually only going to use 50 kW of energy and the other 50 kW is used in the return and exhaust fans? That doesn't make sense to me.
So, which is it? I've received review comments back from LEED stating each of them, for different projects. It appears to be interpreted differently from different LEED Reviewers so apparently they don't know either.
-Jason
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``` | 569 | 2,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-34 | latest | en | 0.939754 |
http://www.jiskha.com/display.cgi?id=1196990787 | 1,498,535,753,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320915.38/warc/CC-MAIN-20170627032130-20170627052130-00475.warc.gz | 567,053,513 | 3,974 | # math, I think I have it now...
posted by .
Is this the correct way to solve this?
5+3x4-2=
3x4=12
+5= 17
-2+15
• math, I think I have it now... -
yes!!!! you got it now.
• math, I think I have it now... -
Hurray!!!
Thanks so much for your help!!!
• math, I think I have it now... -
-9r^4y^2(-3ry^7+2r^3y^4-8r^10)= | 125 | 324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.966591 |
http://www.oxfordmathcenter.com/drupal7/node/199 | 1,540,159,391,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514355.90/warc/CC-MAIN-20181021203102-20181021224602-00035.warc.gz | 515,739,385 | 4,602 | # Exercises - The Collatz Conjecture
1. Suppose we define $f(x)$ in the following way, for positive integers $x$:
$$f(x)= \left\{ \begin{array}{ll} x/2 & \textrm{ if x is even}\\ 3x+1 & \textrm{ if x is odd}\\ \end{array} \right.$$
1. Construct the sequence of values $a_0, a_1, a_2, ...$ so that $a_0 = n$ and $a_{i+1} = f(a_{i})$. Verify that this sequence ultimately gets stuck in the loop 4, 2, 1, 4, 2, 1, ... for the following values of $n$:
1. $n=21$
2. $n=13$
3. $n=31$
2. Find the sequence described in part (a) for other values of $n$. What do you notice?
3. Suppose we use $L(n)$ to denote the position where $1$ first occurs in the sequence so generated and starting with $n$. Show that if $n=8k+4$, then $L(n)=L(n+1)$.
4. Show that if $n=128k+28$, then $L(n)=L(n+1)=L(n+2)$.
◆ ◆ ◆ | 300 | 798 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-43 | latest | en | 0.71471 |
https://us.metamath.org/mpeuni/cantnflt.html | 1,716,061,344,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057494.65/warc/CC-MAIN-20240518183301-20240518213301-00580.warc.gz | 538,361,105 | 20,196 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > cantnflt Structured version Visualization version GIF version
Theorem cantnflt 9113
Description: An upper bound on the partial sums of the CNF function. Since each term dominates all previous terms, by induction we can bound the whole sum with any exponent 𝐴 ↑o 𝐶 where 𝐶 is larger than any exponent (𝐺‘𝑥), 𝑥 ∈ 𝐾 which has been summed so far. (Contributed by Mario Carneiro, 28-May-2015.) (Revised by AV, 29-Jun-2019.)
Hypotheses
Ref Expression
cantnfs.s 𝑆 = dom (𝐴 CNF 𝐵)
cantnfs.a (𝜑𝐴 ∈ On)
cantnfs.b (𝜑𝐵 ∈ On)
cantnfcl.g 𝐺 = OrdIso( E , (𝐹 supp ∅))
cantnfcl.f (𝜑𝐹𝑆)
cantnfval.h 𝐻 = seqω((𝑘 ∈ V, 𝑧 ∈ V ↦ (((𝐴o (𝐺𝑘)) ·o (𝐹‘(𝐺𝑘))) +o 𝑧)), ∅)
cantnflt.a (𝜑 → ∅ ∈ 𝐴)
cantnflt.k (𝜑𝐾 ∈ suc dom 𝐺)
cantnflt.c (𝜑𝐶 ∈ On)
cantnflt.s (𝜑 → (𝐺𝐾) ⊆ 𝐶)
Assertion
Ref Expression
cantnflt (𝜑 → (𝐻𝐾) ∈ (𝐴o 𝐶))
Distinct variable groups: 𝑧,𝑘,𝐵 𝑧,𝐶 𝐴,𝑘,𝑧 𝑘,𝐹,𝑧 𝑆,𝑘,𝑧 𝑘,𝐺,𝑧 𝑘,𝐾,𝑧 𝜑,𝑘,𝑧
Allowed substitution hints: 𝐶(𝑘) 𝐻(𝑧,𝑘)
Proof of Theorem cantnflt
Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 cantnfs.a . . . 4 (𝜑𝐴 ∈ On)
2 cantnflt.c . . . 4 (𝜑𝐶 ∈ On)
3 cantnflt.a . . . 4 (𝜑 → ∅ ∈ 𝐴)
4 oen0 8190 . . . 4 (((𝐴 ∈ On ∧ 𝐶 ∈ On) ∧ ∅ ∈ 𝐴) → ∅ ∈ (𝐴o 𝐶))
51, 2, 3, 4syl21anc 835 . . 3 (𝜑 → ∅ ∈ (𝐴o 𝐶))
6 fveq2 6646 . . . . 5 (𝐾 = ∅ → (𝐻𝐾) = (𝐻‘∅))
7 0ex 5187 . . . . . 6 ∅ ∈ V
8 cantnfval.h . . . . . . 7 𝐻 = seqω((𝑘 ∈ V, 𝑧 ∈ V ↦ (((𝐴o (𝐺𝑘)) ·o (𝐹‘(𝐺𝑘))) +o 𝑧)), ∅)
98seqom0g 8070 . . . . . 6 (∅ ∈ V → (𝐻‘∅) = ∅)
107, 9ax-mp 5 . . . . 5 (𝐻‘∅) = ∅
116, 10syl6eq 2871 . . . 4 (𝐾 = ∅ → (𝐻𝐾) = ∅)
1211eleq1d 2895 . . 3 (𝐾 = ∅ → ((𝐻𝐾) ∈ (𝐴o 𝐶) ↔ ∅ ∈ (𝐴o 𝐶)))
135, 12syl5ibrcom 249 . 2 (𝜑 → (𝐾 = ∅ → (𝐻𝐾) ∈ (𝐴o 𝐶)))
142adantr 483 . . . . . . 7 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → 𝐶 ∈ On)
15 eloni 6177 . . . . . . 7 (𝐶 ∈ On → Ord 𝐶)
1614, 15syl 17 . . . . . 6 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → Ord 𝐶)
17 cantnflt.s . . . . . . . 8 (𝜑 → (𝐺𝐾) ⊆ 𝐶)
1817adantr 483 . . . . . . 7 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → (𝐺𝐾) ⊆ 𝐶)
19 cantnfcl.g . . . . . . . . . 10 𝐺 = OrdIso( E , (𝐹 supp ∅))
2019oif 8972 . . . . . . . . 9 𝐺:dom 𝐺⟶(𝐹 supp ∅)
21 ffn 6490 . . . . . . . . 9 (𝐺:dom 𝐺⟶(𝐹 supp ∅) → 𝐺 Fn dom 𝐺)
2220, 21mp1i 13 . . . . . . . 8 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → 𝐺 Fn dom 𝐺)
23 cantnflt.k . . . . . . . . . 10 (𝜑𝐾 ∈ suc dom 𝐺)
2419oicl 8971 . . . . . . . . . . . . 13 Ord dom 𝐺
25 ordsuc 7507 . . . . . . . . . . . . 13 (Ord dom 𝐺 ↔ Ord suc dom 𝐺)
2624, 25mpbi 232 . . . . . . . . . . . 12 Ord suc dom 𝐺
27 ordelon 6191 . . . . . . . . . . . 12 ((Ord suc dom 𝐺𝐾 ∈ suc dom 𝐺) → 𝐾 ∈ On)
2826, 23, 27sylancr 589 . . . . . . . . . . 11 (𝜑𝐾 ∈ On)
29 ordsssuc 6253 . . . . . . . . . . 11 ((𝐾 ∈ On ∧ Ord dom 𝐺) → (𝐾 ⊆ dom 𝐺𝐾 ∈ suc dom 𝐺))
3028, 24, 29sylancl 588 . . . . . . . . . 10 (𝜑 → (𝐾 ⊆ dom 𝐺𝐾 ∈ suc dom 𝐺))
3123, 30mpbird 259 . . . . . . . . 9 (𝜑𝐾 ⊆ dom 𝐺)
3231adantr 483 . . . . . . . 8 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → 𝐾 ⊆ dom 𝐺)
33 vex 3476 . . . . . . . . . 10 𝑥 ∈ V
3433sucid 6246 . . . . . . . . 9 𝑥 ∈ suc 𝑥
35 simprr 771 . . . . . . . . 9 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → 𝐾 = suc 𝑥)
3634, 35eleqtrrid 2918 . . . . . . . 8 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → 𝑥𝐾)
37 fnfvima 6972 . . . . . . . 8 ((𝐺 Fn dom 𝐺𝐾 ⊆ dom 𝐺𝑥𝐾) → (𝐺𝑥) ∈ (𝐺𝐾))
3822, 32, 36, 37syl3anc 1367 . . . . . . 7 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → (𝐺𝑥) ∈ (𝐺𝐾))
3918, 38sseldd 3947 . . . . . 6 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → (𝐺𝑥) ∈ 𝐶)
40 ordsucss 7511 . . . . . 6 (Ord 𝐶 → ((𝐺𝑥) ∈ 𝐶 → suc (𝐺𝑥) ⊆ 𝐶))
4116, 39, 40sylc 65 . . . . 5 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → suc (𝐺𝑥) ⊆ 𝐶)
42 suppssdm 7821 . . . . . . . . . . 11 (𝐹 supp ∅) ⊆ dom 𝐹
43 cantnfcl.f . . . . . . . . . . . . 13 (𝜑𝐹𝑆)
44 cantnfs.s . . . . . . . . . . . . . 14 𝑆 = dom (𝐴 CNF 𝐵)
45 cantnfs.b . . . . . . . . . . . . . 14 (𝜑𝐵 ∈ On)
4644, 1, 45cantnfs 9107 . . . . . . . . . . . . 13 (𝜑 → (𝐹𝑆 ↔ (𝐹:𝐵𝐴𝐹 finSupp ∅)))
4743, 46mpbid 234 . . . . . . . . . . . 12 (𝜑 → (𝐹:𝐵𝐴𝐹 finSupp ∅))
4847simpld 497 . . . . . . . . . . 11 (𝜑𝐹:𝐵𝐴)
4942, 48fssdm 6506 . . . . . . . . . 10 (𝜑 → (𝐹 supp ∅) ⊆ 𝐵)
50 onss 7483 . . . . . . . . . . 11 (𝐵 ∈ On → 𝐵 ⊆ On)
5145, 50syl 17 . . . . . . . . . 10 (𝜑𝐵 ⊆ On)
5249, 51sstrd 3956 . . . . . . . . 9 (𝜑 → (𝐹 supp ∅) ⊆ On)
5352adantr 483 . . . . . . . 8 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → (𝐹 supp ∅) ⊆ On)
5423adantr 483 . . . . . . . . . . 11 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → 𝐾 ∈ suc dom 𝐺)
5535, 54eqeltrrd 2912 . . . . . . . . . 10 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → suc 𝑥 ∈ suc dom 𝐺)
56 ordsucelsuc 7515 . . . . . . . . . . 11 (Ord dom 𝐺 → (𝑥 ∈ dom 𝐺 ↔ suc 𝑥 ∈ suc dom 𝐺))
5724, 56ax-mp 5 . . . . . . . . . 10 (𝑥 ∈ dom 𝐺 ↔ suc 𝑥 ∈ suc dom 𝐺)
5855, 57sylibr 236 . . . . . . . . 9 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → 𝑥 ∈ dom 𝐺)
5920ffvelrni 6826 . . . . . . . . 9 (𝑥 ∈ dom 𝐺 → (𝐺𝑥) ∈ (𝐹 supp ∅))
6058, 59syl 17 . . . . . . . 8 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → (𝐺𝑥) ∈ (𝐹 supp ∅))
6153, 60sseldd 3947 . . . . . . 7 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → (𝐺𝑥) ∈ On)
62 suceloni 7506 . . . . . . 7 ((𝐺𝑥) ∈ On → suc (𝐺𝑥) ∈ On)
6361, 62syl 17 . . . . . 6 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → suc (𝐺𝑥) ∈ On)
641adantr 483 . . . . . 6 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → 𝐴 ∈ On)
653adantr 483 . . . . . 6 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → ∅ ∈ 𝐴)
66 oewordi 8195 . . . . . 6 (((suc (𝐺𝑥) ∈ On ∧ 𝐶 ∈ On ∧ 𝐴 ∈ On) ∧ ∅ ∈ 𝐴) → (suc (𝐺𝑥) ⊆ 𝐶 → (𝐴o suc (𝐺𝑥)) ⊆ (𝐴o 𝐶)))
6763, 14, 64, 65, 66syl31anc 1369 . . . . 5 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → (suc (𝐺𝑥) ⊆ 𝐶 → (𝐴o suc (𝐺𝑥)) ⊆ (𝐴o 𝐶)))
6841, 67mpd 15 . . . 4 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → (𝐴o suc (𝐺𝑥)) ⊆ (𝐴o 𝐶))
6935fveq2d 6650 . . . . 5 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → (𝐻𝐾) = (𝐻‘suc 𝑥))
70 simprl 769 . . . . . 6 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → 𝑥 ∈ ω)
71 simpl 485 . . . . . 6 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → 𝜑)
72 eleq1 2898 . . . . . . . 8 (𝑥 = ∅ → (𝑥 ∈ dom 𝐺 ↔ ∅ ∈ dom 𝐺))
73 suceq 6232 . . . . . . . . . 10 (𝑥 = ∅ → suc 𝑥 = suc ∅)
7473fveq2d 6650 . . . . . . . . 9 (𝑥 = ∅ → (𝐻‘suc 𝑥) = (𝐻‘suc ∅))
75 fveq2 6646 . . . . . . . . . . 11 (𝑥 = ∅ → (𝐺𝑥) = (𝐺‘∅))
76 suceq 6232 . . . . . . . . . . 11 ((𝐺𝑥) = (𝐺‘∅) → suc (𝐺𝑥) = suc (𝐺‘∅))
7775, 76syl 17 . . . . . . . . . 10 (𝑥 = ∅ → suc (𝐺𝑥) = suc (𝐺‘∅))
7877oveq2d 7149 . . . . . . . . 9 (𝑥 = ∅ → (𝐴o suc (𝐺𝑥)) = (𝐴o suc (𝐺‘∅)))
7974, 78eleq12d 2905 . . . . . . . 8 (𝑥 = ∅ → ((𝐻‘suc 𝑥) ∈ (𝐴o suc (𝐺𝑥)) ↔ (𝐻‘suc ∅) ∈ (𝐴o suc (𝐺‘∅))))
8072, 79imbi12d 347 . . . . . . 7 (𝑥 = ∅ → ((𝑥 ∈ dom 𝐺 → (𝐻‘suc 𝑥) ∈ (𝐴o suc (𝐺𝑥))) ↔ (∅ ∈ dom 𝐺 → (𝐻‘suc ∅) ∈ (𝐴o suc (𝐺‘∅)))))
81 eleq1 2898 . . . . . . . 8 (𝑥 = 𝑦 → (𝑥 ∈ dom 𝐺𝑦 ∈ dom 𝐺))
82 suceq 6232 . . . . . . . . . 10 (𝑥 = 𝑦 → suc 𝑥 = suc 𝑦)
8382fveq2d 6650 . . . . . . . . 9 (𝑥 = 𝑦 → (𝐻‘suc 𝑥) = (𝐻‘suc 𝑦))
84 fveq2 6646 . . . . . . . . . . 11 (𝑥 = 𝑦 → (𝐺𝑥) = (𝐺𝑦))
85 suceq 6232 . . . . . . . . . . 11 ((𝐺𝑥) = (𝐺𝑦) → suc (𝐺𝑥) = suc (𝐺𝑦))
8684, 85syl 17 . . . . . . . . . 10 (𝑥 = 𝑦 → suc (𝐺𝑥) = suc (𝐺𝑦))
8786oveq2d 7149 . . . . . . . . 9 (𝑥 = 𝑦 → (𝐴o suc (𝐺𝑥)) = (𝐴o suc (𝐺𝑦)))
8883, 87eleq12d 2905 . . . . . . . 8 (𝑥 = 𝑦 → ((𝐻‘suc 𝑥) ∈ (𝐴o suc (𝐺𝑥)) ↔ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦))))
8981, 88imbi12d 347 . . . . . . 7 (𝑥 = 𝑦 → ((𝑥 ∈ dom 𝐺 → (𝐻‘suc 𝑥) ∈ (𝐴o suc (𝐺𝑥))) ↔ (𝑦 ∈ dom 𝐺 → (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))))
90 eleq1 2898 . . . . . . . 8 (𝑥 = suc 𝑦 → (𝑥 ∈ dom 𝐺 ↔ suc 𝑦 ∈ dom 𝐺))
91 suceq 6232 . . . . . . . . . 10 (𝑥 = suc 𝑦 → suc 𝑥 = suc suc 𝑦)
9291fveq2d 6650 . . . . . . . . 9 (𝑥 = suc 𝑦 → (𝐻‘suc 𝑥) = (𝐻‘suc suc 𝑦))
93 fveq2 6646 . . . . . . . . . . 11 (𝑥 = suc 𝑦 → (𝐺𝑥) = (𝐺‘suc 𝑦))
94 suceq 6232 . . . . . . . . . . 11 ((𝐺𝑥) = (𝐺‘suc 𝑦) → suc (𝐺𝑥) = suc (𝐺‘suc 𝑦))
9593, 94syl 17 . . . . . . . . . 10 (𝑥 = suc 𝑦 → suc (𝐺𝑥) = suc (𝐺‘suc 𝑦))
9695oveq2d 7149 . . . . . . . . 9 (𝑥 = suc 𝑦 → (𝐴o suc (𝐺𝑥)) = (𝐴o suc (𝐺‘suc 𝑦)))
9792, 96eleq12d 2905 . . . . . . . 8 (𝑥 = suc 𝑦 → ((𝐻‘suc 𝑥) ∈ (𝐴o suc (𝐺𝑥)) ↔ (𝐻‘suc suc 𝑦) ∈ (𝐴o suc (𝐺‘suc 𝑦))))
9890, 97imbi12d 347 . . . . . . 7 (𝑥 = suc 𝑦 → ((𝑥 ∈ dom 𝐺 → (𝐻‘suc 𝑥) ∈ (𝐴o suc (𝐺𝑥))) ↔ (suc 𝑦 ∈ dom 𝐺 → (𝐻‘suc suc 𝑦) ∈ (𝐴o suc (𝐺‘suc 𝑦)))))
9948adantr 483 . . . . . . . . . . 11 ((𝜑 ∧ ∅ ∈ dom 𝐺) → 𝐹:𝐵𝐴)
10020ffvelrni 6826 . . . . . . . . . . . 12 (∅ ∈ dom 𝐺 → (𝐺‘∅) ∈ (𝐹 supp ∅))
10149sselda 3946 . . . . . . . . . . . 12 ((𝜑 ∧ (𝐺‘∅) ∈ (𝐹 supp ∅)) → (𝐺‘∅) ∈ 𝐵)
102100, 101sylan2 594 . . . . . . . . . . 11 ((𝜑 ∧ ∅ ∈ dom 𝐺) → (𝐺‘∅) ∈ 𝐵)
10399, 102ffvelrnd 6828 . . . . . . . . . 10 ((𝜑 ∧ ∅ ∈ dom 𝐺) → (𝐹‘(𝐺‘∅)) ∈ 𝐴)
1041adantr 483 . . . . . . . . . . . 12 ((𝜑 ∧ ∅ ∈ dom 𝐺) → 𝐴 ∈ On)
105 onelon 6192 . . . . . . . . . . . 12 ((𝐴 ∈ On ∧ (𝐹‘(𝐺‘∅)) ∈ 𝐴) → (𝐹‘(𝐺‘∅)) ∈ On)
106104, 103, 105syl2anc 586 . . . . . . . . . . 11 ((𝜑 ∧ ∅ ∈ dom 𝐺) → (𝐹‘(𝐺‘∅)) ∈ On)
10752sselda 3946 . . . . . . . . . . . . 13 ((𝜑 ∧ (𝐺‘∅) ∈ (𝐹 supp ∅)) → (𝐺‘∅) ∈ On)
108100, 107sylan2 594 . . . . . . . . . . . 12 ((𝜑 ∧ ∅ ∈ dom 𝐺) → (𝐺‘∅) ∈ On)
109 oecl 8140 . . . . . . . . . . . 12 ((𝐴 ∈ On ∧ (𝐺‘∅) ∈ On) → (𝐴o (𝐺‘∅)) ∈ On)
110104, 108, 109syl2anc 586 . . . . . . . . . . 11 ((𝜑 ∧ ∅ ∈ dom 𝐺) → (𝐴o (𝐺‘∅)) ∈ On)
1113adantr 483 . . . . . . . . . . . 12 ((𝜑 ∧ ∅ ∈ dom 𝐺) → ∅ ∈ 𝐴)
112 oen0 8190 . . . . . . . . . . . 12 (((𝐴 ∈ On ∧ (𝐺‘∅) ∈ On) ∧ ∅ ∈ 𝐴) → ∅ ∈ (𝐴o (𝐺‘∅)))
113104, 108, 111, 112syl21anc 835 . . . . . . . . . . 11 ((𝜑 ∧ ∅ ∈ dom 𝐺) → ∅ ∈ (𝐴o (𝐺‘∅)))
114 omord2 8171 . . . . . . . . . . 11 ((((𝐹‘(𝐺‘∅)) ∈ On ∧ 𝐴 ∈ On ∧ (𝐴o (𝐺‘∅)) ∈ On) ∧ ∅ ∈ (𝐴o (𝐺‘∅))) → ((𝐹‘(𝐺‘∅)) ∈ 𝐴 ↔ ((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))) ∈ ((𝐴o (𝐺‘∅)) ·o 𝐴)))
115106, 104, 110, 113, 114syl31anc 1369 . . . . . . . . . 10 ((𝜑 ∧ ∅ ∈ dom 𝐺) → ((𝐹‘(𝐺‘∅)) ∈ 𝐴 ↔ ((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))) ∈ ((𝐴o (𝐺‘∅)) ·o 𝐴)))
116103, 115mpbid 234 . . . . . . . . 9 ((𝜑 ∧ ∅ ∈ dom 𝐺) → ((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))) ∈ ((𝐴o (𝐺‘∅)) ·o 𝐴))
117 peano1 7579 . . . . . . . . . . . 12 ∅ ∈ ω
118117a1i 11 . . . . . . . . . . 11 (∅ ∈ dom 𝐺 → ∅ ∈ ω)
11944, 1, 45, 19, 43, 8cantnfsuc 9111 . . . . . . . . . . 11 ((𝜑 ∧ ∅ ∈ ω) → (𝐻‘suc ∅) = (((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))) +o (𝐻‘∅)))
120118, 119sylan2 594 . . . . . . . . . 10 ((𝜑 ∧ ∅ ∈ dom 𝐺) → (𝐻‘suc ∅) = (((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))) +o (𝐻‘∅)))
12110oveq2i 7144 . . . . . . . . . . 11 (((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))) +o (𝐻‘∅)) = (((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))) +o ∅)
122 omcl 8139 . . . . . . . . . . . . 13 (((𝐴o (𝐺‘∅)) ∈ On ∧ (𝐹‘(𝐺‘∅)) ∈ On) → ((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))) ∈ On)
123110, 106, 122syl2anc 586 . . . . . . . . . . . 12 ((𝜑 ∧ ∅ ∈ dom 𝐺) → ((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))) ∈ On)
124 oa0 8119 . . . . . . . . . . . 12 (((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))) ∈ On → (((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))) +o ∅) = ((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))))
125123, 124syl 17 . . . . . . . . . . 11 ((𝜑 ∧ ∅ ∈ dom 𝐺) → (((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))) +o ∅) = ((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))))
126121, 125syl5eq 2867 . . . . . . . . . 10 ((𝜑 ∧ ∅ ∈ dom 𝐺) → (((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))) +o (𝐻‘∅)) = ((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))))
127120, 126eqtrd 2855 . . . . . . . . 9 ((𝜑 ∧ ∅ ∈ dom 𝐺) → (𝐻‘suc ∅) = ((𝐴o (𝐺‘∅)) ·o (𝐹‘(𝐺‘∅))))
128 oesuc 8130 . . . . . . . . . 10 ((𝐴 ∈ On ∧ (𝐺‘∅) ∈ On) → (𝐴o suc (𝐺‘∅)) = ((𝐴o (𝐺‘∅)) ·o 𝐴))
129104, 108, 128syl2anc 586 . . . . . . . . 9 ((𝜑 ∧ ∅ ∈ dom 𝐺) → (𝐴o suc (𝐺‘∅)) = ((𝐴o (𝐺‘∅)) ·o 𝐴))
130116, 127, 1293eltr4d 2926 . . . . . . . 8 ((𝜑 ∧ ∅ ∈ dom 𝐺) → (𝐻‘suc ∅) ∈ (𝐴o suc (𝐺‘∅)))
131130ex 415 . . . . . . 7 (𝜑 → (∅ ∈ dom 𝐺 → (𝐻‘suc ∅) ∈ (𝐴o suc (𝐺‘∅))))
132 ordtr 6181 . . . . . . . . . . . 12 (Ord dom 𝐺 → Tr dom 𝐺)
13324, 132ax-mp 5 . . . . . . . . . . 11 Tr dom 𝐺
134 trsuc 6251 . . . . . . . . . . 11 ((Tr dom 𝐺 ∧ suc 𝑦 ∈ dom 𝐺) → 𝑦 ∈ dom 𝐺)
135133, 134mpan 688 . . . . . . . . . 10 (suc 𝑦 ∈ dom 𝐺𝑦 ∈ dom 𝐺)
136135imim1i 63 . . . . . . . . 9 ((𝑦 ∈ dom 𝐺 → (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦))) → (suc 𝑦 ∈ dom 𝐺 → (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦))))
1371ad2antrr 724 . . . . . . . . . . . . . . . 16 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → 𝐴 ∈ On)
138 eloni 6177 . . . . . . . . . . . . . . . 16 (𝐴 ∈ On → Ord 𝐴)
139137, 138syl 17 . . . . . . . . . . . . . . 15 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → Ord 𝐴)
14048ad2antrr 724 . . . . . . . . . . . . . . . 16 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → 𝐹:𝐵𝐴)
14149ad2antrr 724 . . . . . . . . . . . . . . . . 17 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐹 supp ∅) ⊆ 𝐵)
14220ffvelrni 6826 . . . . . . . . . . . . . . . . . 18 (suc 𝑦 ∈ dom 𝐺 → (𝐺‘suc 𝑦) ∈ (𝐹 supp ∅))
143142ad2antrl 726 . . . . . . . . . . . . . . . . 17 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐺‘suc 𝑦) ∈ (𝐹 supp ∅))
144141, 143sseldd 3947 . . . . . . . . . . . . . . . 16 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐺‘suc 𝑦) ∈ 𝐵)
145140, 144ffvelrnd 6828 . . . . . . . . . . . . . . 15 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐹‘(𝐺‘suc 𝑦)) ∈ 𝐴)
146 ordsucss 7511 . . . . . . . . . . . . . . 15 (Ord 𝐴 → ((𝐹‘(𝐺‘suc 𝑦)) ∈ 𝐴 → suc (𝐹‘(𝐺‘suc 𝑦)) ⊆ 𝐴))
147139, 145, 146sylc 65 . . . . . . . . . . . . . 14 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → suc (𝐹‘(𝐺‘suc 𝑦)) ⊆ 𝐴)
148 onelon 6192 . . . . . . . . . . . . . . . . 17 ((𝐴 ∈ On ∧ (𝐹‘(𝐺‘suc 𝑦)) ∈ 𝐴) → (𝐹‘(𝐺‘suc 𝑦)) ∈ On)
149137, 145, 148syl2anc 586 . . . . . . . . . . . . . . . 16 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐹‘(𝐺‘suc 𝑦)) ∈ On)
150 suceloni 7506 . . . . . . . . . . . . . . . 16 ((𝐹‘(𝐺‘suc 𝑦)) ∈ On → suc (𝐹‘(𝐺‘suc 𝑦)) ∈ On)
151149, 150syl 17 . . . . . . . . . . . . . . 15 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → suc (𝐹‘(𝐺‘suc 𝑦)) ∈ On)
15252ad2antrr 724 . . . . . . . . . . . . . . . . 17 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐹 supp ∅) ⊆ On)
153152, 143sseldd 3947 . . . . . . . . . . . . . . . 16 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐺‘suc 𝑦) ∈ On)
154 oecl 8140 . . . . . . . . . . . . . . . 16 ((𝐴 ∈ On ∧ (𝐺‘suc 𝑦) ∈ On) → (𝐴o (𝐺‘suc 𝑦)) ∈ On)
155137, 153, 154syl2anc 586 . . . . . . . . . . . . . . 15 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐴o (𝐺‘suc 𝑦)) ∈ On)
156 omwordi 8175 . . . . . . . . . . . . . . 15 ((suc (𝐹‘(𝐺‘suc 𝑦)) ∈ On ∧ 𝐴 ∈ On ∧ (𝐴o (𝐺‘suc 𝑦)) ∈ On) → (suc (𝐹‘(𝐺‘suc 𝑦)) ⊆ 𝐴 → ((𝐴o (𝐺‘suc 𝑦)) ·o suc (𝐹‘(𝐺‘suc 𝑦))) ⊆ ((𝐴o (𝐺‘suc 𝑦)) ·o 𝐴)))
157151, 137, 155, 156syl3anc 1367 . . . . . . . . . . . . . 14 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (suc (𝐹‘(𝐺‘suc 𝑦)) ⊆ 𝐴 → ((𝐴o (𝐺‘suc 𝑦)) ·o suc (𝐹‘(𝐺‘suc 𝑦))) ⊆ ((𝐴o (𝐺‘suc 𝑦)) ·o 𝐴)))
158147, 157mpd 15 . . . . . . . . . . . . 13 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → ((𝐴o (𝐺‘suc 𝑦)) ·o suc (𝐹‘(𝐺‘suc 𝑦))) ⊆ ((𝐴o (𝐺‘suc 𝑦)) ·o 𝐴))
159 oesuc 8130 . . . . . . . . . . . . . 14 ((𝐴 ∈ On ∧ (𝐺‘suc 𝑦) ∈ On) → (𝐴o suc (𝐺‘suc 𝑦)) = ((𝐴o (𝐺‘suc 𝑦)) ·o 𝐴))
160137, 153, 159syl2anc 586 . . . . . . . . . . . . 13 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐴o suc (𝐺‘suc 𝑦)) = ((𝐴o (𝐺‘suc 𝑦)) ·o 𝐴))
161158, 160sseqtrrd 3987 . . . . . . . . . . . 12 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → ((𝐴o (𝐺‘suc 𝑦)) ·o suc (𝐹‘(𝐺‘suc 𝑦))) ⊆ (𝐴o suc (𝐺‘suc 𝑦)))
162 eloni 6177 . . . . . . . . . . . . . . . . . 18 ((𝐺‘suc 𝑦) ∈ On → Ord (𝐺‘suc 𝑦))
163153, 162syl 17 . . . . . . . . . . . . . . . . 17 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → Ord (𝐺‘suc 𝑦))
164 vex 3476 . . . . . . . . . . . . . . . . . . . . 21 𝑦 ∈ V
165164sucid 6246 . . . . . . . . . . . . . . . . . . . 20 𝑦 ∈ suc 𝑦
166164sucex 7504 . . . . . . . . . . . . . . . . . . . . 21 suc 𝑦 ∈ V
167166epeli 5444 . . . . . . . . . . . . . . . . . . . 20 (𝑦 E suc 𝑦𝑦 ∈ suc 𝑦)
168165, 167mpbir 233 . . . . . . . . . . . . . . . . . . 19 𝑦 E suc 𝑦
169 ovexd 7168 . . . . . . . . . . . . . . . . . . . . . 22 (𝜑 → (𝐹 supp ∅) ∈ V)
17044, 1, 45, 19, 43cantnfcl 9108 . . . . . . . . . . . . . . . . . . . . . . 23 (𝜑 → ( E We (𝐹 supp ∅) ∧ dom 𝐺 ∈ ω))
171170simpld 497 . . . . . . . . . . . . . . . . . . . . . 22 (𝜑 → E We (𝐹 supp ∅))
17219oiiso 8979 . . . . . . . . . . . . . . . . . . . . . 22 (((𝐹 supp ∅) ∈ V ∧ E We (𝐹 supp ∅)) → 𝐺 Isom E , E (dom 𝐺, (𝐹 supp ∅)))
173169, 171, 172syl2anc 586 . . . . . . . . . . . . . . . . . . . . 21 (𝜑𝐺 Isom E , E (dom 𝐺, (𝐹 supp ∅)))
174173ad2antrr 724 . . . . . . . . . . . . . . . . . . . 20 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → 𝐺 Isom E , E (dom 𝐺, (𝐹 supp ∅)))
175135ad2antrl 726 . . . . . . . . . . . . . . . . . . . 20 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → 𝑦 ∈ dom 𝐺)
176 simprl 769 . . . . . . . . . . . . . . . . . . . 20 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → suc 𝑦 ∈ dom 𝐺)
177 isorel 7056 . . . . . . . . . . . . . . . . . . . 20 ((𝐺 Isom E , E (dom 𝐺, (𝐹 supp ∅)) ∧ (𝑦 ∈ dom 𝐺 ∧ suc 𝑦 ∈ dom 𝐺)) → (𝑦 E suc 𝑦 ↔ (𝐺𝑦) E (𝐺‘suc 𝑦)))
178174, 175, 176, 177syl12anc 834 . . . . . . . . . . . . . . . . . . 19 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝑦 E suc 𝑦 ↔ (𝐺𝑦) E (𝐺‘suc 𝑦)))
179168, 178mpbii 235 . . . . . . . . . . . . . . . . . 18 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐺𝑦) E (𝐺‘suc 𝑦))
180 fvex 6659 . . . . . . . . . . . . . . . . . . 19 (𝐺‘suc 𝑦) ∈ V
181180epeli 5444 . . . . . . . . . . . . . . . . . 18 ((𝐺𝑦) E (𝐺‘suc 𝑦) ↔ (𝐺𝑦) ∈ (𝐺‘suc 𝑦))
182179, 181sylib 220 . . . . . . . . . . . . . . . . 17 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐺𝑦) ∈ (𝐺‘suc 𝑦))
183 ordsucss 7511 . . . . . . . . . . . . . . . . 17 (Ord (𝐺‘suc 𝑦) → ((𝐺𝑦) ∈ (𝐺‘suc 𝑦) → suc (𝐺𝑦) ⊆ (𝐺‘suc 𝑦)))
184163, 182, 183sylc 65 . . . . . . . . . . . . . . . 16 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → suc (𝐺𝑦) ⊆ (𝐺‘suc 𝑦))
18520ffvelrni 6826 . . . . . . . . . . . . . . . . . . . 20 (𝑦 ∈ dom 𝐺 → (𝐺𝑦) ∈ (𝐹 supp ∅))
186175, 185syl 17 . . . . . . . . . . . . . . . . . . 19 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐺𝑦) ∈ (𝐹 supp ∅))
187152, 186sseldd 3947 . . . . . . . . . . . . . . . . . 18 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐺𝑦) ∈ On)
188 suceloni 7506 . . . . . . . . . . . . . . . . . 18 ((𝐺𝑦) ∈ On → suc (𝐺𝑦) ∈ On)
189187, 188syl 17 . . . . . . . . . . . . . . . . 17 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → suc (𝐺𝑦) ∈ On)
1903ad2antrr 724 . . . . . . . . . . . . . . . . 17 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → ∅ ∈ 𝐴)
191 oewordi 8195 . . . . . . . . . . . . . . . . 17 (((suc (𝐺𝑦) ∈ On ∧ (𝐺‘suc 𝑦) ∈ On ∧ 𝐴 ∈ On) ∧ ∅ ∈ 𝐴) → (suc (𝐺𝑦) ⊆ (𝐺‘suc 𝑦) → (𝐴o suc (𝐺𝑦)) ⊆ (𝐴o (𝐺‘suc 𝑦))))
192189, 153, 137, 190, 191syl31anc 1369 . . . . . . . . . . . . . . . 16 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (suc (𝐺𝑦) ⊆ (𝐺‘suc 𝑦) → (𝐴o suc (𝐺𝑦)) ⊆ (𝐴o (𝐺‘suc 𝑦))))
193184, 192mpd 15 . . . . . . . . . . . . . . 15 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐴o suc (𝐺𝑦)) ⊆ (𝐴o (𝐺‘suc 𝑦)))
194 simprr 771 . . . . . . . . . . . . . . 15 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))
195193, 194sseldd 3947 . . . . . . . . . . . . . 14 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐻‘suc 𝑦) ∈ (𝐴o (𝐺‘suc 𝑦)))
196 peano2 7580 . . . . . . . . . . . . . . . . 17 (𝑦 ∈ ω → suc 𝑦 ∈ ω)
197196ad2antlr 725 . . . . . . . . . . . . . . . 16 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → suc 𝑦 ∈ ω)
1988cantnfvalf 9106 . . . . . . . . . . . . . . . . 17 𝐻:ω⟶On
199198ffvelrni 6826 . . . . . . . . . . . . . . . 16 (suc 𝑦 ∈ ω → (𝐻‘suc 𝑦) ∈ On)
200197, 199syl 17 . . . . . . . . . . . . . . 15 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐻‘suc 𝑦) ∈ On)
201 omcl 8139 . . . . . . . . . . . . . . . 16 (((𝐴o (𝐺‘suc 𝑦)) ∈ On ∧ (𝐹‘(𝐺‘suc 𝑦)) ∈ On) → ((𝐴o (𝐺‘suc 𝑦)) ·o (𝐹‘(𝐺‘suc 𝑦))) ∈ On)
202155, 149, 201syl2anc 586 . . . . . . . . . . . . . . 15 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → ((𝐴o (𝐺‘suc 𝑦)) ·o (𝐹‘(𝐺‘suc 𝑦))) ∈ On)
203 oaord 8151 . . . . . . . . . . . . . . 15 (((𝐻‘suc 𝑦) ∈ On ∧ (𝐴o (𝐺‘suc 𝑦)) ∈ On ∧ ((𝐴o (𝐺‘suc 𝑦)) ·o (𝐹‘(𝐺‘suc 𝑦))) ∈ On) → ((𝐻‘suc 𝑦) ∈ (𝐴o (𝐺‘suc 𝑦)) ↔ (((𝐴o (𝐺‘suc 𝑦)) ·o (𝐹‘(𝐺‘suc 𝑦))) +o (𝐻‘suc 𝑦)) ∈ (((𝐴o (𝐺‘suc 𝑦)) ·o (𝐹‘(𝐺‘suc 𝑦))) +o (𝐴o (𝐺‘suc 𝑦)))))
204200, 155, 202, 203syl3anc 1367 . . . . . . . . . . . . . 14 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → ((𝐻‘suc 𝑦) ∈ (𝐴o (𝐺‘suc 𝑦)) ↔ (((𝐴o (𝐺‘suc 𝑦)) ·o (𝐹‘(𝐺‘suc 𝑦))) +o (𝐻‘suc 𝑦)) ∈ (((𝐴o (𝐺‘suc 𝑦)) ·o (𝐹‘(𝐺‘suc 𝑦))) +o (𝐴o (𝐺‘suc 𝑦)))))
205195, 204mpbid 234 . . . . . . . . . . . . 13 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (((𝐴o (𝐺‘suc 𝑦)) ·o (𝐹‘(𝐺‘suc 𝑦))) +o (𝐻‘suc 𝑦)) ∈ (((𝐴o (𝐺‘suc 𝑦)) ·o (𝐹‘(𝐺‘suc 𝑦))) +o (𝐴o (𝐺‘suc 𝑦))))
20644, 1, 45, 19, 43, 8cantnfsuc 9111 . . . . . . . . . . . . . . 15 ((𝜑 ∧ suc 𝑦 ∈ ω) → (𝐻‘suc suc 𝑦) = (((𝐴o (𝐺‘suc 𝑦)) ·o (𝐹‘(𝐺‘suc 𝑦))) +o (𝐻‘suc 𝑦)))
207196, 206sylan2 594 . . . . . . . . . . . . . 14 ((𝜑𝑦 ∈ ω) → (𝐻‘suc suc 𝑦) = (((𝐴o (𝐺‘suc 𝑦)) ·o (𝐹‘(𝐺‘suc 𝑦))) +o (𝐻‘suc 𝑦)))
208207adantr 483 . . . . . . . . . . . . 13 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐻‘suc suc 𝑦) = (((𝐴o (𝐺‘suc 𝑦)) ·o (𝐹‘(𝐺‘suc 𝑦))) +o (𝐻‘suc 𝑦)))
209 omsuc 8129 . . . . . . . . . . . . . 14 (((𝐴o (𝐺‘suc 𝑦)) ∈ On ∧ (𝐹‘(𝐺‘suc 𝑦)) ∈ On) → ((𝐴o (𝐺‘suc 𝑦)) ·o suc (𝐹‘(𝐺‘suc 𝑦))) = (((𝐴o (𝐺‘suc 𝑦)) ·o (𝐹‘(𝐺‘suc 𝑦))) +o (𝐴o (𝐺‘suc 𝑦))))
210155, 149, 209syl2anc 586 . . . . . . . . . . . . 13 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → ((𝐴o (𝐺‘suc 𝑦)) ·o suc (𝐹‘(𝐺‘suc 𝑦))) = (((𝐴o (𝐺‘suc 𝑦)) ·o (𝐹‘(𝐺‘suc 𝑦))) +o (𝐴o (𝐺‘suc 𝑦))))
211205, 208, 2103eltr4d 2926 . . . . . . . . . . . 12 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐻‘suc suc 𝑦) ∈ ((𝐴o (𝐺‘suc 𝑦)) ·o suc (𝐹‘(𝐺‘suc 𝑦))))
212161, 211sseldd 3947 . . . . . . . . . . 11 (((𝜑𝑦 ∈ ω) ∧ (suc 𝑦 ∈ dom 𝐺 ∧ (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)))) → (𝐻‘suc suc 𝑦) ∈ (𝐴o suc (𝐺‘suc 𝑦)))
213212exp32 423 . . . . . . . . . 10 ((𝜑𝑦 ∈ ω) → (suc 𝑦 ∈ dom 𝐺 → ((𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦)) → (𝐻‘suc suc 𝑦) ∈ (𝐴o suc (𝐺‘suc 𝑦)))))
214213a2d 29 . . . . . . . . 9 ((𝜑𝑦 ∈ ω) → ((suc 𝑦 ∈ dom 𝐺 → (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦))) → (suc 𝑦 ∈ dom 𝐺 → (𝐻‘suc suc 𝑦) ∈ (𝐴o suc (𝐺‘suc 𝑦)))))
215136, 214syl5 34 . . . . . . . 8 ((𝜑𝑦 ∈ ω) → ((𝑦 ∈ dom 𝐺 → (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦))) → (suc 𝑦 ∈ dom 𝐺 → (𝐻‘suc suc 𝑦) ∈ (𝐴o suc (𝐺‘suc 𝑦)))))
216215expcom 416 . . . . . . 7 (𝑦 ∈ ω → (𝜑 → ((𝑦 ∈ dom 𝐺 → (𝐻‘suc 𝑦) ∈ (𝐴o suc (𝐺𝑦))) → (suc 𝑦 ∈ dom 𝐺 → (𝐻‘suc suc 𝑦) ∈ (𝐴o suc (𝐺‘suc 𝑦))))))
21780, 89, 98, 131, 216finds2 7588 . . . . . 6 (𝑥 ∈ ω → (𝜑 → (𝑥 ∈ dom 𝐺 → (𝐻‘suc 𝑥) ∈ (𝐴o suc (𝐺𝑥)))))
21870, 71, 58, 217syl3c 66 . . . . 5 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → (𝐻‘suc 𝑥) ∈ (𝐴o suc (𝐺𝑥)))
21969, 218eqeltrd 2911 . . . 4 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → (𝐻𝐾) ∈ (𝐴o suc (𝐺𝑥)))
22068, 219sseldd 3947 . . 3 ((𝜑 ∧ (𝑥 ∈ ω ∧ 𝐾 = suc 𝑥)) → (𝐻𝐾) ∈ (𝐴o 𝐶))
221220rexlimdvaa 3272 . 2 (𝜑 → (∃𝑥 ∈ ω 𝐾 = suc 𝑥 → (𝐻𝐾) ∈ (𝐴o 𝐶)))
222 peano2 7580 . . . . 5 (dom 𝐺 ∈ ω → suc dom 𝐺 ∈ ω)
223170, 222simpl2im 506 . . . 4 (𝜑 → suc dom 𝐺 ∈ ω)
224 elnn 7568 . . . 4 ((𝐾 ∈ suc dom 𝐺 ∧ suc dom 𝐺 ∈ ω) → 𝐾 ∈ ω)
22523, 223, 224syl2anc 586 . . 3 (𝜑𝐾 ∈ ω)
226 nn0suc 7584 . . 3 (𝐾 ∈ ω → (𝐾 = ∅ ∨ ∃𝑥 ∈ ω 𝐾 = suc 𝑥))
227225, 226syl 17 . 2 (𝜑 → (𝐾 = ∅ ∨ ∃𝑥 ∈ ω 𝐾 = suc 𝑥))
22813, 221, 227mpjaod 856 1 (𝜑 → (𝐻𝐾) ∈ (𝐴o 𝐶))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 208 ∧ wa 398 ∨ wo 843 = wceq 1537 ∈ wcel 2114 ∃wrex 3126 Vcvv 3473 ⊆ wss 3913 ∅c0 4269 class class class wbr 5042 Tr wtr 5148 E cep 5440 We wwe 5489 dom cdm 5531 “ cima 5534 Ord word 6166 Oncon0 6167 suc csuc 6169 Fn wfn 6326 ⟶wf 6327 ‘cfv 6331 Isom wiso 6332 (class class class)co 7133 ∈ cmpo 7135 ωcom 7558 supp csupp 7808 seqωcseqom 8061 +o coa 8077 ·o comu 8078 ↑o coe 8079 finSupp cfsupp 8811 OrdIsocoi 8951 CNF ccnf 9102 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1796 ax-4 1810 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2116 ax-9 2124 ax-10 2145 ax-11 2161 ax-12 2177 ax-ext 2792 ax-rep 5166 ax-sep 5179 ax-nul 5186 ax-pow 5242 ax-pr 5306 ax-un 7439 This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-3or 1084 df-3an 1085 df-tru 1540 df-fal 1550 df-ex 1781 df-nf 1785 df-sb 2070 df-mo 2622 df-eu 2653 df-clab 2799 df-cleq 2813 df-clel 2891 df-nfc 2959 df-ne 3007 df-ral 3130 df-rex 3131 df-reu 3132 df-rmo 3133 df-rab 3134 df-v 3475 df-sbc 3753 df-csb 3861 df-dif 3916 df-un 3918 df-in 3920 df-ss 3930 df-pss 3932 df-nul 4270 df-if 4444 df-pw 4517 df-sn 4544 df-pr 4546 df-tp 4548 df-op 4550 df-uni 4815 df-iun 4897 df-br 5043 df-opab 5105 df-mpt 5123 df-tr 5149 df-id 5436 df-eprel 5441 df-po 5450 df-so 5451 df-fr 5490 df-se 5491 df-we 5492 df-xp 5537 df-rel 5538 df-cnv 5539 df-co 5540 df-dm 5541 df-rn 5542 df-res 5543 df-ima 5544 df-pred 6124 df-ord 6170 df-on 6171 df-lim 6172 df-suc 6173 df-iota 6290 df-fun 6333 df-fn 6334 df-f 6335 df-f1 6336 df-fo 6337 df-f1o 6338 df-fv 6339 df-isom 6340 df-riota 7091 df-ov 7136 df-oprab 7137 df-mpo 7138 df-om 7559 df-1st 7667 df-2nd 7668 df-supp 7809 df-wrecs 7925 df-recs 7986 df-rdg 8024 df-seqom 8062 df-1o 8080 df-2o 8081 df-oadd 8084 df-omul 8085 df-oexp 8086 df-er 8267 df-map 8386 df-en 8488 df-dom 8489 df-sdom 8490 df-fin 8491 df-fsupp 8812 df-oi 8952 df-cnf 9103 This theorem is referenced by: cantnflt2 9114 cnfcomlem 9140
Copyright terms: Public domain W3C validator | 16,772 | 24,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-22 | latest | en | 0.244151 |
https://www.chrisvantienhoven.nl/en/qa-items/qa-lines/qa-l10 | 1,511,298,698,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806426.72/warc/CC-MAIN-20171121204652-20171121224652-00683.warc.gz | 757,328,451 | 17,456 | QA-L10 QA-LSD Line
QA-L10 is the line with the Least Sum of Squared Distances of the QA-vertices to this line.
It is the counterpart of QL-P26 in a Quadrilateral which is the point with Least Sum of Squared Distances from this point to the QL-Lines.
It is described by J.L. Coolidge in Ref-25. He proves that the QA-Centroid will lie on this line. Unfortunately his proposed construction doesn’t produce the desired result.
Next construction uses properties of the Complex Plane. See Ref-34, QFG-messages #1585, #1590.
1. Draw a random couple of perpendicular lines intersecting in QA-P1 as x-axis and y-axis.
2. Let the 4 vertices of the QA be represented by complex numbers zi = xi + i yi.
3. The Centroid of the QA-vertices is the point QA-P1 where z = 0.
4. Let the QA-Pz = Centroid of the points represented by zi2 .
5. The searched LSD line is the angle bisector of the x-axis and the line QA-P1.QA-Pz.
The same type of construction can be used for any number of points.
Another construction can be found at [34], QFG-message #1611.
CT-Coefficients:
(Ua + (r-q) W : Ub + (p-r) W : Uc + (q-p) W)
where:
Ua= a2 (r-q) (p2+5 p q+4 q2+5 p r+5 q r+4 r2)
+ b2 (-2 p3-4 p2 q+p q2+3 q3-4 p2 r+2 p q r+6 q2 r-5 p r2+3 q r2)
+ c2 (2 p3+4 p2 q+5 p q2+4 p2 r-2 p q r-3 q2 r-p r2-6 q r2-3 r3)
Ub= a2 (-3 p3-p2 q+4 p q2+2 q3-6 p2 r-2 p q r+4 q2 r-3 p r2+5 q r2)
+ b2 (p-r) (4 p2+5 p q+q2+5 p r+5 q r+4 r2)
+ c2 (-5 p2 q-4 p q2-2 q3+3 p2 r+2 p q r-4 q2 r+6 p r2+q r2+3 r3)
Uc= a2 (3 p3+6 p2 q+3 p q2+p2 r+2 p q r-5 q2 r-4 p r2-4 q r2-2 r3)
+ b2 (-3 p2 q-6 p q2-3 q3+5 p2 r-2 p q r-q2 r+4 p r2+4 q r2+2 r3)
+ c2 (q-p) (4 p2+5 p q+4 q2+5 p r+5 q r+r2)
W=Sqrt[
( a4 (3 p2+4 p q+4 q2+6 p r+4 q r+3 r2) (3 p2+6 p q+3 q2+4 p r+4 q r+4 r2)
+ b4 (4 p2+4 p q+3 q2+4 p r+6 q r+3 r2) (3 p2+6 p q+3 q2+4 p r+4 q r+4 r2)
+ c4 (3 p2+4 p q+4 q2+6 p r+4 q r+3 r2) (4 p2+4 p q+3 q2+4 p r+6 q r+3 r2)
- 2 b2 c2 (p2+3 p q+2 q2+3 p r+q r+2 r2) (4 p2+4 p q+3 q2+4 p r+6 q r+3 r2)
- 2 a2 c2 (2 p2+3 p q+q2+p r+3 q r+2 r2) (3 p2+4 p q+4 q2+6 p r+4 q r+3 r2)
- 2 a2 b2 (2 p2+p q+2 q2+3 p r+3 q r+r2) (3 p2+6 p q+3 q2+4 p r+4 q r+4 r2))]
The algebraic least distance is:
(A0 (T1 + V1 W)) / (2 (T2 + V2 W))
The algebraic largest distance for the perpendicular line through QA-P1 is:
(A0 (T1 - V1 W)) / (2 (T2 - V2 W))
where:
A0 = (a-b-c) (a+b-c) (a-b+c) (a+b+c)
V1 = (p2-p q+q2-p r-q r+r2)
V2 = +a2 (p+q-2 r) (p-2q+r) + b2 (p+q-2 r) (-2 p+q+r) + c2 (-2 p+q+r) (p-2q+r)
T1 =
c2 ( 2 p4 -5 p3 q -5 p2 q2 -5 p q3 +2 q4 +5 p3 r +p2 q r +p q2 r +5 q3 r +4 p2 r2 -2 p q r2 +4 q2 r2 -2 p r3 -2 q r3 -3 r4) +
a2 (-3 p4 -2 p3 q +4 p2 q2 +5 p q3 + 2q4 -2 p3 r -2 p2 q r +p q2 r -5 q3 r +4 p2 r2 +p q r2 -5 q2 r2 +5 p r3 -5 q r3 +2 r4) +
b2 ( 2 p4 +5 p3 q +4 p2 q2 -2 p q3 -3 q4 -5 p3 r +p2 q r -2 p q2 r -2 q3 r -5 p2 r2 +p q r2 +4 q2 r2 -5 p r3 +5 q r3 +2 r4)
T2 =
b4(-8 p4 -12 p3 q -3 p2 q2 -2 p q3 -3 q4 +4 p3 r +18 p2 q r +12 p q2 r -2 q3 r -3 p2 r2 +18 p q r2 -3 q2 r2 +4 p r3 -12 q r3 -8 r4)+
a4(-3 p4 -2 p3 q -3 p2 q2 -12 p q3 -8 q4 -2 p3 r +12 p2 q r +18 p q2 r +4 q3 r -3 p2 r2 +18 p q r2 -3 q2 r2 - 12p r3 +4 q r3 -8 r4)+
c4(-8 p4 +4 p3 q -3 p2 q2 +4 p q3 -8 q4 -12 p3 r +18 p2 q r +18 p q2 r -12 q3 r -3 p2 r2 +12 p q r2 -3 q2 r2 -2 p r3 -2 q r3 -3 r4)+
2 b2 c2 (4 p3 q +3 p2 q2 +3 p q3 +4 q4 +4 p3 r -18 p2 q r -3 p q2 r +q3 r +3 p2 r2 -3 p q r2 -6 q2 r2 +3 p r3 +q r3 +4 r4) +
2 a2 b2 (4 p4 +p3 q -6 p2 q2 +p q3 +4 q4 +3 p3 r -3 p2 q r -3 p q2 r +3 q3 r +3 p2 r2 -18 p q r2 +3 q2 r2 +4 p r3 +4 q r3) +
2 a2 c2 (4 p4 + 3p3 q +3 p2 q2 +4 p q3 +p3 r -3 p2 q r -18 p q2 r +4 q3 r -6 p2 r2 -3 p q r2 +3 q2 r2 +p r3 +3 q r3 +4 r4)
Properties:
• QA-P1 lies on QA-L10.
• There is no line with largest sum of squared distances. However when we look for the line with this property passing through QA-P1, then QA-L10p being the line through QA-P1 perpendicular to QA-L10 will be the line with largest sum of squared distances.
• In a Quadrilateral the triangle formed by the 3 QL-versions of QA-L10 and the triangle formed by the 3 QL-versions of QA-L10p are perspective in a common point of the circumscribed circles of these triangles. | 2,015 | 4,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-47 | latest | en | 0.615981 |
https://estebantorreshighschool.com/equation-help/product-rule-equation.html | 1,618,085,154,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038057476.6/warc/CC-MAIN-20210410181215-20210410211215-00004.warc.gz | 345,464,569 | 30,177 | ## How do you find the product rule?
What is the Product rule? Basically, you take the derivative of f multiplied by g, and add f multiplied by the derivative of g.
## What is product rule in math?
The product rule is a formal rule for differentiating problems where one function is multiplied by another. The rule follows from the limit definition of derivative and is given by. . Remember the rule in the following way. Each time, differentiate a different function in the product and add the two terms together.
## What is product rule in physics?
The product rule states that the derivative of the product of two functions is the product of the derivative of the first function and the second function added with the product of the first function and the derivative of the second function.
## Under what conditions can we apply the product rule?
The product rule is if the two “parts” of the function are being multiplied together, and the chain rule is if they are being composed. For instance, to find the derivative of f(x) = x² sin(x), you use the product rule, and to find the derivative of g(x) = sin(x²) you use the chain rule.
## What is the product formula?
Description. The PRODUCT function multiplies all the numbers given as arguments and returns the product. For example, if cells A1 and A2 contain numbers, you can use the formula =PRODUCT(A1, A2) to multiply those two numbers together.
## What is the formula of U V?
The Product and Quotient Rules are covered in this section. This is used when differentiating a product of two functions. d (uv) = (x² + 1) + x(2x) = x² + 1 + 2x² = 3x² + 1 .
## What is product rule used for?
The product rule is used in calculus when you are asked to take the derivative of a function that is the multiplication of a couple or several smaller functions. In other words, a function f(x) is a product of functions if it can be written as g(x)h(x), and so on.
## How do you differentiate LN?
The steps are as follows:Let y = ln(x).Use the definition of a logarithm to write y = ln(x) in logarithmic form. Treat y as a function of x, and take the derivative of each side of the equation with respect to x.Use the chain rule on the left hand side of the equation to find the derivative.
You might be interested: Line segment equation
## What are the five rules of exponents?
Exponents product rules. Product rule with same base. an ⋅ am = an+m Example: 23 ⋅ 24 = 23+4 = 27 = 2⋅2⋅2⋅2⋅2⋅2⋅2 = 128. Exponents quotient rules. Quotient rule with same base. an / am = anm Example: 25 / 23 = 253 = 22 = 2⋅2 = 4. Exponents power rules. Power rule I. (an) m = a nm Example:
## What is dy dx?
Differentiation allows us to find rates of change. If y = some function of x (in other words if y is equal to an expression containing numbers and x’s), then the derivative of y (with respect to x) is written dy/dx, pronounced “dee y by dee x” .
## What does product mean?
In mathematics, a product is the result of multiplication, or an expression that identifies factors to be multiplied. For example, 30 is the product of 6 and 5 (the result of multiplication), and is the product of and. (indicating that the two factors should be multiplied together).
## What is Leibnitz product rule?
In calculus, the general Leibniz rule, named after Gottfried Wilhelm Leibniz, generalizes the product rule (which is also known as “Leibniz’s rule”). It states that if and are -times differentiable functions, then the product is also -times differentiable and its th derivative is given by.
## What is the difference between chain rule and product rule?
The chain rule is used to differentiate compositions of functions. The product rule is used to differentiate products of function.
### Releated
#### Expectation equation
How do you calculate expectation? The basic expected value formula is the probability of an event multiplied by the amount of times the event happens: (P(x) * n). What is expectation function? Using expectation, we can define the moments and other special functions of a random variable. Definition 2 Let X and Y be random […]
#### Tensile stress equation
What is the formula for tensile stress? Difference Between Tensile Stress And Tensile Strength Tensile stress Tensile strength The formula is: σ = F/A Where, σ is the tensile stress F is the force acting A is the area The formula is: s = P/a Where, s is the tensile strength P is the force […] | 1,064 | 4,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2021-17 | longest | en | 0.939026 |
https://www.physicsforums.com/threads/equation-of-line-parallel-to-plane-and-intersaction-with-other-line.235849/ | 1,579,286,468,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250590107.3/warc/CC-MAIN-20200117180950-20200117204950-00151.warc.gz | 1,037,177,195 | 16,578 | # Equation of line parallel to plane and intersaction with other line
## Homework Statement
Hello!
I have one problem which seems not so difficult:
-Find the equation of line which passes throught the point M(1,0,7), parallel of the plane 3x-y+2z-15=0 and it intersects the line $$\frac{x-1}{4}=\frac{y-3}{2}=\frac{z}{1}$$
## The Attempt at a Solution
The equation of the line will be: $$\frac{x-1}{a_1}=\frac{y}{a_2}=\frac{z-7}{a_3}$$
So we need to find $$\vec{a}(a_1,a_2,a_3)$$ and we need three conditions in the system.
The first condition is $$\vec{a} \circ \vec{n}=0$$ or $$(a_1,a_2,a_3)(3,-1,2)=0$$ or $$3a_1-a_2+2a_3=0$$.
The second condition is the intersection of two lines, and it is:
$$-17a_1+28a_2+12a_3=0$$
Related Precalculus Mathematics Homework Help News on Phys.org
HallsofIvy
Homework Helper
There is no third condition. You cannot determine a1, a2, and a3 uniquely. Any multiple of a given a1, a2, and a3 will also determine the same line. From the two equations you have you can solve for two of the numbers as functions of the third. Choose that third as you please, say equal to 1, and solve for the other 2.
Other way (I think it is very similar to what you did):
r = (1,3,0) + t (4,2,1)
P = (1,0,7)
find direction vector: r - P
and you know n.(r-P) = 0
So, only one condition .. (and only one unknown)
rootX what is r, and what is P?
rootX what is r, and what is P?
r is (1,3,0) + t (4,2,1) ... a line
P is (1,0,7) .. a point
Should work, shouldn't it?
It wouldn't work, since by your opinion you will find point, product of the intersection of line and plane... And we'll need to find parallel vector to the line which satisfies the above conditions... | 549 | 1,695 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2020-05 | longest | en | 0.92057 |
http://www.plustwophysics.com/tag/height/ | 1,544,577,998,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823710.44/warc/CC-MAIN-20181212000955-20181212022455-00244.warc.gz | 461,337,472 | 34,496 | ## Numerical Problems From Optics
1. When an object of height 4cm is placed at 40cm from a mirror the mirror forms the image on the object itself. If the height of the image is equal to the height of the object. Find out the focal
## Height of Antenna and range of transmission
A TV Tower has a height h. Derive expression for the maximum distance upto which signals can be received on earth.
## A body is dropped from a tower of height 96 m. Another body is thrown up from the ground after a second with a velocity of 40 m/s.When and where they will meet?
A body is dropped from a tower of height 96 m. Another body is thrown up from the ground after a second with a velocity of 40 m/s. When and where they will meet? SOUNDARYA asked Answer: For the first body, u=0 m/s Let S=x (from top) x=0.5 gt2 &#...
## The problem of a rock thrown vertically up
A rock is thrown vertically upward from the ground with an initial speed 15m/s. a. how high does it go b. how much time is required for the rock to reach its maximum height? c. what is the rock’s height at t=2.00s? (Posted by Merhawi) Answer: (a) u=15m/s a=-10m/s2 v=0 m/s [...]
## The problem of a rock thrown vertically up
A rock is thrown vertically upward from the ground with an initial speed 15m/s. a. how high does it go b. how much time is required for the rock to reach its maximum height? c. what is the rock’s height at t=2.00s? (Posted by Merhawi) Answer: (a) u=1...
## The problem of a rock thrown vertically up
A rock is thrown vertically upward from the ground with an initial speed 15m/s. a. how high does it go b. how much time is required for the rock to reach its maximum height? c. what is the rock’s height at t=2.00s? (Posted by Merhawi) Answer: (a) u=1...
## A question from Kinematics.
An object thrown vertically up from the ground passes the height 5m twice in an interval of 10s. What is its time of flight? (Divya asked) Answer: The question seems partial. I am assuming that g= 10 m/s2 and the projection is strictly vertical. The o... | 547 | 2,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-51 | latest | en | 0.934403 |
https://newtheory.org/history-of-mathematics/query-from-you-what-is-the-heart-of-algebra-on-the-sat.html | 1,696,055,723,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510603.89/warc/CC-MAIN-20230930050118-20230930080118-00529.warc.gz | 463,919,912 | 20,840 | # Query from you: what is the heart of algebra on the SAT?
Contents
The heart of algebra on the SAT refers to the portion of the math section that focuses on essential algebraic concepts and problem-solving skills, such as linear equations, inequalities, and systems of equations.
The heart of algebra on the SAT is a significant portion of the math section that aims to test students’ abilities to understand essential algebraic concepts and apply problem-solving skills to solve real-world problems. As the SAT is a crucial standardized test for college admission in the United States, a good score in this section can increase a student’s chances of getting into their desired colleges.
According to the College Board, the heart of algebra on the SAT includes concepts like linear equations, inequalities, and systems of equations. Students are tested on their ability to create and solve algebraic equations, interpret graphs and charts, and apply algebraic principles to different scenarios, like word problems and data analysis.
To further understand the significance of mastering algebra skills, Bill Gates once remarked, “Algebra is the key to the universe.” Thus, having strong algebraic abilities can not only boost one’s chances of college acceptance but also prepare them for real-world problem-solving.
Here are some interesting facts related to the heart of algebra on the SAT:
• The heart of algebra section consists of 19 questions out of the total 58 questions in the SAT math section.
• As per the College Board, the heart of algebra questions is usually straightforward and do not require advanced knowledge of algebra.
• The College Board states that the heart of algebra questions are meant to assess students’ ability to think logically and solve problems, which are essential qualities in college.
• The heart of algebra questions are not multiple-choice, and students are expected to write out their answers.
• The heart of algebra section is graded on a scale of 200-800, and a good score is essential for getting into many high-ranked colleges and universities in the US.
IT\\\'S IMPORTANT: Why is it important to improve mathematical proficiency in children?
Below is a table showing the number of questions in each math section of the SAT:
Math Section Number of Questions
Heart of Algebra 19
Problem Solving and Data Analysis 17
This video covers the Heart of Algebra section of the SAT, which includes isolating variables, linear equations, and linear inequalities. The video emphasizes the importance of understanding inverse operations and using them to isolate variables for solving equations. The different forms of linear equations, including slope-intercept form, standard form, and point-slope form, are discussed, as well as methods for solving systems of linear equations using substitution and elimination. Additionally, the video covers graphing linear inequalities and systems of inequalities and emphasizes the importance of validating solutions.
## There are other opinions on the Internet
Heart of Algebra focuses on linear equations, systems of linear equations, and functions. These questions ask you to create equations that represent a situation, solve equations and systems of equations, and make connections between different representations of linear relationships.
Heart of Algebra is one of the 4 sections on the SAT Math Test, and makes up 30% of the math questions. It assesses your ability to solve and create linear equations and inequalities, and systems of equations. The questions may vary in difficulty and may require you to interpret the relationship between equations and graphs. You will receive a subscore out of 15 for this section.
Likewise, What percent of the SAT is heart of Algebra?
Answer will be: Heart of Algebra accounts for the largest part of the SAT math section (33% of the test), so you need to be well prepared for it. In this post, I’ll be discussing this category’s content and question types, working through practice problems, and giving tips on how to ace these questions.
IT\\\'S IMPORTANT: What is math intervention?
One may also ask, How many heart of Algebra questions are on the SAT? As an answer to this: 19 questions
The Heart of Algebra makes up 30% of the math questions you’ll see on the SAT. That works out to 19 questions total – including both multiple choice and open response.
Then, What level of Algebra is on the SAT?
As an answer to this: The SAT is designed to test basic high school math, from courses up to and including Algebra II. Students who wish to demonstrate more advanced skills may consider taking the ACT, which covers trigonometry, or the two SAT Math Subject Tests.
What is the hardest math concept on the SAT?
In reply to that: Linear equations and inequalities and their graphs and systems. Problem Solving and Data Analysis: 29% of test, 17 questions. Ratios, proportions, percentages, and units; analyzing graphical data, probabilities, and statistics. Passport to Advanced Math: 28% of test, 16 questions.
Response will be: The previous version of the SAT had what’s known as a “guessing penalty,” meaning points were deducted for any incorrect answer. However, on the tests you’ll take today you do not lose any points for wrong answers, so you should bubble in a response to every question.
Herein, Does the SAT have calculus questions? Response to this: The Math section is comprised of a single test with two components – a no-calculator portion and a calculator-allowed portion. The SAT also includes an optional Essay. Some schools may require the Essay, so be sure to ask before you take the SAT. SEE SAT PREP OPTIONS
IT\\\'S IMPORTANT: What is the definition of math?
In this manner, How many questions are there in SAT exam? In reply to that: The SAT Reading section contains 52 questions. The SAT Writing & Language section has 44
Accordingly, What is the heart of algebra? Response: Heart of Algebra. Heart of Algebra focuses on linear equations, systems of linear equations, and functions. These questions ask you to create equations that represent a situation, solve equations and systems of equations, and make connections between different representations of linear relationships. Heart of Algebra includes the following types of questions:
Do you answer every question on the SAT?
Response to this: The previous version of the SAT had what’s known as a “guessing penalty,” meaning points were deducted for any incorrect answer. However, on the tests you’ll take today you do not lose any points for wrong answers, so you should bubble in a response to every question.
Beside above, Does the SAT have calculus questions? In reply to that: The Math section is comprised of a single test with two components – a no-calculator portion and a calculator-allowed portion. The SAT also includes an optional Essay. Some schools may require the Essay, so be sure to ask before you take the SAT. SEE SAT PREP OPTIONS
How many questions are there in SAT exam?
Response to this: The SAT Reading section contains 52 questions. The SAT Writing & Language section has 44
What is the heart of algebra?
In reply to that: Heart of Algebra. Heart of Algebra focuses on linear equations, systems of linear equations, and functions. These questions ask you to create equations that represent a situation, solve equations and systems of equations, and make connections between different representations of linear relationships. Heart of Algebra includes the following types of questions:
Rate article | 1,484 | 7,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.943178 |
http://www.tutorsglobe.com/homework-help/algebra/exponential-functions-and-their-graphs-74238.aspx | 1,537,344,617,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155942.15/warc/CC-MAIN-20180919063526-20180919083526-00039.warc.gz | 419,192,683 | 8,400 | #### Exponential Functions and Their Graphs
Exponential Functions and Their Graphs:
Exponential Functions:
Algebraic functions are the functions that can be expressed by using the arithmetic operations and whose values are either rational or the root of rational number. Now, we will be dealing with transcendental functions. The Transcendental functions return values that might not be expressible as the rational numbers or roots of rational numbers.
The algebraic equations can be solved most of the time by hand. The transcendental functions can frequently be solved by hand with a calculator essential if you wish a decimal approximation. Though whenever transcendental and algebraic functions are mixed in an equation, graphical or numerical methods are sometimes the only way to find out the solution.
Simplest exponential function is as: f(x) = ax, a>0, a≠1
The main reasons for the restrictions are simple. When a≤0, then whenever you increase it to a rational power, you might not obtain a real number. Illustration: When a = -2, then (-2)0.5 = sqrt(-2) that is not real. When a = 1, then no matter what x is, the value of f(x) is 1. That is a quite boring function, and it is surely not one-to-one.
Remember that one-to-one functions had some properties which make them desirable. They contain inverses which are also functions. They can be applied to both sides of the equation.
Graphs of Exponential Functions:
The graph of y = 2x is as shown below. Here are a few properties of the exponential function whenever the base is more than 1.
a) The graph passes via the point (0,1).
b) The domain is all the real numbers.
c) The range is y > 0.
d) The graph is rising.
e) Graph is asymptotic to the x-axis as x approaches to the negative infinity.
f) The graph rises devoid of bound as x approaches to the positive infinity.
g) The graph is smooth.
h) Graph is continuous.
What would be the translation if you substituted each and every x with -x? This would be the reflection regarding y-axis. We as well know that whenever we increase a base to the negative power, the one outcome is that the reciprocal of number is taken. Therefore, if we were to graph y = 2-x, the graph would be a reflection regarding y-axis of y = 2x and the function would be equal to y = (1/2)x.
The graph of y=2-x is shown below. The properties of exponential function and its graph whenever the base is between 0 and 1 are specified.
a) The graph passes via the point (0,1).
b) The domain is all the real numbers.
c) The range is y > 0.
d) The graph is reducing.
e) Graph is asymptotic to the x-axis as x approaches to the positive infinity.
f) The graph rises without bound as x approaches to the negative infinity.
g) The graph is continuous.
h) Graph is smooth.
Note that the only differences regard whether the function is decreasing or increasing, and the behavior at the left hand and right hand ends.
Translations of Exponential Graphs:
You can apply what you know regarding translations to help you sketch the graph of exponential functions.
The horizontal translation might affect the increasing or decreasing (when multiplied by a negative), the left hand/right hand behavior of the graph, and the y-intercept, however it won't modify the place of the horizontal asymptote.
The vertical translation might influence the increasing or decreasing (when multiplied by a negative), the y-intercept, and the place of horizontal asymptote. This will not modify whether the graph goes with no bound or is asymptotic (though it might modify where it is asymptotic) to the right or left.
By knowing the characteristics of the fundamental graphs, you can apply such translations to simply sketch the new function.
You must now add up the exponential graph from the front cover of text to the list of such you know.
The natural base e:
Since x rises devoid of bound, the quantity (1+1/x)^x will approach to the transcendental number e.
The limit notation shown above is from calculus. The limit notation is a manner of asking what occurs to the expression as x approaches to the value shown. The limit is dividing line between calculus and algebra. The Calculus is algebra with the concept of limit. People always contain this dread of calculus. The calculus itself is simple. The reason people do not do well in calculus is not since of the calculus, however because they are poor at algebra.
The value for e is around 2.718281828. Here is a slightly more precise, although no more helpful, approximation.
2.71828 18284 59045 23536 02874 71352 66249 77572 47093 69995 95749 66967 62772 40766 30353 54759 45716 82178 52516 64274 27466 39193 20030 59921 81741 35966 29043 57290 03342 95260 59563 07381 32328 62794 34907 63233 82988 07531 95251 01901 15738 34187 93070 21540 89149 93488 41675 09244 76146 06680 82264 80016 84774 11853 74234 54424 37107 53907 77449 92069 55170 27618 38606 26133
Whenever the base e is used, the exponential function becomes f(x) = ex. There is a key on your calculator labeled e^x. On TI-8x calculators, it is on left side as a [2nd] [Ln]. The exponential function with base e is many times abbreviated as exp(). One common place this abbreviation emerges is whenever writing computer programs.
Compounded Interest:
The amount in your saving account can be figured out with exponential functions. Each and every period (I'll suppose monthly), you obtain 1/12 of annual interest rate (r) applied to your account. New amount in the account is 100% of what you started with plus r%/12 of what you started with that. That signifies that you now contain (100%+r%/12) of what you started with. The later month, you will have similar thing, apart from it will be based on what you had at the end of first month.
It is little bit confusing. The resultant formula for compound interest is A = P (1+i)n.
A is the Amount in account. P is the principal you began with. i is the periodic rate, that is the annual percent (written as decimal) r, divided by the number of periods per year, m. n is the number of compounding periods, that is equal to the number of periods per year, m, times the time in years, t. The formula shown above varies slightly from the formula in books, however agrees with the formula that you will use when you go on to Finite Mathematics. In Finite Mathematics, there is a whole chapter on finance and the formulas included.
Continuous Compounding and Growth/Decay:
In old days, continuous compounding of interest is used. You do not find it anymore since it is gives the maximum return on the investment, and banks are in business to make money, just similar to any other for gain institution.
The model for continuous compounding is A = P ert.
Here, A is the Amount, P is the Principal, r is annual percentage rate (written as decimal), t is the time in years and e is the base for the natural logarithms.
Though, the continuous model does make sense for the population growth and radioactive decay. The radioactivity of isotope does not change once a month at the end of month, it is continually modifying.
The exponential model is y = A ekt,
Here, y is the amount present at time t. A is the initial amount present and k is the rate of growth (when positive) or the rate of decay (when negative).
Latest technology based Algebra Online Tutoring Assistance
Tutors, at the www.tutorsglobe.com, take pledge to provide full satisfaction and assurance in Algebra help via online tutoring. Students are getting 100% satisfaction by online tutors across the globe. Here you can get homework help for Algebra, project ideas and tutorials. We provide email based Algebra help. You can join us to ask queries 24x7 with live, experienced and qualified online tutors specialized in Algebra. Through Online Tutoring, you would be able to complete your homework or assignments at your home. Tutors at the TutorsGlobe are committed to provide the best quality online tutoring assistance for Algebra Homework help and assignment help services. They use their experience, as they have solved thousands of the Algebra assignments, which may help you to solve your complex issues of Algebra. TutorsGlobe assure for the best quality compliance to your homework. Compromise with quality is not in our dictionary. If we feel that we are not able to provide the homework help as per the deadline or given instruction by the student, we refund the money of the student without any delay. | 1,934 | 8,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2018-39 | longest | en | 0.928527 |
https://zh.wikipedia.org/wiki/Draft:%E7%BA%BF%E6%80%A7%E5%8F%98%E5%8F%82%E6%95%B0%E6%8E%A7%E5%88%B6 | 1,513,328,830,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948567785.59/warc/CC-MAIN-20171215075536-20171215095536-00345.warc.gz | 853,027,317 | 18,211 | # 草稿:线性变参数控制
## 增益规划
#### 经典增益规划方法的缺点
• 增益规划有個重大的缺点:在设计操作点外的操作条件,不一定可以获得足够的性能,甚至有些连稳定性都无法保证。
• 多变量控制器的规划常常是乏味且费时的工作,尤其是在航空航天的控制中更是如此。在航空航天中,为了在更高的气动包线英语Flight envelope上获得更好的性能,参数的改变十分巨大。
• 另一点很重要的是,選定的规划變數反映了在飛行條件變化時,飛行動力學上的變化,有可能利用增益规划將線性鲁棒控制的方法論整合到非線性控制設計中,不過設計過程不會明確的提到全域穩定性、鲁棒性以及其性能特性。
## 線性變參數系統
### 參數相依系統
${\displaystyle {\dot {x}}=f(x,u,t)}$
${\displaystyle {\dot {x}}=f(x(t),u(t),t),x(t_{0})}$
${\displaystyle x(t_{0})=x_{0},u(t_{0})=u_{0}}$
The state variables describe the mathematical "state" of a dynamical system and in modeling large complex nonlinear systems if such state variables are chosen to be compact for the sake of practicality and simplicity, then parts of dynamic evolution of system are missing. The state space description will involve other variables called exogenous variables whose evolution is not understood or is too complicated to be modeled but affect the state variables evolution in a known manner and are measurable in real-time using sensors. When a large number of sensors are used, some of these sensors measure outputs in the system theoretic sense as known, explicit nonlinear functions of the modeled states and time, while other sensors are accurate estimates of the exogenous variables. Hence, the model will be a time varying, nonlinear system, with the future time variation unknown, but measured by the sensors in real-time. In this case, if ${\displaystyle w(t),w}$ denotes the exogenous variable , and ${\displaystyle x(t)}$ denotes the modeled state, then the state equations are written as
${\displaystyle {\dot {x}}=f(x(t),w(t),{\dot {w}}(t),u(t))}$
The parameter ${\displaystyle w}$ is not known but its evolution is measured in real time and used for control. If the above equation of parameter dependent system is linear in time then it is called Linear Parameter Dependent systems. They are written similar to Linear Time Invariant form albeit the inclusion in time variant parameter.
${\displaystyle {\dot {x}}=A(w(t))x(t)+B(w(t))u(t)}$
${\displaystyle y=C(w(t))x(t)+D(w(t))u(t)}$
Parameter-dependent systems are linear systems, whose state-space descriptions are known functions of time-varying parameters. The time variation of each of the parameters is not known in advance, but is assumed to be measurable in real time. The controller is restricted to be a linear system, whose state-space entries depend causally on the parameter’s history. There exist three different methodologies to design a LPV controller namely,
1. which relies on the for bounds on performance and robustness.
2. Single Quadratic Lyapunov Function (SQLF)
3. Parameter Dependent Quadratic Lyapunov Function (PDQLF) to bound the achievable level of performance.
These problems are solved by reformulating the control design into finite-dimensional, convex feasibility problems which can be solved exactly, and infinite-dimensional convex feasibility problems which can be solved approximately . This formulation constitutes a type of gain scheduling problem and contrast to classical gain scheduling, this approach address the effect of parameter variations with assured stability and performance.
## 參考資料
1. ^ J. Balas, Gary. Linear Parameter-Varying Control And Its Application to Aerospace Systems (PDF). ICAS. [2002].
2. ^ Wu, Fen. Control of Linear Parameter Varying systems. Univ. of California, Berkeley. [1995].
## 參考資料
• Briat, Corentin. Linear Parameter-Varying and Time-Delay Systems - Analysis, Observation, Filtering & Control. Springer Verlag Heidelberg. 2015. ISBN 978-3-662-44049-0.
• Roland, Toth. Modeling and Identification of Linear Parameter-Varying Systems. Springer Verlag Heidelberg. 2010. ISBN 978-3-642-13812-6.
• Javad, Mohammadpour; Carsten, W. Scherer (编). Control of Linear Parameter Varying Systems with Applications. Springer Verlag New York. 2012. ISBN 978-1-4614-1833-7. | 1,056 | 3,859 | {"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} | 3.03125 | 3 | CC-MAIN-2017-51 | latest | en | 0.687255 |
http://stackoverflow.com/questions/16837538/n-bytes-sized-integers-with-different-memory-representation | 1,455,051,556,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701157472.18/warc/CC-MAIN-20160205193917-00301-ip-10-236-182-209.ec2.internal.warc.gz | 209,141,370 | 18,201 | # n-bytes sized integers with different memory representation
I'm dealing with a lot of memory represented custom integer types soon and keeping this portable, always results in the same operations. I'm think of build some sort of custom Integer class but I was wondering if such a thing already exists? For example to support something like this:
``````char * buffer_ptr = //....
UInteger<5> d( buffer_ptr, E_Type_BigEndian );
d = 20;
uint64 e = 1234567890;
d += e
``````
The result would be a memory representation of a 5-Byte/40Bit unsigned Integer in BigEndian supporting assignment and maybe operations with standard host order types.
Or does boost or something offer help with that?
Thank you very much!
-
It wouldn't be terribly hard to support "X bit" (or "X byte") integers for the full complement of math operations, using single byte operations, and simply either starting at the top or bottom byte to do this.
However, if you want some semblance of performance, you will not want to do that.
I worked on a project which did essentially this for integers of arbitrary bitness. However, to achieve reasonable performance for math operations (e.g. int3 x = 2; x += 5; ...), it used a 32-bit integer (64-bit int for larger than 32 bit) to perform all simple math operations, and only at the end, when the value is transferred out of the variable itself does any surplus bits get masked off. This will almost certainly improve the calculation performance by a factor of 10-100, depending on the operation and the sequences used.
As for big/little endian, again, I would use native format for the internal representation, and just convert it when it's transferred out (although I'm not sure that's a "normal" conversion - if you have "
`````` Uinteger<3> a(..., E_Type_BigEndian);
a = 0x12345678
uint64 b = a;
``````
Surely `b` should not contain `0x78563412;`?
- | 446 | 1,882 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2016-07 | latest | en | 0.936454 |
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# 4.NBT.3 Practice, Homework, and Activities 4th Grade Common Core Math
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### PRODUCT DESCRIPTION
4.NBT.3 Practice, Homework, and Activities contains practice sheets, homework sheets that are similar to the practice sheets, and activities that you can use to teach the standard.
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Rounding is a difficult concept for some students to master. I found that when students had a stronger understanding of number sense they did not rely on rounding rules. So I created place value rounding mats which has students blend number sense with roudning rules. Maybe this is something your students would benefit from as well.
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Thanks! I hope this helps your students learn the concept of rounding.
This is for single classroom use only. If you intend to share purchase multiple license fee for each person. Thanks! (©NLarocca2015)
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https://deepchecks.com/question/what-is-the-best-model-for-time-series-forecasting/ | 1,718,656,307,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861737.17/warc/CC-MAIN-20240617184943-20240617214943-00073.warc.gz | 168,648,713 | 9,670 | # What is the best model for time series forecasting?
Times Series Methods are the many approaches used to analyze data over a certain period.
The key is to pick the right forecasting approach, taking into account the specifics of the time-series data.
### Smoothing-based Model
Data smoothing is a statistical method used in time series forecasting that eliminates anomalies to better see a trend. There will always be an element of chance in any set of data compiled over time. As a result of smoothing, cyclical, and trending patterns may be seen through the noise in the data.
### Moving-average Model
The moving-average model (MA model), often referred to as the moving-average process, is a popular technique for modeling univariate time series in the field of time series prediction. The output variable is stated to be linearly dependent on the current and various past values of a random component within the context of the moving-average model.
The MA model, along with the autoregressive (AR) model, is a specific example and fundamental aspect of the more general ARMA and ARIMA time series models, which have a more intricate stochastic structure.
The finite MA model, in contrast to the AR model, remains inert at all times.
### Exponential Smoothing Model
Using the exponential window function, exponential smoothing is a method for reducing noise in time series data. If you want to make a conclusion based on certain assumptions you already have about the data, then look into using exponential smoothing. Single, double, and triple are all subcategories of the broader category of it.
### Modeling with Moving Averages vs. Exponential Smoothing
Unlike the ordinary moving average that uses a constant weighting of all prior data, exponential functions utilize ever-decreasing weights based on how far back in time the observations are.
Observation weights in moving averages are all equal to 1/N. However, with exponential smoothing, the weights assigned to the data depend on one or more smoothing parameters, which must be computed.
Testing. CI/CD. Monitoring. | 406 | 2,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-26 | latest | en | 0.907078 |
https://mathematics-monster.com/symbols/Integral.html | 1,721,755,564,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518059.67/warc/CC-MAIN-20240723163815-20240723193815-00792.warc.gz | 345,045,042 | 6,377 | Integral
The Mathematical Symbol "Integral (∫)"
homesitemapsymbolsintegral
The ∫ Symbol in Mathematics: Integral
The realm of mathematics is teeming with symbols that encapsulate deep and intricate concepts. One of the most recognizable and foundational symbols is the ∫ symbol, denoting the concept of integration. This article will delve into its meaning, core applications, and offer some illustrative examples.
Usage
The ∫ symbol signifies the operation of integration, a foundational concept in calculus. Integration can be viewed as the reverse process of differentiation and is instrumental in finding areas under curves, solving differential equations, and describing accumulated quantities, among other applications.
Examples
• Example 1: Basic Integration:
The expression $$∫ x^2 dx$$ represents the integral of $$x^2$$ with respect to $$x$$. The result is $$\frac{x^3}{3} + C$$, where $$C$$ is an arbitrary constant.
• Example 2: Definite Integral:
The expression $$∫_0^1 x^2 dx$$ denotes the definite integral of $$x^2$$ from 0 to 1. This computes the area under the curve $$y = x^2$$ between those two bounds, resulting in a value of $$\frac{1}{3}$$.
In summary, the ∫ symbol stands as a beacon of calculus, representing the integration operation. Its applications span a wide array of mathematical and real-world problems, making it indispensable in the study and application of mathematics.
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Codes for the ∫ Symbol
The Symbol ∫ Alt Code Alt 8747 HTML Code ∫ HTML Entity ∫ CSS Code \222B Hex Code ∫ Unicode U+222B
How To Insert the ∫ Symbol
(Method 1) Copy and paste the symbol.
The easiest way to get the ∫ symbol is to copy and paste it into your document.
Bear in mind that this is a UTF-8 encoded character. It must be encoded as UTF-8 at all stages (copying, replacing, editing, pasting), otherwise it will render as random characters or the dreaded �.
(Method 2) Use the "Alt Code."
If you have a keyboard with a numeric pad, you can use this method. Simply hold down the Alt key and type 8747. When you lift the Alt key, the symbol appears. ("Num Lock" must be on.)
(Method 3) Use the HTML Decimal Code (for webpages).
HTML TextOutput
<b>My symbol: ∫</b>My symbol: ∫
(Method 4) Use the HTML Entity Code (for webpages).
HTML TextOutput
<b>My symbol: ∫</b>My symbol: ∫
(Method 5) Use the CSS Code (for webpages).
CSS and HTML TextOutput
<style>
span:after {
content: "\222B";}
</style>
<span>My symbol:</span>
My symbol: ∫
(Method 6) Use the HTML Hex Code (for webpages and HTML canvas).
HTML TextOutput
<b>My symbol: ∫</b>My symbol: ∫
On the assumption that you already have your canvas and the context set up, use the Hex code in the format 0x222B to place the ∫ symbol on your canvas. For example:
JavaScript Text
const x = "0x"+"E9"
ctx.fillText(String.fromCodePoint(x), 5, 5);
Output
(Method 7) Use the Unicode (for various, e.g. Microsoft Office, JavaScript, Perl).
The Unicode for ∫ is U+222B. The important part is the hexadecimal number after the U+, which is used in various formats. For example, in Microsoft Office applications (e.g. Word, PowerPoint), do the following:
TypeOutput
222B
[Hold down Alt]
[Press x]
(The 222B turns into ∫. Note that you can omit any leading zeros.)
In JavaScript, the syntax is \uXXXX. So, our example would be \u222B. (Note that the format is 4 hexadecimal characters.)
JavaScript TextOutput
let str = "\u222B"
document.write("My symbol: " + str)
My symbol: ∫
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Infinity (∞) | 1,123 | 4,428 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-30 | latest | en | 0.842457 |
https://www.informit.com/articles/article.aspx?p=2216986&seqNum=3 | 1,723,781,207,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00330.warc.gz | 608,973,005 | 39,737 | # Programming: Principles and Practice Using C++: Vectors and Arrays
This chapter is from the book
## 18.3 Copying
Consider again our incomplete vector:
```class vector {
int sz; // the size
double* elem; // a pointer to the elements
public:
vector(int s) // constructor
:sz{s}, elem{new double[s]} { /* . . . */ } // allocates memory
~vector() // destructor
{ delete[] elem; } // deallocates memory
// . . .
};```
Let’s try to copy one of these vectors:
```void f(int n)
{
vector v(3); // define a vector of 3 elements
v.set(2,2.2); // set v[2] to 2.2
vector v2 = v; // what happens here?
// . . .
}```
Ideally, v2 becomes a copy of v (that is, = makes copies); that is, v2.size()==v.size() and v2[i]==v[i] for all is in the range [0:v.size()). Furthermore, all memory is returned to the free store upon exit from f(). That’s what the standard library vector does (of course), but it’s not what happens for our still-far-too-simple vector. Our task is to improve our vector to get it to handle such examples correctly, but first let’s figure out what our current version actually does. Exactly what does it do wrong? How? And why? Once we know that, we can probably fix the problems. More importantly, we have a chance to recognize and avoid similar problems when we see them in other contexts.
The default meaning of copying for a class is “Copy all the data members.” That often makes perfect sense. For example, we copy a Point by copying its coordinates. But for a pointer member, just copying the members causes problems. In particular, for the vectors in our example, it means that after the copy, we have v.sz==v2.sz and v.elem==v2.elem so that our vectors look like this:
That is, v2 doesn’t have a copy of v’s elements; it shares v’s elements. We could write
```v.set(1,99); // set v[1] to 99
v2.set(0,88); // set v2[0] to 88
cout << v.get(0) << ' ' << v2.get(1);```
The result would be the output 88 99. That wasn’t what we wanted. Had there been no “hidden” connection between v and v2, we would have gotten the output 0 0, because we never wrote to v[0] or to v2[1]. You could argue that the behavior we got is “interesting,” “neat!” or “sometimes useful,” but that is not what we intended or what the standard library vector provides. Also, what happens when we return from f() is an unmitigated disaster. Then, the destructors for v and v2 are implicitly called; v’s destructor frees the storage used for the elements using
`delete[] elem;`
and so does v2’s destructor. Since elem points to the same memory location in both v and v2, that memory will be freed twice with likely disastrous results (§17.4.6).
### 18.3.1 Copy constructors
So, what do we do? We’ll do the obvious: provide a copy operation that copies the elements and make sure that this copy operation gets called when we initialize one vector with another.
Initialization of objects of a class is done by a constructor. So, we need a constructor that copies. Unsurprisingly, such a constructor is called a copy constructor. It is defined to take as its argument a reference to the object from which to copy. So, for class vector we need
vector(const vector&);
This constructor will be called when we try to initialize one vector with another. We pass by reference because we (obviously) don’t want to copy the argument of the constructor that defines copying. We pass by const reference because we don’t want to modify our argument (§8.5.6). So we refine vector like this:
```class vector {
int sz;
double* elem;
public:
vector(const vector&) ; // copy constructor: define copy
// . . .
};```
The copy constructor sets the number of elements (sz) and allocates memory for the elements (initializing elem) before copying element values from the argument vector:
```vector:: vector(const vector& arg)
// allocate elements, then initialize them by copying
:sz{arg.sz}, elem{new double[arg.sz]}
{
copy(arg.elem,arg.elem+sz,elem); // std::copy(); see §B.5.2
}```
Given this copy constructor, consider again our example:
vector v2 = v;
This definition will initialize v2 by a call of vector’s copy constructor with v as its argument. Again given a vector with three elements, we now get
Given that, the destructor can do the right thing. Each set of elements is correctly freed. Obviously, the two vectors are now independent so that we can change the value of elements in v without affecting v2 and vice versa. For example:
```v.set(1,99); // set v[1] to 99
v2.set(0,88); // set v2[0] to 88
cout << v.get(0) << ' ' << v2.get(1);```
This will output 0 0.
vector v2 = v;
we could equally well have said
vector v2 {v};
When v (the initializer) and v2 (the variable being initialized) are of the same type and that type has copying conventionally defined, those two notations mean exactly the same thing and you can use whichever notation you like better.
### 18.3.2 Copy assignments
We handle copy construction (initialization), but we can also copy vectors by assignment. As with copy initialization, the default meaning of copy assignment is memberwise copy, so with vector as defined so far, assignment will cause a double deletion (exactly as shown for copy constructors in §18.3.1) plus a memory leak. For example:
```void f2(int n)
{
vector v(3); // define a vector
v.set(2,2.2);
vector v2(4);
v2 = v; // assignment: what happens here?
// . . .
}```
We would like v2to be a copy of v (and that’s what the standard library vector does), but since we have said nothing about the meaning of assignment of our vector, the default assignment is used; that is, the assignment is a memberwise copy so that v2’s sz and elem become identical to v’s sz and elem, respectively. We can illustrate that like this:
When we leave f2(), we have the same disaster as we had when leaving f() in §18.3 before we added the copy constructor: the elements pointed to by both v and v2 are freed twice (using delete[]). In addition, we have leaked the memory initially allocated for v2’s four elements. We “forgot” to free those. The remedy for this copy assignment is fundamentally the same as for the copy initialization (§18.3.1). We define an assignment that copies properly:
```class vector {
int sz;
double* elem;
public:
vector& operator=(const vector&) ; // copy assignment
// . . .
};
vector& vector::operator=(const vector& a)
// make this vector a copy of a
{
double* p = new double[a.sz]; // allocate new space
copy(a.elem,a.elem+a.sz,elem); // copy elements
delete[] elem; // deallocate old space
elem = p; // now we can reset elem
sz = a.sz;
return *this; // return a self-reference (see §17.10)
}
```
Assignment is a bit more complicated than construction because we must deal with the old elements. Our basic strategy is to make a copy of the elements from the source vector:
``` double* p = new double[a.sz]; // allocate new space
copy(a.elem,a.elem+a.sz,elem); // copy elements```
Then we free the old elements from the target vector:
`delete[] elem; // deallocate old space`
Finally, we let elem point to the new elements:
```elem = p; // now we can reset elem
sz = a.sz;```
We can represent the result graphically like this:
We now have a vector that doesn’t leak memory and doesn’t free (delete[]) any memory twice.
When implementing the assignment, you could consider simplifying the code by freeing the memory for the old elements before creating the copy, but it is usually a very good idea not to throw away information before you know that you can replace it. Also, if you did that, strange things would happen if you assigned a vector to itself:
```vector v(10);
v = v; // self-assignment```
Please check that our implementation handles that case correctly (if not with optimal efficiency).
### 18.3.3 Copy terminology
Copying is an issue in most programs and in most programming languages. The basic issue is whether you copy a pointer (or reference) or copy the information pointed to (referred to):
• Shallow copy copies only a pointer so that the two pointers now refer to the same object. That’s what pointers and references do.
• Deep copy copies what a pointer points to so that the two pointers now refer to distinct objects. That’s what vectors, strings, etc. do. We define copy constructors and copy assignments when we want deep copy for objects of our classes.
Here is an example of shallow copy:
```int* p = new int{77};
int* q = p; // copy the pointer p
*p = 88; // change the value of the int pointed to by p and q```
We can illustrate that like this:
In contrast, we can do a deep copy:
```int* p = new int{77};
int* q = new int{*p}; // allocate a new int, then copy the value pointed to by p
*p = 88; // change the value of the int pointed to by p```
We can illustrate that like this:
Using this terminology, we can say that the problem with our original vector was that it did a shallow copy, rather than copying the elements pointed to by its elem pointer. Our improved vector, like the standard library vector, does a deep copy by allocating new space for the elements and copying their values. Types that provide shallow copy (like pointers and references) are said to have pointer semantics or reference semantics (they copy addresses). Types that provide deep copy (like string and vector) are said to have value semantics (they copy the values pointed to). From a user perspective, types with value semantics behave as if no pointers were involved — just values that can be copied. One way of thinking of types with value semantics is that they “work just like integers” as far as copying is concerned.
### 18.3.4 Moving
If a vector has a lot of elements, it can be expensive to copy. So, we should copy vectors only when we need to. Consider an example:
```vector fill(istream& is)
{
vector res;
for (double x; is>>x; ) res.push_back(x);
return res;
}
void use()
{
vector vec = fill(cin);
// ... use vec ...
}```
Here, we fill the local vector res from the input stream and return it to use(). Copying res out of fill() and into vec could be expensive. But why copy? We don’t want a copy! We can never use the original (res) after the return. In fact, res is destroyed as part of the return from fill(). So how can we avoid the copy? Consider again how a vector is represented in memory:
We would like to “steal” the representation of res to use for vec. In other words, we would like vec to refer to the elements of res without any copy.
After moving res’s element pointer and element count to vec, res holds no elements. We have successfully moved the value from res out of fill() to vec. Now, res can be destroyed (simply and efficiently) without any undesirable side effects:
We have successfully moved 100,000 doubles out of fill() and into its caller at the cost of four single-word assignments.
How do we express such a move in C++ code? We define move operations to complement the copy operations:
```class vector {
int sz;
double* elem;
public:
vector(vector&& a); // move constructor
vector& operator=(vector&&); // move assignment
// . . .
};```
The funny && notation is called an “rvalue reference.” We use it for defining move operations. Note that move operations do not take const arguments; that is, we write (vector&&) and not (const vector&&). Part of the purpose of a move operation is to modify the source, to make it “empty.” The definitions of move operations tend to be simple. They tend to be simpler and more efficient than their copy equivalents. For vector, we get
```vector::vector(vector&& a)
:sz{a.sz}, elem{a.elem} // copy a’s elem and sz
{
a.sz = 0; // make a the empty vector
a.elem = nullptr;
}
vector& vector::operator=(vector&& a) // move a to this vector
{
delete[] elem; // deallocate old space
elem = a.elem; // copy a’s elem and sz
sz = a.sz;
a.elem = nullptr; // make a the empty vector
a.sz = 0;
return *this; // return a self-reference (see §17.10)
}```
By defining a move constructor, we make it easy and cheap to move around large amounts of information, such as a vector with many elements. Consider again:
```vector fill(istream& is)
{
vector res;
for (double x; is>>x; ) res.push_back(x);
return res;
}```
The move constructor is implicitly used to implement the return. The compiler knows that the local value returned (res) is about to go out of scope, so it can move from it, rather than copying.
The importance of move constructors is that we do not have to deal with pointers or references to get large amounts of information out of a function. Consider this flawed (but conventional) alternative:
```vector* fill2(istream& is)
{
vector* res = new vector;
for (double x; is>>x; ) res->push_back(x);
return res;
}
void use2()
{
vector* vec = fill(cin);
// ... use vec ...
delete vec;
}```
Now we have to remember to delete the vector. As described in §17.4.6, deleting objects placed on the free store is not as easy to do consistently and correctly as it might seem.
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This web site contains links to other sites. Please be aware that we are not responsible for the privacy practices of such other sites. We encourage our users to be aware when they leave our site and to read the privacy statements of each and every web site that collects Personal Information. This privacy statement applies solely to information collected by this web site. | 4,882 | 23,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-33 | latest | en | 0.830127 |
https://wikis.hu-berlin.de/mmint/Basics:_The_Sample_Space,_Events_and_Probabilities/en | 1,606,295,234,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141181482.18/warc/CC-MAIN-20201125071137-20201125101137-00560.warc.gz | 556,303,997 | 6,553 | # The Sample Space, Events and Probabilities
The set of all possible outcomes of an exeriment is called the sample space which we will denote by $S$. Consider the process of rolling a die. The set of possible outcomes is the set $S=\{1,2,3,4,5,6\}$ Each element of $\ S$ is a basic outcome. However, one might be interested in whether the number thrown is even, whether it is a three or a seven, and so on. Thus we need to be able to speak of various combinations of basic outcomes, that is subsets of $S$. An event is defined to be a subset of the set of possible outcomes $S$. We will denote an event using the symbol $E$. Events which consist of only one element, such as ’a two was thrown’ are called simple events or elementary events. Simple events are by definition not divisible into more basic events, as each of them includes one and only one possible outcome. Example:Rolling a single die once results in the occurrence of one of the simple events {1}, {2}, {3}, {4}, {5}, {6}.As we have indicated, the sample space $S$ is {1,2,3,4,5,6}. Example:Tossing a coin twice.Sample space: $S=\{TT,TH,HT,HH\}$.Simple events: {$TT$},{$TH$},{$HT$ },{$HH$}, $T\equiv$Tail , $H\equiv$Head.This specification also holds if two coins are tossed once. It will be convenient to be able to combine events in various ways, in order that we can make statements such as ”one of these two events happened” or ” both events occured”. For example, one might want to say that ”either a 2 or 4 was thrown”. or ”an even number larger than 3 was thrown. Since events are sets, (in particular subsets of the set $S$), we may draw upon the conventional tools of set theory. | 440 | 1,662 | {"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": 15, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2020-50 | latest | en | 0.947268 |
http://www.utteraccess.com/forum/lofiversion/index.php/t125006.html | 1,369,472,886,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705884968/warc/CC-MAIN-20130516120444-00035-ip-10-60-113-184.ec2.internal.warc.gz | 787,095,970 | 2,325 | Full Version: COUNTIF Query
Kell
Hi,
I have got a column containing dates in the format dd/mm/yyyy. I want to be able to do a count of these dates for each month i.e. 12 dates in September, 3 in October etc. so that I can count how many entries were made in September, October etc.
I can get it to count just the days:
=COUNTIF(A1:A50,"01/09/2002")
but I do not know how I would count how many entries are made for that month, not day?
Any help will be appreciated.
Thanks
Kel
MarkM
OK, ever heard of array formula's?? Well here goes:
type the following formula but don't press enter:
=count(if(month(a1:a50)=9,a1:a50))
Now hold down Ctrl + Shift and now hit enter. The formula should now have "{" type brackets around it to look like this:
{=count(if(month(a1:a50)=9,a1:a50))}
This should work. Bare in mind that if you need to change the formula for whatever reason, you need to do the whole Ctrl + Shift + Enter thing again.
Look in help if you need to understand how this works. One thing you should remember is that if there are any blank cells in a range of dates Excel treats them as 01/01/1900, so if you are looking at dates occuring in January, and you have blank cells in the range, the formula will over-count.
HTH
kountry
Good answer and thanks for the advice on the pitfalls of testing a whole batch of dates.
Couldn't you also add an if statement to the Countif function which excludes any dates with Year(Date)=1900? Would that work?
Cheers
MarkM
Yes, of course, I was just leaving that up to the poster to work out for herself! | 391 | 1,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.953008 |
https://exercism.io/tracks/javascript/exercises/grains/solutions/5ab72ca94e7cb07512281472 | 1,590,919,320,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413097.49/warc/CC-MAIN-20200531085047-20200531115047-00450.warc.gz | 338,382,867 | 7,670 | # brunnock's solution
## to Grains in the JavaScript Track
Published at Jul 13 2018 · 0 comments
Instructions
Test suite
Solution
#### Note:
This solution was written on an old version of Exercism. The tests below might not correspond to the solution code, and the exercise may have changed since this code was written.
Calculate the number of grains of wheat on a chessboard given that the number on each square doubles.
There once was a wise servant who saved the life of a prince. The king promised to pay whatever the servant could dream up. Knowing that the king loved chess, the servant told the king he would like to have grains of wheat. One grain on the first square of a chess board. Two grains on the next. Four on the third, and so on.
There are 64 squares on a chessboard.
Write code that shows:
• how many grains were on each square, and
• the total number of grains
## For bonus points
Did you get the tests passing and the code clean? If you want to, these are some additional things you could try:
• Optimize for speed.
Then please share your thoughts in a comment on the submission. Did this experiment make the code better? Worse? Did you learn anything from it?
## Setup
Go through the setup instructions for JavaScript to install the necessary dependencies:
http://exercism.io/languages/javascript/installation
## Running the test suite
The provided test suite uses Jasmine. You can install it by opening a terminal window and running the following command:
``````npm install -g jasmine
``````
Run the test suite from the exercise directory with:
``````jasmine grains.spec.js
``````
In many test suites all but the first test have been marked "pending". Once you get a test passing, activate the next one by changing `xit` to `it`.
## Source
JavaRanch Cattle Drive, exercise 6 http://www.javaranch.com/grains.jsp
## Submitting Incomplete Solutions
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
### big-integer.spec.js
``````var BigInt = require('./big-integer');
describe('The big-integer module\'s returned object', function () {
var bigI;
beforeEach(function () {
bigI = BigInt(42);
});
afterEach(function () {
bigI = null;
});
it('is not a number', function () {
expect(typeof 42).toBe('number');
expect(typeof bigI).not.toBe('number');
expect(typeof bigI).toBe('object');
});
it('can be compared to a stringified number by calling \'.toString()\'',
function () {
expect(bigI).not.toBe(42);
expect(bigI).not.toBe('42');
expect(bigI.toString()).toBe('42');
// NOTE:
// The '==' operator calls '.toString()' here in order to compare.
expect(bigI == '42').toBe(true);
// While the line above is easier to write and read, we will use the
// 'expect(bigI.toString()).toBe(expected)' way so that test failure
// "Expected '84' to be '42'." instead of
// "Expected false to be true."
});
it('is immutable', function () {
expect(bigI.toString()).toBe('42');
bigI.subtract(10);
expect(bigI.toString()).toBe('42');
});
expect(bigI.toString()).toBe('84');
});
it('can perform power operations', function () {
bigI = BigInt(10);
bigI = bigI.pow(2);
expect(bigI.toString()).toBe('100');
});
// The "Methods" section of the README is especially useful:
//
// https://github.com/peterolson/BigInteger.js#methods
});``````
### grains.spec.js
``````/**
* In JavaScript, integers beyond +/- 9007199254740991 cannot be accurately
* represented. To see this in action, console.log() out the expected number
* of grains on square #64:
*
* console.log(9223372036854775808);
* // => 9223372036854776000
* // ^^^^
*
* This is because, in JavaScript, integers are represented as 64-bit floating
*
* https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number/MAX_SAFE_INTEGER
* http://stackoverflow.com/questions/307179/what-is-javascripts-highest-integer-value-that-a-number-can-go-to-without-losin
*
* So, an accurate solution to this problem requires the use of a
* "big integer" type. There are multiple ways to use big integer types.
* We have provided you with BigInteger.js. You can read more about it here:
*
* https://github.com/peterolson/BigInteger.js
* ^--- The "Methods" section of the README will be especially helpful.
*
* https://github.com/peterolson/BigInteger.js/blob/master/spec/spec.js
* ^--- Tests are a good way to understand, in addition to the README.
*
* To get you started, this folder has a file of the big-integer module.
* See its tests in this folder for a quick primer on how to use it! ( :
*/
var Grains = require('./grains');
describe('Grains', function () {
var grains = new Grains();
it('square 1', function () {
expect(grains.square(1)).toBe('1');
});
xit('square 2', function () {
expect(grains.square(2)).toBe('2');
});
xit('square 3', function () {
expect(grains.square(3)).toBe('4');
});
xit('square 4', function () {
expect(grains.square(4)).toBe('8');
});
xit('square 16', function () {
expect(grains.square(16)).toBe('32768');
});
xit('square 32', function () {
expect(grains.square(32)).toBe('2147483648');
});
xit('square 64', function () {
expect(grains.square(64)).toBe('9223372036854775808');
});
xit('total', function () {
expect(grains.total()).toBe('18446744073709551615');
});
});``````
``````var Grains = function() {
this.square = function(exp) {
return Math.pow(2,exp-1);
}
this.total = function() {
var total=0;
for (var x=1; x<65; ++x) {
total += this.square(x);
}
}
}
module.exports=Grains;`````` | 1,384 | 5,521 | {"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} | 3.359375 | 3 | CC-MAIN-2020-24 | latest | en | 0.921143 |
http://www.riddlesandanswers.com/puzzles-brain-teasers/may-temperature-may-reach-as-high-as-2000-degrees-celcius-who-am-i-riddles/ | 1,579,952,731,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251672440.80/warc/CC-MAIN-20200125101544-20200125130544-00408.warc.gz | 257,330,076 | 26,497 | # MAY TEMPERATURE MAY REACH AS HIGH AS 2000 DEGREES CELCIUS WHO AM I RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS
#### Popular Searches
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## Solving May Temperature May Reach As High As 2000 Degrees Celcius Who Am I Riddles
Here we've provide a compiled a list of the best may temperature may reach as high as 2000 degrees celcius who am i puzzles and riddles to solve we could find.
Our team works hard to help you piece fun ideas together to develop riddles based on different topics. Whether it's a class activity for school, event, scavenger hunt, puzzle assignment, your personal project or just fun in general our database serve as a tool to help you get started.
Here's a list of related tags to browse:
The results compiled are acquired by taking your search "may temperature may reach as high as 2000 degrees celcius who am i" and breaking it down to search through our database for relevant content.
Browse the list below:
## 2000 Degrees Celsius
Hint:
The Thermosphere
Did you answer this riddle correctly?
YES NO
Solved: 44%
## Aztecs And Mayas Riddle
Hint:
Mexico
Did you answer this riddle correctly?
YES NO
Solved: 60%
## Brought On The Mayflower Riddle
Hint:
Pilgrims and smallpox!
Did you answer this riddle correctly?
YES NO
Solved: 54%
## Whats The Temperature?
Hint:
273.15 degrees Celsius. When making conversions like this you must convert to the Kelvin scale since it is the only one that works on a scale beginning with 0.
Did you answer this riddle correctly?
YES NO
Solved: 42%
## The Highest Altitude Layer
Hint:
The Thermosphere
Did you answer this riddle correctly?
YES NO
Solved: 89%
## Temperature Drop Riddle
Hint:
Troposphere
Did you answer this riddle correctly?
YES NO
Solved: 67%
## Higher Than A House
Hint:
Any, Houses can't jump
Did you answer this riddle correctly?
YES NO
Solved: 27%
## Jumping Higher Than A House
Hint:
All of them. Houses can't jump!
Did you answer this riddle correctly?
YES NO
Solved: 46%
## High In The Sky Riddle
Hint:
The Tower of Babel
Did you answer this riddle correctly?
YES NO
Solved: 67%
## High Tide Boat Riddle
Hint:
None. The boat is floating on the water, so as the tide rises, so does the ladder.
Did you answer this riddle correctly?
YES NO
Solved: 60% | 612 | 2,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-05 | latest | en | 0.862398 |
https://exercism.io/tracks/go/exercises/bob/solutions/87ad5cbc83664ec2a8b897239f4e17e8 | 1,576,052,906,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540530452.95/warc/CC-MAIN-20191211074417-20191211102417-00237.warc.gz | 354,576,766 | 7,161 | artemkorsakov's solution
to Bob in the Go Track
Published at Jan 31 2019 · 0 comments
Instructions
Test suite
Solution
Bob is a lackadaisical teenager. In conversation, his responses are very limited.
Bob answers 'Sure.' if you ask him a question.
He answers 'Whoa, chill out!' if you yell at him.
He answers 'Calm down, I know what I'm doing!' if you yell a question at him.
He says 'Fine. Be that way!' if you address him without actually saying anything.
He answers 'Whatever.' to anything else.
Bob's conversational partner is a purist when it comes to written communication and always follows normal rules regarding sentence punctuation in English.
Running the tests
To run the tests run the command `go test` from within the exercise directory.
If the test suite contains benchmarks, you can run these with the `--bench` and `--benchmem` flags:
``````go test -v --bench . --benchmem
``````
Keep in mind that each reviewer will run benchmarks on a different machine, with different specs, so the results from these benchmark tests may vary.
Further information
For more detailed information about the Go track, including how to get help if you're having trouble, please visit the exercism.io Go language page.
Source
Inspired by the 'Deaf Grandma' exercise in Chris Pine's Learn to Program tutorial. http://pine.fm/LearnToProgram/?Chapter=06
Submitting Incomplete Solutions
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
bob_test.go
``````package bob
import "testing"
func TestHey(t *testing.T) {
for _, tt := range testCases {
actual := Hey(tt.input)
if actual != tt.expected {
msg := `
ALICE (%s): %q
BOB: %s
Expected Bob to respond: %s`
t.Fatalf(msg, tt.description, tt.input, actual, tt.expected)
}
}
}
func BenchmarkHey(b *testing.B) {
for i := 0; i < b.N; i++ {
for _, tt := range testCases {
Hey(tt.input)
}
}
}``````
cases_test.go
``````package bob
// Source: exercism/problem-specifications
// Commit: ca79943 Bob: Fix grammatical error in testdata (#1319)
// Problem Specifications Version: 1.4.0
var testCases = []struct {
description string
input string
expected string
}{
{
"stating something",
"Tom-ay-to, tom-aaaah-to.",
"Whatever.",
},
{
"shouting",
"WATCH OUT!",
"Whoa, chill out!",
},
{
"shouting gibberish",
"FCECDFCAAB",
"Whoa, chill out!",
},
{
"Does this cryogenic chamber make me look fat?",
"Sure.",
},
{
"asking a numeric question",
"You are, what, like 15?",
"Sure.",
},
{
"fffbbcbeab?",
"Sure.",
},
{
"talking forcefully",
"Let's go make out behind the gym!",
"Whatever.",
},
{
"using acronyms in regular speech",
"It's OK if you don't want to go to the DMV.",
"Whatever.",
},
{
"forceful question",
"WHAT THE HELL WERE YOU THINKING?",
"Calm down, I know what I'm doing!",
},
{
"shouting numbers",
"1, 2, 3 GO!",
"Whoa, chill out!",
},
{
"no letters",
"1, 2, 3",
"Whatever.",
},
{
"question with no letters",
"4?",
"Sure.",
},
{
"shouting with special characters",
"ZOMG THE %^*@#\$(*^ ZOMBIES ARE COMING!!11!!1!",
"Whoa, chill out!",
},
{
"shouting with no exclamation mark",
"I HATE THE DMV",
"Whoa, chill out!",
},
{
"statement containing question mark",
"Ending with ? means a question.",
"Whatever.",
},
{
"non-letters with question",
":) ?",
"Sure.",
},
{
"prattling on",
"Wait! Hang on. Are you going to be OK?",
"Sure.",
},
{
"silence",
"",
"Fine. Be that way!",
},
{
"prolonged silence",
" ",
"Fine. Be that way!",
},
{
"alternate silence",
"\t\t\t\t\t\t\t\t\t\t",
"Fine. Be that way!",
},
{
"multiple line question",
"\nDoes this cryogenic chamber make me look fat?\nNo.",
"Whatever.",
},
{
"starting with whitespace",
" hmmmmmmm...",
"Whatever.",
},
{
"ending with whitespace",
"Okay if like my spacebar quite a bit? ",
"Sure.",
},
{
"other whitespace",
"\n\r \t",
"Fine. Be that way!",
},
{
"non-question ending with whitespace",
"This is a statement ending with whitespace ",
"Whatever.",
},
}``````
``````package bob
import (
"regexp"
"strings"
)
func Hey(remark string) string {
var remarkWithoutWhitespaces = regexp.MustCompile("(\\s+)").ReplaceAllString(remark, "");
if len(remarkWithoutWhitespaces) == 0 {
return "Fine. Be that way!"
}
var isQuestion = strings.HasSuffix(remarkWithoutWhitespaces, "?")
var isYell = (remarkWithoutWhitespaces == strings.ToUpper(remarkWithoutWhitespaces)) && (regexp.MustCompile(".*([A-Z]+).*").MatchString(remarkWithoutWhitespaces))
if isQuestion && isYell {
return "Calm down, I know what I'm doing!"
}
if isQuestion {
return "Sure."
}
if isYell {
return "Whoa, chill out!"
}
return "Whatever."
}`````` | 1,242 | 4,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-51 | latest | en | 0.822977 |
http://www-4.unipv.it/offertaformativa/portale/corso.php?lingua=2&idAttivitaFormativa=342065&modulo=0&anno=2021/2022 | 1,675,374,615,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500041.18/warc/CC-MAIN-20230202200542-20230202230542-00063.warc.gz | 43,419,931 | 2,986 | FUNDAMENTALS OF MECHANICS OF MATERIALS AND STRUCTURES
Stampa
Enrollment year
2020/2021
2021/2022
Regulations
DM270
ICAR/08 (CONSTRUCTION SCIENCE)
Department
DEPARTMENT OF ELECTRICAL,COMPUTER AND BIOMEDICAL ENGINEERING
Course
INDUSTRIAL ENGINEERING
Curriculum
Meccanica
Year of study
Period
2nd semester (07/03/2022 - 17/06/2022)
ECTS
6
Lesson hours
54 lesson hours
Language
Italian
Activity type
WRITTEN AND ORAL TEST
Teacher
MORGANTI SIMONE (titolare) - 5 ECTS
ALAIMO GIANLUCA - 1 ECTS
Prerequisites
Calculus 1, Physics 1, Mathematical Physics, Algebra
Learning outcomes
Understanding and assimilation of basic concepts related to the foundations of continuum mechanics for general 3D elastic solids and elementary mechanics of deformable one-dimensional structures. Acquisition of operational capabilities necessary to solve statically determinate and indeterminate beams of basic type using different approaches, as well as the schematic design and verification of beams with general loading conditions.
Course contents
Basic concepts: force, moment, couple, vector/tensor operations, notations
Beam equilibrium: Kinematics and statics of the straight beam, internal actions and diagrams.
Euler-Bernoulli beam theory, consistency, equilibrium, constitutive law. Elastic-linear-isotropic law and formulation of the elastic problem.
Strain state. Consistency of the deformable continuum and kinematics relations. The finite-deformation tensor (Green-Lagrange). Hypothesis of "small displacements": small deformation tensor. Principal deformations and invariants. Volume and shape change. Internal consistency.
Stress state: General aspects of the structural problem. Force and stress. the Cauchy stress tensor. Principal directions and invariants. 2D and 3D stress states. The Mohr stress representation. Equilibrium conditions.
Constitutive law. Stress-strain relations and experimental evidence. Elasticity, anelasticity, failure. Elastic law: energy aspects, existence and uniqueness of the elastic response. Elastic-linear-isotropic law: elastic constants. Elastic limit and failure-yield criteria. The elastic problem.
Formulation of the problem and uniqueness of the solution.
Position of the problem of De Saint Venant. Axial action and bending. Torsion. Shear: approximate treatment.
Teaching methods
Lectures: 32 hours
Practical classes: 22 hours | 540 | 2,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.808976 |
http://plus.maths.org/content/os/issue8/features/proof2/index | 1,418,970,993,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802768276.101/warc/CC-MAIN-20141217075248-00135-ip-10-231-17-201.ec2.internal.warc.gz | 224,643,310 | 18,559 | # The origins of proof II : Kepler's proofs
Issue 8
May 1999
As we explained in The Origins of Proof, Part I in Issue 7 of PASS Maths, the concept of a "proof" was developed in the field of geometry by the Greeks. The Pythagoreans and Euclid were among the mathematicians who developed the idea of abstract deduction. But during the Renaissance the philosophy of nature increasingly came to rely upon mathematics to help to explain the Universe and its workings.
Astronomy had previously been a branch of mathematics, that is, the business of astronomers was only to provide the means of calculating the positions of planets (as seen against the background of the fixed stars). Kepler brought astronomy into natural philosophy by proposing to find the actual paths of the planets in relation to the Sun which, following the theory put forward by Nicolaus Copernicus (1473 - 1543) in 1543, Kepler believed to be in the centre of the Universe and the body around which the planets (including the Earth) revolved. Thus, mathematician though he was, Kepler was doing what we should now call Physics. To find the paths of the planets he would need to use innovative forms of mathematics; and he tried to make his proofs as obviously sound as possible by spelling the steps out in detail and emphasising the physical basis for his mathematical procedures. Of course, his proofs are not what we would now call "proofs" in Mathematics, as they were based on data collected from observations - but they are mathematically rigorous and they introduce the crucially important notion of observational error. From then on astronomers learned to ask not whether a theory was right or wrong but whether it fitted the observations to within the known margin of error. Kepler's methods influenced the thinkers of the time, and are an important step in the development of proof as we know it in Science today.
Figure 1: Portrait of Kepler, engraving by Jakob van Heyden.
Johannes Kepler (1571 - 1630) is now chiefly remembered as a mathematical astronomer, in particular for discovering three laws that describe the motion of the planets. In their modern forms, these are
1. The path of each planet is an ellipse with the Sun at one of its foci.
2. As the planet moves along its path, a line joining the planet to the Sun sweeps out equal areas in equal times.
3. For any two planets, the ratio of the squares of their periods of revolution about the Sun is the same as the ratio of the cubes of their mean distances from the Sun.
These are usually known as "Kepler's three laws of planetary motion". It was from the third law that Isaac Newton (1642 - 1727) was to deduce the existence of an inverse square law of gravitation (in Mathematical Principles of Natural Philosophy or, to give it its Latin title, Philosophiae Naturalis Principia Mathematica, London, 1687).
Figure 2: Diagram to show what is now known as Kepler's second law (the area law). The darkly shaded areas are equal, and are swept out in equal times. The planet moves faster when nearer the Sun.
Kepler published the first two laws in 1609 in a work aptly titled New Astronomy. The third appeared ten years later in a book about cosmology, called The Harmony of the World. Both of these works are now available in English translation (see the bibliography) but neither is exactly an easy read. Indeed Kepler himself admits as much in the New Astronomy, saying in the second paragraph of his Introduction:
I myself, a professional mathematician, on re-reading my own work find it strains my mental powers to recall to mind from the figures the meanings of the demonstrations, meanings which I myself originally put into the figures and the text from my mind. But when I attempt to remedy the obscurity of the material by putting in extra words, I see myself falling into the opposite fault of becoming chatty in something mathematical.
[trans. JVF]
It is good to know one has the author's permission to find the mathematical reasoning difficult to follow.
The New Astronomy was printed as a large folio. In an effort to make it look attractive, the publisher added some decorations to the diagrams. Modern readers may not find the effect totally reassuring. Indeed, it seems that on the whole Kepler's contemporaries also found the work difficult. The reason for this was to do with how Kepler set about proving that planetary motion could be described by the two laws. This method was new. It was, in fact, new to Kepler as well as to his readers. So we shall start by having a look at what he was taught about mathematics.
# Kepler's early years and his education
Kepler was the eldest child of an innkeeper's daughter and a mercenary soldier. His father never returned from a military campaign he joined when his son was about six. So Kepler's childhood was spent in his grandfather's inn. He tells us he used to help serve customers there - which no doubt provided practice in arithmetic.
However, Kepler also got a conventional education, first at schools and then at the University of Tübingen. In consequence, his ideas about what made for a good proof were very solidly based on Euclid. That is, we may expect to find series of definitions, followed by what Kepler calls axioms (but Euclid called Common Notions and Postulates - see The Origins of Proof, Part I in Issue 7), followed in turn by deductive proofs. In most of his works, Kepler does his best to set all arguments out in this way. No doubt he thought it was clear, as well as visibly rigorous. All the same, he follows the habits of his time in not always filling in all the details.
The mathematical methods that Kepler employed in his proofs were generally also closely based on Euclid. And that includes the proofs in the New Astronomy.
# The orbit of Mars and the first two laws
Kepler was not studying planetary motion in general; he was engaged on the specific task of finding an orbit of Mars - that is, a mathematical model for its motion. This process took about five years, and there are 987 surviving folio pages of arithmetic. So it is hardly surprising that, while Kepler's full title for what we are calling the New Astronomy mentions "commentaries on the motion of the planet Mars", in his more personal reflections Kepler - obviously mindful of Mars being the Ancient Roman god of war - refers to his construction of the orbit as "my war with Mars".
Figure 3: Nicolaus Copernicus, De revolutionibus orbium coelestium (On the revolutions of the celestial spheres). The Copernican system of the world, in which the Sun is at rest in the centre of the Universe. Copernicus shows only the order of the spheres. There is no attempt to show their relative sizes.
As we know, it ended Kepler 1: Mars 0. And that turned out to be the final in regard to the motion of the planets. In 1687, Isaac Newton used Kepler's three laws to help him give a mathematical explanation of the Solar system. That is the happy ending. Kepler had done his bit towards proving that Copernicus was right to say the Earth went round the Sun (see figure 3). Kepler had been a Copernican ever since his student days at Tübingen.
## Constructing an orbit from circles
However, on the practical level, in considering the orbit of Mars, there was a serious problem of proof. And it was a new problem, too.
Until then, scholars who attempted to explain the workings of nature ("natural philosophers") had taken it for granted that for a motion to be eternal, as the motions of the planets obviously were, they must be uniform. Their being uniform meant that no point in time and no position in the path was special, so there was no need to look for a beginning or an end. Here we have a relic of the Ancient, pagan, idea that the Universe had neither beginning nor end in time (though it was considered finite in spatial extent). In fact, since the orbits showed a strict periodicity, the idea was that the motion of each body should be uniform and circular.
However, this satisfying idea - which we know about from the fourth century BC onwards - was rapidly overtaken by astronomical observation, and by the time of Claudius Ptolemy (who lived in Alexandria and is known to have made astronomical observations in the period 129 - 141 AD) astronomers constructed the orbits of planets by combining circles, the centre of one riding on another, and there were various compromises made about what counted as uniform motion. All the same, natural philosophers continued to believe they had got the basic principles right.
Figure 4: Tycho Brahe, De mundi aetherei recentioribus phaenomenis (Recent phenomena in the aetherial world), Uraniborg, 1599. The system of the world which Tycho believed was correct, in which the Earth is at rest in the centre of the Universe.
## Tycho Brahe's observations
Kepler had a huge number of observations to work with. These had been accumulated by Tycho Brahe (1546 - 1601). Tycho wanted to reform astronomy, but seems to have been unclear how this was to be done - except that it would involve a lot of observations and proving the truth of his own description of the planetary system, in which planets circle the Sun, which in turn moves round the stationary Earth (see figure 4). One is (I think) entitled to wonder whether one reason why Tycho went on making more and more observations was because that put off the day when he had to think about how to use them. As it happened, he simply decided to hand that part of the work over to assistants. And in the event he left it too late: he did not live to see Kepler's success. In fact, he might not have been entirely pleased about it, because Tycho did not believe Copernicus was right, whereas Kepler's laws shout at us that Kepler did.
Figure 5: Tycho Brahe, Astronomiae instauratae mechanica (Apparatus for the foundation of astronomy), 1598, mural quadrant (constructed in 1582). As the name implies, the instrument is part of the wall. The scale, radius 1.94 metres, is made of brass, and inside it there is an elaborate painting showing Tycho at work in his observatory (which has an alchemical laboratory in the basement). Two assistants are shown beside the quadrant helping to record the time of the observations. One reads the clock and the other writes down the reading.
Click here for a higher resolution image of the mural quadrant.
Kepler first encountered this mass of planetary observations when he came to Prague as Tycho's assistant, in 1600. As we shall see, he must have discussed with Tycho how accurate the observations were. Tycho seems to have told him that they could be relied upon to five minutes of arc. (One degree of arc is divided into sixty minutes, each of which is divided into sixty seconds.) In fact, Tycho's observing instruments (see figure 5) were exceedingly well-designed, and neither expense nor trouble was spared in their construction and in improving their performance. Modern checks - comparing Tycho's observed positions with positions of planets found by modern calculations (computer-assisted of course) - show that Tycho was right about how good they were. The limit is about that imposed by the resolving power of the human eye. (The telescope came into use for astronomy only after Tycho's death.)
## Kepler's calculations
Kepler's account of his calculations in the New Astronomy is certainly not a full diary of what he actually did, but there is good reason to suppose that it does, in outline, follow the actual progress of his reasoning. We know, therefore, that the law we call the second (sometimes also called the "area law") was in fact arrived at first.
### The area law
As his calculations progressed, Kepler was able to use exact geometrical methods to find areas. But at first finding the areas was a difficult task, and Kepler's method was approximate, effectively adding up areas of triangles with their vertex at the Sun and very small vertical angles, as shown in figure 6. Kepler used triangles whose vertical angle was one minute of arc. His method is best near the apsides, that is the positions at which the planet is nearest to or furthest from the Sun, which (once he knows the shape of the path) will turn out to be the ends of the major axis of the ellipse. The calculations are made more difficult by having to allow for the motion of the Earth, which is, of course, not in the same plane as the motion of Mars. (The planes of the two orbits are inclined to one another at about 1 degree 50 minutes.)
Figure 6: Finding the area swept out by a line joining a planet to the Sun. The points M1, M2 etc. represent positions of Mars. S is the Sun.
Kepler's method of finding areas is, as he says, similar to that used by Archimedes (c.287 - 212 BC) to find the area of a circle. But it is not exactly the same, and unlike Archimedes' procedure, Kepler's is not completely rigorous. However, Kepler's method does, also, look a little like integration, and it did in fact mark the beginning of a technique called "the calculus of indivisibles" which is now seen as an ancestor of modern calculus.
Anyway, by Chapter 40 of his book, Kepler had decided the area law was good enough to go on with. He used it as a way of measuring time.
### The ellipse
The shape of the curve also gave trouble. Essentially, Kepler had to try different shapes, using a certain number of observations to find the curve, then use the curve to find some more positions, for times when he had observations available, and then check whether these calculated positions agreed with the observed ones. He arrived at the ellipse at the end of Chapter 58:
My argument was as in Chapters 49, 50 and 56: The Circle of Chapter 43 is wrong because it is too large, and the ellipse of Chapter 45 is too small. And the amounts by which they respectively exceed and fall short are the same. Now between the circle and the ellipse there is no other intermediary except a different ellipse. Therefore the path of the planet is an Ellipse; ...
[New Astronomy, Chapter 58]
This passage marks sudden death for about two thousand years of using circles in astronomy. However, Kepler simply carries on with his work. He is, moreover, aware that the ellipse is an idealisation. In a letter to a friend he had said
the most true path of the planet [Mars] is an ellipse, ... or certainly so close to an ellipse that the difference is insensible.
[Kepler to David Fabricius, 11 October 1605]
That is, he recognised that he was always working with approximate values - even if he was looking for an exact geometrical figure to describe the path.
## Error and approximation
On the other hand, he knew the size of the errors in Tycho's observations, and used this to establish how closely theory must agree with observation. This raises another problem, one which Kepler himself had not noticed. He assumed that the error in values calculated from an orbit based on observations would be the same as the errors in the original observations. This agreement is in fact crucial to his whole proceeding.
We now know it is not true that the calculated orbit will necessarily have the same accuracy as the observations used to construct it. So-called "chaotic" behaviour of quite simple functions shows that small errors in input can become huge variations in output. However, Kepler was lucky. Well, perhaps not merely lucky: he was, after all, carrying out all these calculations by hand, so he could see that they were, as we would now put it, "well behaved".
The New Astronomy has seventy chapters (337 pages in the original edition). By the end, Kepler has - in the opinion of competent historians - established the elements of an orbit of Mars that agrees with the modern one so closely that in making the comparison allowance has to be made for the small changes that have occurred in the orbit since Kepler's time. (These changes, called "secular changes", are mainly due to the gravitational pull of Jupiter.)
## Assumptions and arithmetic
Kepler's first two laws were discovered in the course of his finding the orbit of a planet - a standard task for an astronomer, though, as we have seen, in this case there was the complicating factor that Kepler had a huge number of observations at his disposal. He naturally wanted to make the best possible use of them. In order to do so, he made as few assumptions as possible about the shape of the path. The repetitious mass of Kepler's constructions and computations is daunting, indeed so daunting that it is hard for anyone to understand how he could have kept his head as he worked his way along the line of reasoning. But he did. And the method of repetitious trial and error, converging toward an acceptably accurate answer, has the advantage that errors of calculation will show up at the next stage. A computer would have helped, though.
In fact, the analogy with the iterative procedures of programs written for electronic computers did tempt a twentieth-century astronomer to try a machine simulation of Kepler's procedure. He incautiously entitled the resulting paper "The computer versus Kepler" and concluded that Kepler had been clumsy, needing more iterations than the machine had. However, a few years later the same astronomer published another paper, called "The computer versus Kepler revisited", in which it was confessed that there had been a mistake, and Kepler had in fact done at least as well as the computer.
In Kepler's day, astronomers took it for granted that they would need to do a lot of arithmetic, and assistants were sometimes employed to help with the routine work. However, it was exactly at this time that people were making the first attempts to build machines to carry out arithmetical operations automatically. An abacus does not qualify because you have to push the beads (or move the counters).
Figure 7: Wilhelm Schickard's calculating machine, a sketch by Schickard enclosed with his letter to Kepler dated 7 March 1624.
Among the first of these machines for carrying out arithmetic was one designed by Kepler's friend Wilhelm Schickard (1582 - 1635), who was a professor of mathematics at Tübingen (see figure 7). Some letters survive in which Kepler and Schickard discuss this project, and it seems rather likely that Kepler contributed some ideas to the design. The machine, like most others of the time, never got much beyond a working prototype. However, it is interesting, and novel, in being designed to lose numbers from the right when answers came out with too many figures to be accommodated in the machine. Perhaps that idea was more likely to occur to an astronomer? Astronomical calculations very often involve working with long numbers.
# The third law
Kepler's third law, which relates the periods of planets to the sizes of their orbits, is really a law describing the structure of the Solar system. It was in his first book on cosmology (1596) that Kepler had first mentioned the desirability of finding such a relationship. So it is appropriate that the law did indeed first appear in his next work on cosmology. But that was many years later, in 1619, and the law only just made it: Kepler discovered it only after the book was almost finished. As he explains
...if you want the exact moment in time, it [the correct form of the law] was conceived mentally on 8th March in this year one thousand six hundred and eighteen, but submitted to calculation in an unlucky way, and therefore rejected as false, and finally returning on the 15th of May and adopting a new line of attack, stormed the darkness of my mind. So strong was the support from the combination of my labour of seventeen years on the observations of Brahe and the present study, which conspired together, that at first I believed I was dreaming, and assuming my conclusion among my basic premises. But it is absolutely certain and exact that the proportion between the periodic times of any two planets is precisely the sesquialterate proportion of their mean distances..."
[Harmony of the World, Linz, 1619]
At the end of the final (fifth) book of the Harmony of the World Kepler notes that the work was finished on 27 May 1618, and the fifth book was revised "while it was being printed", the revisions being completed on 19 February 1619. The revisions were of course occasioned by the discovery of the Third Law, which plays a significant part in the final version.
As the quotation above makes clear, checking that the squares of the periods were proportional to the cubes of the mean radii of the orbits (that is, the semi-major axes of the orbital ellipses) was merely a matter of arithmetic. Kepler had by then found orbits for all the planets. So the proof was simple - even if Kepler was (it seems) a bit accident-prone in handling the numbers.
Kepler's letters survive in quantity, but all the same he has left no clue how he came to the eventual correct form of the law. It seems likely that he simply tried lots of different formulae, and the one that worked was the one he published. Perhaps the rather long delay before making the revisions indicates that he was racking his brains (unsuccessfully) for an explanation of why the law should take this particular form? It is simple, but not quite simple enough to be explained merely aesthetically.
## Proofs and physics
The proofs of Kepler's first two laws are mathematical, but they are dealing with facts that belong to physics. However, this does not rule out Kepler's preferred method of proof, which is exactly that of Euclid: giving definitions and axioms and then proceeding by deduction. All mathematicians of the time, and most philosophers, were agreed that this method was the most reliable one.
However, neat and obviously rigorous methods of deductive proof cannot be applied if one is working with messy numbers derived from observation - or from experiment, for that matter. Even when everyone knows the points one is plotting ought to lie on a straight line, the prudent cheat will give the supposed results an air of reality by putting in some "scatter". On the more serious level: one of the most important things one learns from doing experiments or making observations of one's own is that one usually needs to know rather a lot about the acceptable form of the "answer" in order to deduce it from what one measures. It is because of this that any kind of experimental discovery is usually contested. Sometimes, of course the sceptics are right - as they were about those famous "canals" on Mars.
As Kepler was determined to make the best possible use of the huge number of observations, he used a method that involved repeated trial and error. Historically, this is interesting precisely because we know he otherwise strongly preferred to model himself on Euclid. The method of "try it and check" has in fact got a background in Medieval and Renaissance algebra. But Kepler's methods were not known in astronomy which used geometry, generally regarded (at the time) as intellectually more respectable than algebra. Since Euclid was concerned with geometry, Kepler was inviting direct comparison.
On the other hand, although at the time astronomers were known as "mathematicians", their work did not actually involve proof in quite the same way as Euclid's did. Astronomers were expected to predict, say, positions of the Moon that were then found to agree, or not agree, with observation. Kepler's work is different in that it sets out to find a new geometrical description of the orbit of Mars, and of the structure of the Solar system, and thus gets into proving general laws that are somewhat like theorems.
For a deductive proof of Kepler's laws the world had to wait for Isaac Newton. He took the laws as geometrical facts, and could afford to ignore the numerical vicissitudes that had attended Kepler's proofs.
# The proof of the pudding...
With hindsight - which historians all have a lot of - it is clear that Kepler has proved the truth of his three laws, though Newton, who did not actually read Kepler, did remark unkindly that he he knew the orbits were not circular and guessed them to be elliptical. This is not fair. In calculating orbits for the other planets, Kepler did check that the laws applied to them as well.
But the details of Kepler's work on the other planets remained in manuscript, and the mathematics of the New Astronomy was declared extremely difficult by Kepler's contemporaries - a judgement with which it is easy to sympathise. (Newton, who was working much later, had no occasion to go back to Kepler's own books.) What in effect served for proof that Kepler's laws were correct was that the tables he based on them, and on Tycho's observations, proved to be accurate over a long period of time.
Figure 8: Johannes Kepler, Tabulae Rudolphinae, Ulm, 1628, frontispiece. Kepler designed this frontispiece himself. It shows astronomy in the form of a temple, the pillars being the work of past astronomers. Kepler is seen at work in one of the reliefs on the base. Another shows a map of the island of Hveen, in Copenhagen Sound, where Tycho made most of his observations.
Click here for a higher resolution image of the frontispiece.
When Kepler succeeded Tycho as Imperial Mathematician to the Holy Roman Emperor Rudolf II, he took on the task of compiling astronomical tables, which would bear Rudolf's name and thus bring him lasting fame. Kepler's Rudolphine Tables, published in Ulm in 1628, fulfilled this obligation to the letter.
The Rudolphine Tables were the first set of astronomical tables to continue to be reliable for twenty years, for thirty years... As the years mounted up, and Kepler's tables still gave good agreement with observation, their continued accuracy inclined astronomers to accept Kepler's first two laws, and, since these laws both related the planets to the Sun, to share Kepler's belief in the Copernican theory. That wrote QED under something Kepler considered an important truth.
# Bibliography
Both of the works discussed here have been published in English translation.
Johannes Kepler, New Astronomy, trans. William H. Donahue, Cambridge: Cambridge University Press, 1992. [Translation of Johannes Kepler, Astronomia nova..., Heidelberg, 1609]
Johannes Kepler, The Harmony of the World, translated into English with an introduction and notes, by E. J. Aiton, A. M. Duncan and J. V. Field, Memoirs of the American Philosophical Society, vol. 209, Philadelphia, 1997. [Translation of Johannes Kepler, Harmonices mundi libri V, Linz, 1619]
On Tycho's instruments, see:
J. V. Field, "What is scientific about a scientific instrument?", Nuncius, III.2, 1988, 3-26.
## About the author
Dr J.V. Field, who is a historian of science, has worked on Kepler for many years, but started out as a mathematician, and in the mid 1960s helped to design the optical system of the Anglo-Australian telescope, using Cambridge University's Edsac II and Titan computers.
Dr. Field is an Honorary Visiting Research Fellow in the Department of History of Art, Birkbeck College(University of London).
For further information, see : | 5,778 | 27,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2014-52 | latest | en | 0.973243 |
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David Sheard sets the third puzzle. Can you solve it?
The day is finally here: issue 11 is out now! To help you celebrate launch day in style, we’ve set a puzzle hunt: throughout the day, we are posting a series of puzzles. The answers to these puzzles form clues to the four-digit code for the door to let you into the Chalkdust issue 11 secret backstage lounge.
The third puzzle is set by David Sheard. You can see the puzzle below, or download it as a pdf.
Be sure to come back at 3pm when the fourth puzzle will be posted.
## David’s puzzle
The digits 1 to 9 must be entered into the 9×9 grid below following normal sudoku rules: each digit must appear exactly once in each row, column, and 3×3 block. Additionally, the grid also contains thermometers and inequalities.
In each grey thermometer, the number in the circular bulb is the smallest number, and the numbers increase as you move away from the bulb. The sums of the digits on a diagonal indicated by an arrow must satisfy the inequality shown by the arrow (if the sum of the digits on a diagonal is written at the start of the arrow, then the resulting inequality must be true).
In this smaller example puzzle, the digits 1 to 4 must appear in each row, column, and 2×2 block exactly once. Starting from the grey bulbs, the numbers increase along each thermometer; and the numbers in the cells highlighted in blue in the solution add to more than or equal to 7. The sums of the digits on the other
indicated diagonals satisfy the inequalities written by their arrows.
The puzzle (click to enlarge)
The digit in the cell shaded blue is a factor of the code.
David is a final year PhD student at UCL studying geometric group theory. When not doing maths he can usually be found singing or playing the flute.
• ### My favourite LaTeX package
The Chalkdust editors share some of their favourites
• ### In conversation with Ulrike Tillmann
Sophie Maclean and David Sheard speak to a very top(olog)ical mathematician!
Solitons
• ### Chalkdust issue 11 puzzle hunt #5
Adam Townsend sets the fifth and final puzzle. Can you solve it?
• ### Chalkdust issue 11 puzzle hunt #4
Humbug sets the fourth puzzle. Can you solve it?
• ### Chalkdust issue 11 puzzle hunt #2
TD Dang sets the second puzzle. Can you solve it? | 538 | 2,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2022-33 | latest | en | 0.912486 |
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### The Perforated Cube
##### Stage: 4 Challenge Level:
A cube is made from smaller cubes, 5 by 5 by 5, then some of those cubes are removed. Can you make the specified shapes, and what is the most and least number of cubes required ?
### Cubic Net
##### Stage: 4 and 5 Challenge Level:
This is an interactive net of a Rubik's cube. Twists of the 3D cube become mixes of the squares on the 2D net. Have a play and see how many scrambles you can undo!
### Bands and Bridges: Bringing Topology Back
##### Stage: 2 and 3
Lyndon Baker describes how the Mobius strip and Euler's law can introduce pupils to the idea of topology.
### Spotting the Loophole
##### Stage: 4 Challenge Level:
A visualisation problem in which you search for vectors which sum to zero from a jumble of arrows. Will your eyes be quicker than algebra?
### Making Tracks
##### Stage: 4 Challenge Level:
A bicycle passes along a path and leaves some tracks. Is it possible to say which track was made by the front wheel and which by the back wheel?
### Building Tetrahedra
##### Stage: 4 Challenge Level:
Can you make a tetrahedron whose faces all have the same perimeter?
### Doesn't Add Up
##### Stage: 4 Challenge Level:
In this problem we are faced with an apparently easy area problem, but it has gone horribly wrong! What happened?
### Around and Back
##### Stage: 4 Challenge Level:
A cyclist and a runner start off simultaneously around a race track each going at a constant speed. The cyclist goes all the way around and then catches up with the runner. He then instantly turns. . . .
### Hypotenuse Lattice Points
##### Stage: 4 Challenge Level:
The triangle OMN has vertices on the axes with whole number co-ordinates. How many points with whole number coordinates are there on the hypotenuse MN?
### Sea Defences
##### Stage: 2 and 3 Challenge Level:
These are pictures of the sea defences at New Brighton. Can you work out what a basic shape might be in both images of the sea wall and work out a way they might fit together?
### Tetrahedra Tester
##### Stage: 3 Challenge Level:
An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length?
### Platonic Planet
##### Stage: 4 Challenge Level:
Glarsynost lives on a planet whose shape is that of a perfect regular dodecahedron. Can you describe the shortest journey she can make to ensure that she will see every part of the planet?
### Jam
##### Stage: 4 Challenge Level:
A game for 2 players
### Something in Common
##### Stage: 4 Challenge Level:
A square of area 3 square units cannot be drawn on a 2D grid so that each of its vertices have integer coordinates, but can it be drawn on a 3D grid? Investigate squares that can be drawn.
### Floating in Space
##### Stage: 4 Challenge Level:
Two angles ABC and PQR are floating in a box so that AB//PQ and BC//QR. Prove that the two angles are equal.
### Like a Circle in a Spiral
##### Stage: 2, 3 and 4 Challenge Level:
A cheap and simple toy with lots of mathematics. Can you interpret the images that are produced? Can you predict the pattern that will be produced using different wheels?
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### Contact
##### Stage: 4 Challenge Level:
A circular plate rolls in contact with the sides of a rectangular tray. How much of its circumference comes into contact with the sides of the tray when it rolls around one circuit?
### Charting Success
##### Stage: 3 and 4 Challenge Level:
Can you make sense of the charts and diagrams that are created and used by sports competitors, trainers and statisticians?
### Mystic Rose
##### Stage: 4 Challenge Level:
Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes.
### Speeding Boats
##### Stage: 4 Challenge Level:
Two boats travel up and down a lake. Can you picture where they will cross if you know how fast each boat is travelling?
### Summing Squares
##### Stage: 4 Challenge Level:
Discover a way to sum square numbers by building cuboids from small cubes. Can you picture how the sequence will grow?
### When the Angles of a Triangle Don't Add up to 180 Degrees
##### Stage: 4 and 5
This article outlines the underlying axioms of spherical geometry giving a simple proof that the sum of the angles of a triangle on the surface of a unit sphere is equal to pi plus the area of the. . . .
### Just Opposite
##### Stage: 4 Challenge Level:
A and C are the opposite vertices of a square ABCD, and have coordinates (a,b) and (c,d), respectively. What are the coordinates of the vertices B and D? What is the area of the square?
### Double Trouble
##### Stage: 4 Challenge Level:
Simple additions can lead to intriguing results...
### Corridors
##### Stage: 4 Challenge Level:
A 10x10x10 cube is made from 27 2x2 cubes with corridors between them. Find the shortest route from one corner to the opposite corner.
### Problem Solving, Using and Applying and Functional Mathematics
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Charting More Success
##### Stage: 3 and 4 Challenge Level:
Can you make sense of the charts and diagrams that are created and used by sports competitors, trainers and statisticians?
### Steel Cables
##### Stage: 4 Challenge Level:
Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions?
### Baravelle
##### Stage: 2, 3 and 4 Challenge Level:
What can you see? What do you notice? What questions can you ask?
### Triangles in the Middle
##### Stage: 3, 4 and 5 Challenge Level:
This task depends on groups working collaboratively, discussing and reasoning to agree a final product.
### Star Gazing
##### Stage: 4 Challenge Level:
Find the ratio of the outer shaded area to the inner area for a six pointed star and an eight pointed star.
### Clocking Off
##### Stage: 2, 3 and 4 Challenge Level:
I found these clocks in the Arts Centre at the University of Warwick intriguing - do they really need four clocks and what times would be ambiguous with only two or three of them?
### Building Gnomons
##### Stage: 4 Challenge Level:
Build gnomons that are related to the Fibonacci sequence and try to explain why this is possible.
### Penta Colour
##### Stage: 4 Challenge Level:
In how many different ways can I colour the five edges of a pentagon red, blue and green so that no two adjacent edges are the same colour?
### Thinking Through, and By, Visualising
##### Stage: 2, 3 and 4
This article is based on some of the ideas that emerged during the production of a book which takes visualising as its focus. We began to identify problems which helped us to take a structured view. . . .
### Lost on Alpha Prime
##### Stage: 4 Challenge Level:
On the 3D grid a strange (and deadly) animal is lurking. Using the tracking system can you locate this creature as quickly as possible?
### One and Three
##### Stage: 4 Challenge Level:
Two motorboats travelling up and down a lake at constant speeds leave opposite ends A and B at the same instant, passing each other, for the first time 600 metres from A, and on their return, 400. . . .
### Picturing Triangle Numbers
##### Stage: 3 Challenge Level:
Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### Clocked
##### Stage: 3 Challenge Level:
Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours? | 2,411 | 10,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.891203 |
https://stats.stackexchange.com/questions/61106/estimation-of-simultaneous-equations-model | 1,719,026,319,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862249.29/warc/CC-MAIN-20240622014659-20240622044659-00290.warc.gz | 483,359,961 | 40,300 | Estimation of simultaneous equations model
My model is as follows: \begin{aligned} y&=a_0 + a_1x_1 + a_2x_2 \\ x_2 &= b_0 + b_1x_1 + b_2z \end{aligned}
I'm only interested in the effect of $x_2$ on $y$. More precisely, I want to see the combined effect of:
1. $x_2$ directly, and
2. $x_2$ implicitly (through $x_1$)
After estimating the coefficients, I have the following procedure:
1. solve equation 2 for $x_1$
2. substitute the resulting term in equation 1.
3. look at the "new" $x_2$ coefficient: $(a_1/b_1 + a_2)$
The model comes from theory and there's no possibility to include other variables as instruments.
I tried multiple equations OLS with HAC standard errors and got some good results, but I'm not at all convinced that this is the right way.
• Welcome to the site, @user26594. Can you clarify your question? We need something more specific than you're looking for some advice or comments. What is it you find unsatisfactory about your current results? Commented Jun 6, 2013 at 22:23
• Perhaps I'm being dense, but can't you just collapse your model into $y=a_0 + a_1x_1 + a_2z$ and go from there? Commented Jun 6, 2013 at 23:04
• The results were good and in line with Theory, but I have concerns that my estimates are biased (because of the endogeneity of x. @Matt Krause: no, this would just give me the direct effect of $x_1$ Commented Jun 7, 2013 at 7:49
The way you have your model set up (which are usually intended to reflect the causality in these setups), it looks like it's $x_1$ that has an implicit and explicit effect on $z$, not $x_2$ ($x_1$ affects $x_2$ which affects $z$, and $x_1$ also has a direct effect on $z$).
$\quad\quad\quad\quad\quad$
• Thx for your response. The variable $z$ is exogenous and not so important for the model, but of course it also effects my estimated coefficients. My concerns are the unbiasedness of my estimates for $a_1$, $b_1$ and $a_2$ . The mediation mehthods (never heard of it before) look interesting, especially for checking whether this the causal effects are as expected. Commented Jun 7, 2013 at 8:16
• Why does $x_1$ has an effect on $z$? Could you not use something related to two-staged-least-squares with $x_1$ and $z$ being the IVs for $x_2$? If you dont want to use any instruments there is a approach called Rank-Order Instrument Variables: Rummery, Vella, Verbeek - Estimating the Returns to Education for Australian Youth via Rank-Order Instrumental Variables Commented Jun 7, 2013 at 8:34 | 715 | 2,479 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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-2024-26 | latest | en | 0.936268 |
https://shoppingandbenefits.com/get-discounts/when-a-bond-is-issued-at-a-discount-the-carrying-value-of-the-bond-at-maturity-will-be.html | 1,642,988,196,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304345.92/warc/CC-MAIN-20220123232910-20220124022910-00710.warc.gz | 497,729,256 | 19,483 | # When a bond is issued at a discount the carrying value of the bond at maturity will be?
Contents
The carrying value of a bond is not equal to the bond payable amount unless the bond was issued at par. When a bond is issued at a premium, the carrying value is higher than the face value of the bond. When a bond is issued at a discount, the carrying value is less than the face value of the bond.
## How is the carrying value of a bond issued at a discount calculated?
The carrying value equals the face value of the bond plus the remaining premium to be amortized. Use the equation \$1,000 + \$64 = \$1,064. Calculate the carrying value of a bond sold at a discount using the same method. Subtract the unamortized discount from the face value.
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## What is the carrying value of bonds at maturity?
The carrying value of bonds at maturity will always equal their par value. In other words, par value (nominal, principal, par or face amount), the amount on which the issuer pays interest, and which, most commonly, has to be repaid at the end of the term.
## When bonds are issued at a discount below face amount the carrying value and the corresponding interest expense increase over time?
When bonds are issued at a discount (below face amount), the carrying value and the corresponding interest expense increase over time. Interest expense on bonds payable is calculated as the: Carrying value times the market interest rate. A bond issue with a face amount of \$500,000 bears interest at the rate of 7%.
## What does it mean when a bond is issued at a discount?
Key Takeaways. Bond discount is the amount by which the market price of a bond is lower than its principal amount due at maturity. A bond issued at a discount has its market price below the face value, creating a capital appreciation upon maturity since the higher face value is paid when the bond matures.
## What is a bond carrying value?
The carrying value of a bond refers to the net amount between the bond’s face value plus any un-amortized premiums or minus any amortized discounts. … Premiums and discounts are amortized over the life of the bond, therefore book value equals par value at maturity.
## When bonds are issued at a discount what happens to the carrying value and interest expense?
if bonds are issued at a discount, over the life of the bonds, interest expense will: Decrease.
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## When bonds are issued the carrying book value is always the par value of the bond?
Definition: The carrying value of a bond is the par value or face value of that bond plus any unamortized premiums or less any unamortized discounts. The net amount between the par value and the premium or discount is called the carrying value because it is reported on the balance sheet.
## How would the carrying value of bonds payable change over time for bonds issued at a discount?
When bonds are issued at a discount, what happens to the carrying value and interest expense over the life of the bonds? Carrying value and interest expense increase. Carrying value and interest expense decrease.
## How do you find the carrying value?
To calculate the carrying value or book value of an asset at any point in time, you must subtract any accumulated depreciation, amortization, or impairment expenses from its original cost.
## When a bond is issued at a discount at what value is it reported on the balance sheet?
The premium or the discount on bonds payable that has not yet been amortized to interest expense will be reported immediately after the par value of the bonds in the liabilities section of the balance sheet.
## When bonds are issued at a discount the borrower must pay more at maturity than the amount originally received true or false?
When bonds are issued at a discount, the borrower must pay more at maturity than the amount originally received. The account Discount on Bonds Payable actually represents interest expense and will be amortized over the life of the bond.
## When a bond is issued at a discount is the market interest rate higher or lower than the contract interest rate on the bond?
A discount bond is offered at a lower price than the prevailing market rate. Buying the bond at a discount means that investors pay a price lower than the face value of the bond. However, it does not necessarily mean it offers better returns than other bonds. Let take an example of a bond with a \$1,000 face value.
## Why bonds are issued at discount and premium?
So, when interest rates fall, bond prices rise as investors rush to buy older higher-yielding bonds and as a result, those bonds can sell at a premium. Conversely, as interest rates rise, new bonds coming on the market are issued at the new, higher rates pushing those bond yields up. … So, those bonds sell at a discount.
## When a bond sells at a discount quizlet?
The right to receive \$1,000 at maturity. -Allocates a portion of the total discount to interest expense each interest period. -Increases the market value of the Bonds Payable. -Decreases the Bonds Payable account.
## When bonds are issued at a premium?
When a bond is issued at a premium, that means that the bond is sold for an amount greater than the bond’s face value. This generally means that the bond’s contract rate is greater than the market rate. | 1,123 | 5,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.938154 |
http://www.clubdetirologrono.com/multiplication-worksheets-2-times-tables/rontavstudio-24-new-2-times-table-worksheet-graphics-grahapada-com-multiplication-worksheets-5-10-tables-multi/ | 1,566,773,681,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330907.46/warc/CC-MAIN-20190825215958-20190826001958-00210.warc.gz | 226,384,623 | 12,402 | # Rontavstudio 24 New 2 Times Table Worksheet Graphics Grahapada Com Multiplication Worksheets 5 10 Tables Multi
Category : Math Worksheet.
Topic : Multiplication worksheets 2 and 3 times tables.
Author : Ferreira Craig.
Published : Fri, Jul 12th 2019 05:11 AM.
Format : jpg/jpeg.
Thus, the math worksheets which you get for your kids should include interesting word problems that help them with the practical application of the lessons they learn. It should also present the same problem in a variety of ways to ensure that a child grasp of a subject is deeper and comprehensive. There are several standard exercises which train students to convert percentages, decimals and fractions. Converting percentage to decimals for example is actually as simple as moving the decimal point two places to the left and losing the percent sign "%." Thus 89% is equal to 0.89. Expressed in fraction, that would be 89/100. When you drill kids to do this often enough, they learn to do conversion almost instinctively.
I believe in the importance of mathematics in our daily lives and it is critical that we nurture our kids with a proper math education. Mathematics involves pattern and structure; it is all about logic and calculation. Understanding of these math concepts are also needed in understanding science and technology. Learning math is quite difficult for most kids. As a matter of fact, it causes stress and anxiety to parents. How much stress our kids go through?
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##### Pattern worksheets for kindergarten
People have to start from the ground, then first step, second, third and so on to reach their destination floor. Exactly the same way students have to start from Kindergarten, then grade one, grade two and three and so on to reach their math destination. Also, if some of the steps are broken in the staircase, it is still hard to reach the desired floor using those steps. Same way, if you are missing some of the basic concepts from elementary grades, math for you is still hard. Now, the kindergarten, first grade and second grade are like first couple of the steps of the stairs. You can learn this level of math easily, as you can jump enough to take yourself to second or third step of the stairs easily. As it is very hard to reach sixth or seventh step of a stairs by jumping from the ground, exactly the same way to learn grade five or higher grade math is very hard (or impossible) without having the good knowledge of the kindergarten to grade three or grade four math.
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Let us discuss some tangible advantages of Mathematics in today world. One should also be aware of the wide importance of Mathematics, and the way in which it is advancing at a spectacular rate. Mathematics is about pattern and structure; it is about logical analysis, deduction, calculation within these patterns and structures. When patterns are found, often in widely different areas of science and technology, the mathematics of these patterns can be used to explain and control natural happenings and situations. Mathematics has a pervasive influence on our everyday lives, and contributes to the wealth of the individual. | 743 | 3,664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-35 | latest | en | 0.937165 |
http://www.chegg.com/homework-help/signal-processing-and-linear-systems-1st-edition-chapter-4.3-solutions-9780195219173 | 1,448,723,312,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398453553.36/warc/CC-MAIN-20151124205413-00355-ip-10-71-132-137.ec2.internal.warc.gz | 347,787,583 | 15,154 | View more editions
Signal Processing and Linear Systems
# TEXTBOOK SOLUTIONS FOR Signal Processing and Linear Systems 1st Edition
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PROBLEM
Chapter: Problem:
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 4
• Step 2 of 4
• Anonymous
the second part of the sum should be j/(pi*t)
• Anonymous
switching the j allows you to flip the sign, because you multiply by -(j^2)
• Anonymous
If you multiply by -(j^2) you are fine to do that because that's the same as 1 but you would then cancel a j on the bottom and be left with a j on top(two negatives cancel and you get the addition. First commenter is correct, second is wrong and so is the displayed answer.
• Step 3 of 4
• Step 4 of 4
Corresponding Textbook
Signal Processing and Linear Systems | 1st Edition
9780195219173ISBN-13: 0195219171ISBN: B.P.LathiAuthors: | 270 | 997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2015-48 | latest | en | 0.844749 |
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# Please Ensure that part e) is completed The standing wave properties of an ear canal are...
Please Ensure that part e) is completed
The standing wave properties of an ear canal are often modelled as a tube with one end open and one end closed. This is shown in the following diagram for a tube of length L = 2.1 cm. The fundamental mode for the sound-pressure standing wave is indicated.The standing wave properties of an ear canal are often modelled as a tube with one end open and one end closed. This is shown in the following diagram for a tube of length L = 2.1 cm. The fundamental mode for the sound-pressure standing wave is indicated.
a) What is the fundamental frequency?
b) Redraw the tube 3 times. Sketch in the standing wave patterns for the 2nd, 3rd and 5th harmonics.
c) What is the wavelength of the 5th harmonic?
d) What is the fundamental frequency if the ear was filled with water?
e) Consider a second tube that is open-open with length L2.
(i) Draw the standing wave pattern for the 3rd harmonic.
(ii) What is L2 so that the 3rd harmonic frequency equals the second harmonic frequency of the ear canal? (assume air-filled for both)
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Coins can be redeemed for fabulous gifts. | 282 | 1,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.942023 |
https://gateoverflow.in/94397/gate1988-11 | 1,563,928,939,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195530246.91/warc/CC-MAIN-20190723235815-20190724021815-00136.warc.gz | 413,744,759 | 19,944 | 556 views
A number of processes could be in a deadlock state if none of them can execute due to non-availability of sufficient resources. Let $P_i, 0 \leq i \leq 4$ represent five processes and let there be four resources types $r_j, 0 \leq j \leq 3$. Suppose the following data structures have been used.
Available: A vector of length $4$ such that if Available $[i]=k$, there are k instances of resource type $r_j$ available in the system.
Allocation. A $5 \times 4$ matrix defining the number of each type currently allocated to each process. If Allocation $[i,j]=k$ then process $p_i$ is currently allocated $k$ instances of resource type $r_j$.
Max. A $5 \times 4$ matrix indicating the maximum resource need of each process. If $Max[i,j]=k$ then process $p_i$, may need a maximum of $k$ instances of resource type $r_j$ in order to complete the task.
Assume that system allocated resources only when it does not lead into an unsafe state such that resource requirements in future never cause a deadlock state. Now consider the following snapshot of the system.$$\overset{\Large{\text{Allocation}}}{\begin{array}{llll} & r_{0} & r_{1} & r_{2} & r_{3} \\ p_{0} & 0 & 0 & 1 & 2 \\ p_{1} & 1 & 0 & 0 & 0 \\ p_{2} & 1 & 3 & 5 & 4 \\ p_{3} & 0 & 6 & 3 & 2 \\ p_{4} & 0 & 0 & 1 & 4 \\ \end{array}}\quad \qquad \overset{\Large{\text{Max}}}{\begin{array}{llll} r_{0} & r_{1} & r_{2} & r_{3} \\ 0 & 0 & 1 & 2 \\ 1 & 7 & 5 & 0 \\ 2 & 3 & 5 & 6 \\ 0 & 6 & 5 & 2 \\ 0 & 6 & 5 & 6 \\ \end{array}} \qquad \quad \overset{\Large{\text{Available}}}{\begin{array}{llll} r_{0} & r_{1} & r_{2} & r_{3} \\ 1 & 5 & 2 & 0 \\ \end{array}}$$Is the system currently in a safe state? If yes, explain why.
edited | 556 views
0
Let me know if i m wrong...
Here, we are asked to "Avoid Deadlock" and Bankers Algorithm is the algorithm for this.
The crux of the algorithm is to allocate resources to a process only if there exist a safe sequence after the allocation. i.e., after allocating the requested resources there exist a sequence of execution of the processes such that deadlock would not happen. There can be multiple safe sequences but we need to get any one of them to say that a state is safe.
Now coming to the given question, first lets make the $\text{NEED}$ matrix which shows the future need of all the processes and can be obtained by $\text{Max} - \text{Allocation}.$$\overset{\Large{\text{Max}}}{\begin{array}{llll} &r_{0} & r_{1} & r_{2} & r_{3} \\ p_0& 0 & 0 & 1 & 2 \\ p_1& 1 & 7 & 5 & 0 \\ p_2&2 & 3 & 5 & 6 \\ p_3& 0 & 6 & 5 & 2 \\ p_4& 0 & 6 & 5 & 6 \\ \end{array}} - \overset{\Large{\text{Allocation}}}{\begin{array}{llll} & r_{0} & r_{1} & r_{2} & r_{3} \\ p_{0} & 0 & 0 & 1 & 2 \\ p_{1} & 1 & 0 & 0 & 0 \\ p_{2} & 1 & 3 & 5 & 4 \\ p_{3} & 0 & 6 & 3 & 2 \\ p_{4} & 0 & 0 & 1 & 4 \\ \end{array}} = \overset{\Large{\text{Need}}}{\begin{array}{llll} &r_{0} & r_{1} & r_{2} & r_{3} \\ p_0& 0 & 0 & 0 & 0 \\ p_1& 0 & 7 & 5 & 0 \\ p_2&1 & 0 & 0 & 2 \\ p_3& 0 & 0 & 2 & 0 \\ p_4& 0 & 6 & 4 & 2 \\ \end{array}}.$$ Since$P_0$does not require any more resource we can finish this first releasing$1$instance of$r_2$and$2$instances of$r_3.$Thus our$\text{Available}$vector becomes $$\begin{bmatrix}1&5&2&0\end{bmatrix}+\begin{bmatrix}0&0&1&2\end{bmatrix} =\begin{bmatrix}1&5&3&2\end{bmatrix}.$$ Now, either$p_2$or$p_3$can finish as both their requirements are not greater than the$\text{Available}$vector. Say,$p_2$finishes. It releases$\begin{bmatrix}2&3&5&6\end{bmatrix}$and our$\text{Available}$becomes $$\begin{bmatrix}1&5&3&2\end{bmatrix}+\begin{bmatrix}2&3&5&6\end{bmatrix} =\begin{bmatrix}3&8&8&8\end{bmatrix}.$$ Now, any of$p_1,p_3,p_4\$ can finish and so we do not need to proceed further to determine that the state is safe. One of the possible safe sequence is $$p_0-p_2-p_1-p_3-p_4.$$
Yes System is in Safe state .
Safe sequence is P0,P2,P1,P3,P4
Need Matrix:-
0 0 0 0
0 7 5 0
1 0 0 2
0 0 2 0
0 6 4 2
P0---->available 1 5 3 2
P2---->available 2 8 8 6
P1---->P3---->P4
yes it is safe state and safe sequence is P0->P2->P1->P3->P4->
1
2 | 1,553 | 4,099 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-30 | longest | en | 0.780687 |
https://plainmath.org/inferential-statistics/92901-jason-estimates-that-his-car-loses-12-o | 1,718,943,614,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862036.35/warc/CC-MAIN-20240621031127-20240621061127-00845.warc.gz | 402,524,847 | 20,248 | sailorlyts14eh
2022-09-03
Jason estimates that his car loses 12% of its value every year. The initial value is 12,000. Which best describes the graph of the function that represents the value of the car after X years ?
barquegese2
Every year, the car's value gets multiplied by 0.88, so the equation that gives the value, y, of the car after x years is
$y=12000\left(0.88{\right)}^{x}$
graph{$12000\left(0.88{\right)}^{x}\left[-5,20,-5000,15000\right]$}
Do you have a similar question? | 154 | 490 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 14, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-26 | latest | en | 0.889706 |
http://a4a.learnport.org/forum/topics/linear-functions-5-day-plan-1?xg_source=activity | 1,531,862,983,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589902.8/warc/CC-MAIN-20180717203423-20180717223423-00328.warc.gz | 7,193,271 | 12,148 | # Linear Functions-5 day plan
Linear Functions-5 day plan for Algebra for All
Carrie Weber
I was not able to attach all documents, it only allowed for 3 items. Missing is the Frayer model, which can easily be found online, and the TI-84 Scavenger Hunt which can be found on this network.
Teaching in Michigan, I have included the MMC content expectations that will be introduced, although some not in their entirety:
A2.1.3 Represent functions in symbols, graphs, tables, diagrams, or words and translate among representations.
A2.1.7 Identify and interpret the key features of a function from its graph or its formula(e), (e.g., slope, intercept(s), asymptote(s), maximum and minimum value(s),
symmetry, and average rate of change over an interval).
A2.4.1 Write the symbolic forms of linear functions (standard [i.e., Ax + By = C, where B ≠ 0], point-slope, and slope-intercept) given appropriate information and convert between forms.
A2.4.2 Graph lines (including those of the form x = h and y = k) given appropriate information.
A2.4.3 Relate the coefficients in a linear function to the slope and x- and y-intercepts of its graph.
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Attachments: | 284 | 1,173 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-30 | latest | en | 0.897438 |
https://edu-spot.com/rational-numbers/ | 1,721,490,777,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515300.51/warc/CC-MAIN-20240720144323-20240720174323-00625.warc.gz | 192,838,357 | 14,148 | # Rational Numbers
## Ex 1.1
#### 1. Using appropriate properties find.
(i) -2/3 × 3/5 + 5/2 – 3/5 × 1/6
##### Solution:
-2/3 × 3/5 + 5/2 – 3/5 × 1/6
= -2/3 × 3/5– 3/5 × 1/6+ 5/2 (by commutativity)
= 3/5 (-2/3 – 1/6)+ 5/2
= 3/5 ((- 4 – 1)/6)+ 5/2
= 3/5 ((–5)/6)+ 5/2 (by distributivity)
= – 15 /30 + 5/2
= – 1 /2 + 5/2
= 4/2
= 2
(ii) 2/5 × (- 3/7) – 1/6 × 3/2 + 1/14 × 2/5
Solution:
##### 2/5 × (- 3/7) – 1/6 × 3/2 + 1/14 × 2/5
= 2/5 × (- 3/7) + 1/14 × 2/5 – (1/6 × 3/2) (by commutativity)
= 2/5 × (- 3/7 + 1/14) – 3/12
= 2/5 × ((- 6 + 1)/14) – 3/12
= 2/5 × ((- 5)/14)) – 1/4
= (-10/70) – 1/4
= – 1/7 – 1/4
= (– 4– 7)/28
= – 11/28
2. Write the additive inverse of each of the following
Solution:
(i) 2/8
Additive inverse of 2/8 is – 2/8
(ii) -5/9
Additive inverse of -5/9 is 5/9
(iii) -6/-5 = 6/5
Additive inverse of 6/5 is -6/5
(iv) 2/-9 = -2/9
Additive inverse of -2/9 is 2/9
(v) 19/-16 = -19/16
Additive inverse of -19/16 is 19/16
3. Verify that: -(-x) = x for.
(i) x = 11/15
(ii) x = -13/17
Solution:
(i) x = 11/15
We have, x = 11/15
The additive inverse of x is – x (as x + (-x) = 0)
Then, the additive inverse of 11/15 is – 11/15 (as 11/15 + (-11/15) = 0)
The same equality 11/15 + (-11/15) = 0, shows that the additive inverse of -11/15 is 11/15.
Or, – (-11/15) = 11/15
i.e., -(-x) = x
(ii) -13/17
We have, x = -13/17
The additive inverse of x is – x (as x + (-x) = 0)
Then, the additive inverse of -13/17 is 13/17 (as 11/15 + (-11/15) = 0)
The same equality (-13/17 + 13/17) = 0, shows that the additive inverse of 13/17 is -13/17.
Or, – (13/17) = -13/17,
i.e., -(-x) = x
4. Find the multiplicative inverse of the
(i) -13 (ii) -13/19 (iii) 1/5 (iv) -5/8 × (-3/7) (v) -1 × (-2/5) (vi) -1
Solution:
(i) -13
Multiplicative inverse of -13 is -1/13
(ii) -13/19
Multiplicative inverse of -13/19 is -19/13
(iii) 1/5
Multiplicative inverse of 1/5 is 5
(iv) -5/8 × (-3/7) = 15/56
Multiplicative inverse of 15/56 is 56/15
(v) -1 × (-2/5) = 2/5
Multiplicative inverse of 2/5 is 5/2
(vi) -1
Multiplicative inverse of -1 is -1
5. Name the property under multiplication used in each of the following.
(i) -4/5 × 1 = 1 × (-4/5) = -4/5
(ii) -13/17 × (-2/7) = -2/7 × (-13/17)
(iii) -19/29 × 29/-19 = 1
Solution:
(i) -4/5 × 1 = 1 × (-4/5) = -4/5
Here 1 is the multiplicative identity.
(ii) -13/17 × (-2/7) = -2/7 × (-13/17)
The property of commutativity is used in the equation
(iii) -19/29 × 29/-19 = 1
Multiplicative inverse is the property used in this equation.
6. Multiply 6/13 by the reciprocal of -7/16
Solution:
Reciprocal of -7/16 = 16/-7 = -16/7
According to the question,
6/13 × (Reciprocal of -7/16)
6/13 × (-16/7) = -96/91
7. Tell what property allows you to compute 1/3 × (6 × 4/3) as (1/3 × 6) × 4/3
Solution:
1/3 × (6 × 4/3) = (1/3 × 6) × 4/3
Here, the way in which factors are grouped in a multiplication problem, supposedly, does not change the product. Hence, the Associativity Property is used here.
8. Is 8/9 the multiplication inverse of
? Why or why not?
Solution:
= -7/8
[Multiplicative inverse ⟹ product should be 1]
According to the question,
8/9 × (-7/8) = -7/9 ≠ 1
Therefore, 8/9 is not the multiplicative inverse of
.
9. If 0.3 the multiplicative inverse of
? Why or why not?
Solution:
= 10/3
0.3 = 3/10
[Multiplicative inverse ⟹ product should be 1]
According to the question,
3/10 × 10/3 = 1
Therefore, 0.3 is the multiplicative inverse of
.
10. Write
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.
Solution:
(i)The rational number that does not have a reciprocal is 0. Reason:
0 = 0/1
Reciprocal of 0 = 1/0, which is not defined.
(ii) The rational numbers that are equal to their reciprocals are 1 and -1.
Reason:
1 = 1/1
Reciprocal of 1 = 1/1 = 1 Similarly, Reciprocal of -1 = – 1
(iii) The rational number that is equal to its negative is 0.
Reason:
Negative of 0=-0=0
11. Fill in the blanks.
(i) Zero has reciprocal.
(ii) The numbers and are their own reciprocals
(iii) The reciprocal of – 5 is .
(iv) Reciprocal of 1/x, where x ≠ 0 is .
(v) The product of two rational numbers is always a .
(vi) The reciprocal of a positive rational number is .
Solution:
(i) Zero has no reciprocal.
(ii) The numbers -1 and 1 are their own reciprocals
(iii) The reciprocal of – 5 is -1/5.
(iv) Reciprocal of 1/x, where x ≠ 0 is x.
(v) The product of two rational numbers is always a rational number.
(vi) The reciprocal of a positive rational number is positive.
## Ex 1.2
1. Represent these numbers on the number line.
(i) 7/4
(ii) -5/6
Solution:
(i) 7/4
Divide the line between the whole numbers into 4 parts. i.e., divide the line between 0 and 1 to 4 parts, 1 and 2 to 4 parts and so on.
Thus, the rational number 7/4 lies at a distance of 7 points away from 0 towards positive number line.
(ii) -5/6
Divide the line between the integers into 4 parts. i.e., divide the line between 0 and -1 to 6 parts, -1 and -2 to 6 parts and so on. Here since the numerator is less than denominator, dividing 0 to – 1 into 6 part is sufficient.
Thus, the rational number -5/6 lies at a distance of 5 points, away from 0, towards negative number line
2. Represent -2/11, -5/11, -9/11 on a number line.
Solution:
Divide the line between the integers into 11 parts.
Thus, the rational numbers -2/11, -5/11, -9/11 lies at a distance of 2, 5, 9 points away from 0, towards negative number line respectively.
3. Write five rational numbers which are smaller than 2.
Solution:
The number 2 can be written as 20/10
Hence, we can say that, the five rational numbers which are smaller than 2 are:
2/10, 5/10, 10/10, 15/10, 19/10
4. Find the rational numbers between -2/5 and ½.
Solution:
Let us make the denominators same, say 50.
-2/5 = (-2 × 10)/(5 × 10) = -20/50
½ = (1 × 25)/(2 × 25) = 25/50
Ten rational numbers between -2/5 and ½ = ten rational numbers between -20/50 and 25/50
Therefore, ten rational numbers between -20/50 and 25/50 = -18/50, -15/50, -5/50, -2/50, 4/50, 5/50, 8/50, 12/50, 15/50, 20/50
5. Find five rational numbers between.
(i) 2/3 and 4/5
(ii) -3/2 and 5/3
(iii) ¼ and ½
Solution:
(i) 2/3 and 4/5
Let us make the denominators same, say 60
i.e., 2/3 and 4/5 can be written as:
2/3 = (2 × 20)/(3 × 20) = 40/60
4/5 = (4 × 12)/(5 × 12) = 48/60
Five rational numbers between 2/3 and 4/5 = five rational numbers between 40/60 and 48/60
Therefore, Five rational numbers between 40/60 and 48/60 = 41/60, 42/60, 43/60, 44/60, 45/60
(ii) -3/2 and 5/3
Let us make the denominators same, say 6
i.e., -3/2 and 5/3 can be written as:
-3/2 = (-3 × 3)/(2× 3) = -9/6
5/3 = (5 × 2)/(3 × 2) = 10/6
Five rational numbers between -3/2 and 5/3 = five rational numbers between -9/6 and 10/6
Therefore, Five rational numbers between -9/6 and 10/6 = -1/6, 2/6, 3/6, 4/6, 5/6
(iii) ¼ and ½
Let us make the denominators same, say 24.
i.e., ¼ and ½ can be written as:
¼ = (1 × 6)/(4 × 6) = 6/24
½ = (1 × 12)/(2 × 12) = 12/24
Five rational numbers between ¼ and ½ = five rational numbers between 6/24 and 12/24
Therefore, Five rational numbers between 6/24 and 12/24 = 7/24, 8/24, 9/24, 10/24, 11/24
6. Write five rational numbers greater than -2.
Solution:
-2 can be written as – 20/10
Hence, we can say that, the five rational numbers greater than -2 are
-10/10, -5/10, -1/10, 5/10, 7/10
7. Find ten rational numbers between 3/5 and ¾,
Solution:
Let us make the denominators same, say 80.
3/5 = (3 × 16)/(5× 16) = 48/80
3/4 = (3 × 20)/(4 × 20) = 60/80
Ten rational numbers between 3/5 and ¾ = ten rational numbers between 48/80 and 60/80
Therefore, ten rational numbers between 48/80 and 60/80 = 49/80, 50/80, 51/80, 52/80, 54/80, 55/80, 56/80, 57/80, 58/80, 59/80 | 3,100 | 7,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-30 | latest | en | 0.817943 |
http://alex.state.al.us/all.php?std_id=53783 | 1,498,720,988,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323889.3/warc/CC-MAIN-20170629070237-20170629090237-00044.warc.gz | 12,860,579 | 8,677 | Narrow Results:
Lesson Plans (2) Multimedia (2) Informational Materials (1) Interactives/Games (1)
## ALEX Lesson Plans
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Subject: Mathematics (4 - 5), or Technology Education (3 - 5)
Title: Please Excuse My Dear Aunt Sally
Description: As a hands-on lesson, the students will recognize the correct usage of the order of operations. In groups, the students will work cooperatively to discover the importance of following step-by-step instructions and apply that knowledge in solving algebraic equations by correctly using the order of operations. This lesson plan was created by exemplary Alabama Math Teachers through the AMSTI project.
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Subject: Mathematics (3 - 5), or Technology Education (3 - 5)
Title: Let's Go Shopping!
Description: This technology-based lesson provides students practice in collecting and organizing data through the use of grocery store circulars.
## ALEX Multimedia
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Lattice Multiplication
Overview:
"Lattice Multiplication" is another strategy to help multiplication make sense. Fourth grade students demonstrate and explain the steps of lattice multiplication.
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Finding Prime Using Eratosthenes' Seive
Overview:
This podcast was created by a group of 5th grade girls after they investigated Prime Numbers. The podcast gives background information about Eratosthenes and prime numbers. Then the students take the audience through the process of finding prime numbers using the Seive of Eratosthenes.
## Thinkfinity Informational Materials
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Subject: Mathematics
Title: Number and Operations Web Links
Description: This collection of Web links, reviewed and presented by Illuminations, offers teachers and students information about and practice in concepts related to arithmetic. Users can read the Illuminations Editorial Board's review of each Web site, or choose to link directly to the sites.
Thinkfinity Partner: Illuminations | 419 | 2,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-26 | latest | en | 0.860781 |
https://anyassignment.com/chemistry/bio-lab-report-osmosis-assignment-55939/ | 1,566,065,145,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313436.2/warc/CC-MAIN-20190817164742-20190817190742-00212.warc.gz | 361,589,950 | 14,972 | # Bio lab report osmosis Assignment
Words: 1356
It may also be used to describe a physical process in which any solvent moves across a comparable membrane (permeable to the solvent, but not the solute) separating two solutions of different concentrations. Osmosis can be made to do work. The osmotic pressure is defined to be the minimum pressure required to maintain an equilibrium, with no Nett movement of solvent. Osmotic pressure is a colligating property, meaning that the osmotic pressure depends on the molar concentration of the solute but not on its identity.
Osmosis, unlike diffusion, requires a force to work. This force is supplied by the solute’s interaction with the membrane. Solute particles move randomly due to Brownian motion. If they move towards pores in the membrane, they are repelled, and in being repelled, acquire momentum directed away from the membrane. The momentum is rapidly transferred to surrounding water molecules, driving them away from the membrane as well.
order now
The osmotic gradient is the difference in concentration between two solutions on either side of a comparable membrane, and is used to tell the difference in percentages of the concentration of a specific particle dissolved in a solution. Usually the osmotic gradient s used while comparing solutions that have a comparable membrane between them allowing water to diffuse between the two solutions, toward the hypersonic solution (the solution with the higher concentration).
This would effect the final reading of the percentage of weight change because the cut potato cubes would be having a either a higher mass or lower mass depending whether the potato cube was cut slightly bigger than 1 CM or slightly smaller than lack respectively. The increase in weight for the potato cubes as hon.. In the table for example during trial 1 was different than in trial 2 for the distilled water. This would end up having an average and a less precise reading of the final weight change, hence having a less precise percentage of weight change. The second random error could have possibly been the time to end the stopwatch at exactly fifteen minutes for each and every one potato cube immersed in the salt solution. This would end up in most probably having an excess of salt solution in the potato cube as the chances of having a less salt solution is lower.
This is because the delayed time of reaction after the stopwatch ad shown fifteen minutes may have caused excess salt solution to enter the potato cube causing the final mass of the potato cube to increase more than what it should have been, thus increasing the average weight change of the potato cubes and the percentage change. This was most probably shown in the trial 2 of the 2% salt solution. Hence, the reading of the trial 2 was larger than the reading of the trial 1 of the 2% salt solution. The third random error was that the measuring cylinder used to measure the salt solution of different percentages was washed with distilled water and was not dried properly. This would result n the salt solution being diluted by the distilled water causing the percentage of the salt solution to decrease. This WOUld then effect the results of weight change because the salt solution is then not what it is supposed to be.
It would have been lower than expected and the percentage of weight change of potato cubes would have been lower The last random error that was possible in this experiment was the drying of the potato cube was not constant before and after it was immersed into the salt solution. This would have left the readings to be not the same for every trial of different percentage of salt solution. The tissue paper used was not given the same drying capabilities for each of the potato cubes. Drying capabilities meaning the length the tissue paper was placed on the potato CUbes to absorb the remaining salt solution, the tissue paper was reused and didn’t having the absorbing capabilities of what it should have been having and many more factors.
This would then change the data for each and every trial for different percentages leaving the mass obtained to not be precise, hence, causing the percentage of weight change to be different for each and every trial and this would effect the precision of the experiment. Some of the Seibel systematic errors in this experiment are the temperature of the room was kept at a cold temperature with the air-conditioning. This would effect the rate of which the the water diffuses to the potato cubes from the salt solution to potato cubes or the potato cubes to salt solution. The final percentage of weight change would be effected as well as the average weight change of the of the potato CUbes.
This is because the osmosis process may have been slowed down because of the rate of water diffusion decrease which would decrease all the readings of the final weight of the potato cubes Lastly, the final systematic error is that the salt solution used for all were not washed correctly down to its percentage. For example, the original stock of salt solution given to the lab was 5% and the one that was washed down was not exactly to 0. 5%, and 3%. The readings could be higher or lower depending on the way the original salt solution was washed. This means that the reading of the percentage of weight change would change in all directions because the 1% of the salt solution could have been 2% and the reading would have increased causing the average weight change to increase and the percentage of the weight change to increase. Some f the strengths in this experiment is the use of a digital stopwatch to measure the time taken.
This helps a lot because the time taking would be easier and it would have reduced the chances of having a random error. Another strength is use a peeler to peel the potato skin off before cutting it. This helps by not wasting time and having the chance of cutting ourselves using a knife. Some of the weakness of this experiment is the environment where the experiment was done. There was presence of air molecules blowing from the air-conditioning and this would have maybe dried the potato cubes that came out of the salt solutions quicker than what it should have. The ways to improve this experiment is based on the weakness which is to turn off the air-conditioning during the experiment.
This would help drastically because there will be no external air forces drying the potato cubes faster than what it should be because blowing air is a factor that effects rate of loss of water. The significance of this experiment is to show that the water moves from a higher concentration gradient to a lower concentration gradient and that the salt solution depending on the percentage would effect the mass of the potato cubes because the salt solution either moved into the potato bubs or the water in the potato cubes moved out and into the salt solution. This is also to show that cells do osmosis to regulate the osmotic pressure in themselves and they do this on a regular basis to maintain the turgidity of the cell.
Cell to cell diffusion of water is controlled by this process called osmosis. Some of the precautions that were taken in this experiment was the usage of lab coats all the time during the experiment. This is to ensure that even a harmless solution like a salt solution wouldn’t come in contact with our skin or dirty our clothing. The knife was used carefully because it was sharp and so that no cuts were gotten on our fingers. All the apparatus and material were handled with care so that we do not break them. Overall, this experiment was a success. My lab partner, Elian and I managed to work together well to accomplish this experiment successfully. This included us splitting the task equally.
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https://wattstoamps.com/amps-to-watts | 1,695,489,543,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506528.19/warc/CC-MAIN-20230923162848-20230923192848-00192.warc.gz | 676,054,025 | 28,385 | Home » Amps to Watts
# Amps to Watts
Welcome to amps to watts, our website about the conversion from the current in amperes to the power in watts. The electric current, denoted I, measured in the unit ampere (A), is the flow of an electric charge. The electric power, denoted P, measured in the unit watt (W), is the work done per unit of time. In the equation P = I × V, V stands for the voltage (electric potential) in volts. Making use of this equation, we are going to transform amperes to watts.
Circuit
## Convert Amps to Watts
As follows from the equation P(W) = I(A) × V(V), in order to determine the power P, in addition to I, the voltage V must be known. To convert amps to watts, use these formulas:
• DC: P(W) = I(A) × V(V)
• AC, Single Phase: P(W) = I(A) × PF × V(V)
• AC, Three Phase, Line to Line Voltage: P(W) = I(A) × √3 × PF × VL-L(V)
• AC, Three Phase, Line to Neutral Voltage: P(W) = I(A) × 3 × PF × VL-0(V)
DC, in our context, stands for direct current, and, along the same lines, AC means alternating current; retail customers usually receive their electric power in the latter form.
In contrast, anything which can run on batteries works on DC, e.g. a laptop. Thus, power supplies also produce an unidirectional flow of electric charge. A power inverter changes DC to AC.
The power factor PF is the fraction of real power drawn in W divided by the supplied power measured in volt-amperes (VA); its mathematical definition can be found further below.
Let’s look at how to convert amps to watts by means of an example:
What is the power consumption in W if the current in amperes (often shortened to amps) is 2 A and the voltage supply is 220 V, PF=0.6?
• DC: P(W) = 2 A × 220 V = 440 W
• AC, Single Phase: P(W) = 2 A × 0.6 × 220 V = 264 W
• AC, Three Phase, Line to Line Voltage: P(W) = 2 A × √3 × 0.6 × 220 V = 457.26 W
• AC, Three Phase, Line to Neutral Voltage: P(W) = 2 A × 3 × 0.6 × 220 V = 792 W
When converting amperes to watts, avoid pen and pencil. Also, instead or applying the formula feeding a calculator, we recommended to you utilizing our converter at the top of this article.
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## About our Amps to Watts Calculator
Employing our amps to watts calculator at the beginning of this article is easy: Enter the current in amps, then insert the volts. If you have a direct current, then you already see the result Else, choose from the drop-down menu.
Provided that you have an alternating current, then you must also specify the number of phases, the power factor PF, and the voltage type in case you have three phases.
Home appliances are plugged into an AC single phase system. Only in cases of AC with three phases the voltage type (line to line or line to neutral voltage) must be specified.
The PF is usually defined as cos (phase angle between voltage and current), and thus takes on a value in the interval [0, 1]. Real power is also called active power and true power.
Here you can find all information about watts to amps, such as a power to current calculator and the formulas. Read on to learn everything about the amp to watt conversion.
## Amps to Watts Conversion
Reading our article till this very line, you know how to conduct an amps to watts conversion using our calculator or the amp to watt formula which corresponds to your electric system.
Therefore, you can answer these FAQs below without difficulties:
• How many watts in an amp?
• How many watts in an ampere?
• How to convert amps to watts?
• How much watt in an amp?
• How many watts per amp?
To learn more about James Watt and André-Marie Ampère, or the units named after these inventors, visit the websites listed in the reference section at the bottom of our home page.
Our frequent conversions in this category include, for example:
In the final paragraph below, there is the summary of amperes to watts, along with a method for looking frequent current to power conversions up.
## Amperes to Watts
You have now made it to the concluding paragraph of our amperes to watts article, summarized by this image:
The bottom line of amps to watt is that both, the amps as well as the volts, must be known.
In the absence of either, the electric current or the electric potential, the watts cannot be calculated as there is no direct amp to watt relationship.
To locate frequent conversions you can use the search form in the sidebar of our site. There, supposed that x denotes your current, enter, for example, x amps into watts.
In the same fashion, conversions and terms including x amps to watts converter, x amps to watts formula, and x amps watts can be located. Try it out now inserting x amp to watts.
Here’s everything about amps to kilowatts.
For all feedback including questions and comments on ampere to watt, fill in the form below or send us an email with a meaningful title such as, for instance, converting amps to watts.
If you are pleased with our information on how many watts in an amp, then help us attracting more visitors by pressing the social buttons on the left of this page or at the bottom.
Make sure to bookmark us, and many thanks for visiting wattstoamps.com!
Submitting... | 1,255 | 5,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-40 | longest | en | 0.912023 |
https://www.jiskha.com/questions/62757/if-f-x-x-2-2x-find-f-x-2-f-2-2-the-answer-is-x-2-but-l-got-x-2-2x | 1,550,930,440,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249501174.94/warc/CC-MAIN-20190223122420-20190223144420-00360.warc.gz | 866,143,959 | 5,979 | # algebra
if f(x) = x^2 - 2x, find f(x+2) - f(2) / 2
but l got: x^2 - 2x - 8 = 0
which is x = 4 or x = -2
1. đ 0
2. đ 0
3. đ 66
1. =[(x+2)^2 - 2(x+2)] - [(2)^2 - 2(2)]/2
=x^2+4x+4 - 2x-4 - (4-4)/2
=x^2+2x - 0
=x(x+2)
= 0 or -2
I have the x+2 part of your "correct answer", but I also got the "x" in there. That's the answer I came up with, if you don't understand how I arrive my answer, let me know.
1. đ 0
2. đ 0
posted by Jake1214
2. wait, i mistyped. it should be:
if f(x) = x^2 - 2x, find f(x+2) - f(2) / x
divided by x, not 2.
sorry...
1. đ 0
2. đ 0
posted by dave
3. Hmm... that should be the same answer because f(2) = 0 and 0 divide by any number is 0 and "x" cannot be zero.
1. đ 0
2. đ 0
posted by Jake1214
4. well i redid it and i got this:
= x^2 + 4x + 4 - 2x - 4 - 2^2 - 4 / 2
= x^2 + 2x / x
divide the x and i get "x+2"
is it correct??
1. đ 0
2. đ 0
posted by dave
5. Where is the /x comes from on your x^2+2x/x... you were /2 up there on the above step... and even if the problem is f(2)/x... your f(2) = 0, that mean you get 0/x which equals 0, thus leaving you only x^2+2x on the left side, which equals x(x+2).
Basically, I disagree with what you did up there because you can't just go from:
x^2 + 2x - 0/x to (x^2 + 2x)/x
1. đ 0
2. đ 0
posted by Jake1214
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More Similar Questions | 1,678 | 4,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-09 | latest | en | 0.913472 |
https://askworksheet.com/3rd-grade-equivalent-fractions-worksheet/ | 1,679,847,387,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945473.69/warc/CC-MAIN-20230326142035-20230326172035-00171.warc.gz | 152,649,459 | 21,223 | # 3rd Grade Equivalent Fractions Worksheet
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Grade 4 Fractions Worksheet Color Equivalent Fractions Fractions Worksheets Fractions 4th Grade Fractions | 744 | 4,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-14 | latest | en | 0.862514 |
http://www.sabernomics.com/sabernomics/index.php/2005/03/introducing-the-sabernomics-simple-projection-system/ | 1,441,043,630,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644066275.44/warc/CC-MAIN-20150827025426-00001-ip-10-171-96-226.ec2.internal.warc.gz | 688,542,278 | 11,340 | ## Introducing the Sabernomics Simple Projection System
Well, I didn’t start out trying to do this, but have developed a projection system. Last week, lost in an attempt to solve some question that I have since forgotten, I accidentally came close to generating projections for both hitters and pitchers based solely on player statistics from the previous season. I decided to put a little more effort in to complete it for hitters. Pitchers will come later. While my method is simple, I was surprised at the predictive power that it appears to have employing data only from the previous seasons of hitters. Here are the factors I used to generate the predictions:
Walk rate
Strikeout rate
Home run rate
Extra-base-hits per hit
Batting average on balls in play
I also controlled for the age, league, and park of the player. I used these factors in some form or another to estimate the batting average, on-base percentage, and isolated power for each player. From these I generated SLG and OPS. Now, this system is very simple.
I have only projected seasons for players with a minimum of 150 plate appearances in 2004.
I assume that every player is playing for the same team as last year (adjusting for new teams is a pain, I may do this later).
I project players who stayed on the same team for the entire season, because it’s a pain to look at split seasons (sorry, no Beltran projection).
I do not look at minor league stats.
I only project the big-4 stats: AVG, OBP, SLG, and OPS.
I would like to fix all of these problems, but time is scarce right now. Hopefully, soon I can make changes at a latet time.
So, how does the model predict? Well, using player seasons from 1998-2004 the predicted OPS explains about 50% of the variance of the actual OPS. The root-mean-squared-error (RMSE) is 0.086, which means two-thirds of the observations lie within 86 OPS “points” of the prediction. In the 2004 Baseball Prospectus, Nate Silver compares six projection systems for 2003. The systems ranged from explaining 42-50% of the variance of OPS with RMSEs from 0.085-0.098. When I apply my system to 2003 it explains 53% of the variance with an RMSE of 0.086. In fairness to these other systems, I am only looking at players with more than 150 PAs in the major leagues, not minor leaguers, foreign players, or previously injured players. But, I think that it is interesting that the system seems to be projecting similar numbers.
One of the most obvious weaknesses of the system looks to actually be a strength, which I did not expect to find. looking only at one previous season seems to tell us a lot about a player, even a player who deviates from his career norm. Look at Chipper Jones’s 2004 actual, 2005 projected, and career lines:
```Year AVG OBP SLG OPS
2004 .248 .362 .485 .847
2005 .289 .386 .543 .928
Career .304 .401 .537 .937
```
As Braves fans know, Chipper had a horrible year last year based on his career stats, which we assumed was a product of his injury. But, now I’m not so sure he didn’t just have some bad luck. The system I employed has no way of knowing that Chipper was a career .937 OPS guy. But, from the information that I included from last year, and knowing his age and park, it concluded Chipper would do 80 points better in 2005. In fact, PECOTA (which I will not post here) projects Chipper’s numbers to be worse than mine, even though it includes Chipper’s past good performances. Now, that doesn’t mean PECOTA is wrong, but it is interesting that the system I developed has projected such a huge jump for a 33 year-old player that is consistent with his career output.
Anyway, here are the 2005 SSPS projections. If you have any thoughts or suggestions, please feel free to pass them along to me. I may or may not make updates, but I hope to post pitcher projections shortly.
### 8 Responses “Introducing the Sabernomics Simple Projection System”
1. Anonymous says:
Looking forward to more on this, JC.
2. Anonymous says:
Your system really likes Adam Dunn (.600+ SLG) and hates Ichiro (292/343/372). Obviously this is self-selection, but does this system prefer the TTO mashers vs. slap hitters (Ichiro, and maybe Nomar [who has no projection despite 354 PAs between BOS/CHC] to a lesser extent)?
It’d be interesting to see what your system projected for Darin Erstad after his 2000 season.
3. Anonymous says:
I assume that JC could run the projection on Erstad 2000 to generate an expected Erstad 2001 and compare it with real 2001. However that’s probably not a fair test, since both Erstad’s 2000 and 2001 seasons were likely used to generate the model. How much out-of-sample testing have you used, JC?
If this system regresses hitter H/BIP rates heavily (as some on primer have suggested) I guess it would penalize line drive hitters heavily since they would be more likely to sustain high H/BIP rates long term. Still, very interesting stuff, as always!
4. David Rugge says:
I like your hitting projections a lot more than the pitching projections. The big weakness of the pitching projections is that it only predicts ERA, not innings pitched, so one does not get a good idea of how long the pitcher will maintain that ERA.
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http://mathhelpforum.com/pre-calculus/12805-perpendicular-lines.html | 1,481,315,446,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542765.40/warc/CC-MAIN-20161202170902-00212-ip-10-31-129-80.ec2.internal.warc.gz | 168,974,342 | 9,693 | 1. ## Perpendicular Lines
1. Geometry. Floor plans for a building have the four corners of a room located at the points (2, 3), (11, 6), (-3, 18), and (8, 21). Determine whether the side through the points (2, 3) and (11, 6) is parallel to the side through the points (-3, 18) and (8, 21).
For the floor plans given in #1, determine whether the sidethrough the points (2, 3) and (11, 6) is perpendicular to the side through the points
(2, 3) and (-3, 18).
2. Parallel vs. Perpendicular:
Both involve slopes, and the rules are as follows:
If two lines (slopes) are parallel, then they are congruent (equal). In other words, if the slopes of the lines are m1 and m2:
m1 = m2
If two lines (slopes) are perpenducular, then they are oposite reciprocals of each other. In other words, for m1 and m2:
m1 = -1/m2
Therefore, all we need to do is find the slopes between the points you specified and see if the slopes match up to the rules above.
Slope is m = (y2 - y1)/(x2 - x1)
1)
For the points (2,3) and (11,6)
m1 = (6-3)/(11-2) = 3/9 = 1/3
For the points (-3,18) and (8,21)
m2 = (21-18)/(8-(-3)) = 3/(8+3) = 3/11
m1 does not equal m2 since 1/3 does not equal 3/11. The two slopes are not parallel.
2)
For the points (2,3) and (11,6) (same as above)
m1= 1/3
For the points (2,3) and (-3,8)
m2 = (8-3)/(-3-2) = 5/(-1) = -5
m1 does not equal -1/m2 since 1/3 does not equal (-1)/(-5) [Note: m2 would have to equal -3 or m1 would have to equal +1/5 for the two angles to be perpedicular]. The two slopes are not perpendicular. | 534 | 1,529 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2016-50 | longest | en | 0.898708 |
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Algebra is commonly used in formulas when we do not know at least one of the numbers, or when one of the numbers can change. As the values for the base and height can be changed, we can use.
#### What is algebra? - BBC Bitesize.
Students of our Edexcel Level 2 and Level 3 Awards in Algebra will develop a thorough knowledge and understanding of concepts in algebra, and a strong foundation in mathematical techniques. Level 2 and Level 3 Awards in Algebra have been extended for a further 3 years until 31 August 2022, so you can still teach from September 2019.
#### Algebra 2: Coursework Description - LearningPath.org.
ALGEBRA. Home. 1 1. Algebraic expressions. The four operations and their signs. The function of parentheses. Terms versus factors. Powers and exponents. The order of operations. Evaluating algebraic expressions. 1 2. Signed numbers: Positive and negative. The absolute value and the algebraic sign. Subtracting a larger number from a smaller. The.
## Solution
The coursework for Algebra 2 will be a continuation of work completed in Algebra 1. It is a high school course in math. It takes the basic concepts in algebra learned in Algebra 1 and expands on them. Students will take the skills learned in Algebra 1 and use them to learn additional skills in Algebra 2. Reasons to complete Algebra 2 Coursework.
Learn about algebra from top-rated math teachers. Whether you’re interested in learning basic pre-algebra skills, or algebra I and II, including logic gates and Boolean algebra, Udemy has a course to help you better understand algebra concepts.
## Results
With one of our advanced algebra classes, you can easily learn about linear equations, inequalities, functions, exponential, sequences, series, probability, and trigonometry. We also have training courses that focus entirely on algebra functions, expressions, and equations helping you to gain a much better grasp of algebraic calculations.
#### Algebra Course - Your Favourite Teacher.
Algebra I is one of the most critical courses that students take in high school. Not only does it introduce them to a powerful reasoning tool with applications in many different careers, but algebra is the gateway to higher education.
#### Algebra 2 Tutors In Kirkland QC - The Knowledge Roundtable.
Algebra Course; Current Status. Not Enrolled. Price. Free Every 0 days. Get Started. orLogin. This Algebra course covers topics in factorising expressions, inequalities, and sequences. Please watch the films to access the topics and quizzes for this course. Course Content. Expand All.
#### Course Catalogue - Introduction to Linear Algebra (MATH08057).
Linear Algebra courses from top universities and industry leaders. Learn Linear Algebra online with courses like Mathematics for Machine Learning and Mathematics for Machine Learning: Linear Algebra.
#### How to Learn Algebra (with Pictures) - wikiHow.
An undergraduate degree in mathematics provides an excellent basis for graduate work in mathematics or computer science, or for employment in such mathematics-related fields as systems analysis, operations research, or actuarial science.
#### Algebra: A Complete Course - VideoText.
Welcome to the Mathematical Institute course management portal. Here you will find overviews, synopses and material for the courses available on your degree. Courses are classified into undergraduate and postgraduate sections - please use the tabs above to browse the appropriate section.
#### A Lecture Course in Geometric Algebra.
Algebra Courses. Title Examinable Term Number of Lectures; Commutative Algebra: Part III Examinable: Michaelmas: 24: Finite Dimensional Lie and Associate Algebras: Part III Examinable: Michaelmas: 24: Profinite Groups: Part III Examinable: Lent: 24: Representation Theory of Symmetric Groups.
Top Essay Writing Services Discounts Best essay services discount codes Essay Discounts Top Custom writing services discounts | 1,062 | 5,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-17 | longest | en | 0.863779 |
https://brainbell.com/tutors/A+/Hardware/Lesson_Summary_c5.htm | 1,708,834,988,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474581.68/warc/CC-MAIN-20240225035809-20240225065809-00564.warc.gz | 151,861,960 | 12,023 | # Testing a Coil
Since an inductor is simply a wire coil, it can be tested for continuity in much the same way as a transformer is tested.
Visually inspect the wire for deterioration. If it shows signs of breakage or burned areas, it should be replaced. If the wire looks good, follow up with a conductivity test. Turn the system power off. Disconnect one lead to the coil (this might require a soldering iron) and connect one meter lead to each end of the coil. A null or low reading indicates continuity. A reading of high or infinite resistance indicates a lack of continuity. Replace the coil.
## Lesson Summary
The following points summarize the main elements of this lesson:
• Ohm's Law is the basic formula for power calculations.
• There are two types of power: alternating current (AC) and direct current (DC).
• A multimeter is used to measure volts, amps, continuity, and resistance.
• A computer technician should be familiar with the various electronic components found on a power-supply circuit board.
• In a new installation, a computer technician should never assume that the power supplied in the wall outlets is correct. Always test and be sure.
• Be careful with large electrolytic capacitors. They can store electrical charges and must be discharged before it is safe to work with them.
# Electrostatic Discharge (ESD)
This lesson discusses a phenomenon that can damage or ruin sensitive electronic equipment-electrostatic discharge (ESD), sometimes referred to as static electricity. Fortunately, it is one of the easiest things to protect against.
After this lesson, you will be able to:
• Define ESD (electrostatic discharge).
• Avoid ESD.
Estimated lesson time: 15 minutes | 352 | 1,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.909492 |
https://kr.mathworks.com/matlabcentral/cody/players/17044932/solved | 1,590,374,393,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347387155.10/warc/CC-MAIN-20200525001747-20200525031747-00459.warc.gz | 428,540,673 | 18,899 | Cody
# Gökhan Topuz
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#### Problem 8. Add two numbers
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#### Problem 11. Back and Forth Rows
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Tags matrices
#### Problem 13. Remove all the consonants
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#### Problem 5. Triangle Numbers
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#### Problem 7. Column Removal
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#### Problem 2. Make the vector [1 2 3 4 5 6 7 8 9 10]
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Tags basic, colon, x=1:10
#### Problem 4. Make a checkerboard matrix
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#### Problem 42651. Vector creation
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#### Problem 1024. Doubling elements in a vector
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#### Problem 3076. Create a vector
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Tags vector, uab
#### Problem 2631. Flip the vector from right to left
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#### Problem 605. Whether the input is vector?
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#### Problem 1107. Find max
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Tags find, max, vector
#### Problem 624. Get the length of a given vector
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#### Problem 135. Inner product of two vectors
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#### Problem 247. Arrange Vector in descending order
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#### Problem 6. Select every other element of a vector
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#### Problem 3. Find the sum of all the numbers of the input vector
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#### Problem 672. Longest run of consecutive numbers
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#### Problem 44960. Rescale Scores
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#### Problem 44950. Calculate Inner Product
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#### Problem 44952. Find MPG of Lightest Cars
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#### Problem 44948. Calculate a Damped Sinusoid
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#### Problem 44951. Verify Law of Large Numbers
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#### Problem 44947. Find the Oldest Person in a Room
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#### Problem 44944. Convert from Fahrenheit to Celsius
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#### Problem 1. Times 2 - START HERE
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Tags intro, math, basics
1 – 33 of 33 | 738 | 2,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.728334 |
http://www.jiskha.com/display.cgi?id=1315603861 | 1,496,116,422,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463613780.89/warc/CC-MAIN-20170530031818-20170530051818-00280.warc.gz | 674,398,111 | 3,934 | # College Math
posted by on .
Sales at a computer retail store were \$1,000,000 in 1982 and \$1,790,000 in 1987
let x=0 represent 1982
find the equation giving year sales. estimate sales in 1990.
Could you guys help me with this.
• College Math - ,
sales=slope*time+constant
slope= 1.790/5
sales= 1.790/5 * (year-1982)+1,000,000
year is in the format of like 1990
sales in 1990=1.790/6(1990-1982)+1000000
• College Math - ,
For some reason this is not correct :/
It is not in one of the multiple choices | 167 | 516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-22 | latest | en | 0.909516 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=156542 | 1,371,633,932,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708546926/warc/CC-MAIN-20130516124906-00091-ip-10-60-113-184.ec2.internal.warc.gz | 309,414,499 | 951 | ```Question 205335
Determine whether the the two planes; 4x+y-z = 3 and 2x - 5y + 3z =2 are parallel, orthogonal, conincident or none of these.
-------
4x + y - z = 3
Its normal vector, A, is 4i + j - k
r = sqrt(16+1+1) = 3sqrt(2)
--------
2x - 5y + 3z =2
Its normal vector, B, is 2i - 5j + 3k
r = sqrt(4+25+9) = sqrt(38)
-------------------
Finding the angle between them will clarify it.
A dot B = |A||B|cos
A dot B = 8-5-3 = 0
Since the magnitudes of both A and B are non-zero, the cos = 0.
--> 90º orthogonal
``` | 202 | 516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2013-20 | latest | en | 0.759095 |
https://stats.stackexchange.com/questions/55992/convergence-of-batch-gradient-descent-in-logistic-regression | 1,582,724,847,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146342.41/warc/CC-MAIN-20200226115522-20200226145522-00376.warc.gz | 546,287,518 | 32,318 | # Convergence of batch gradient descent in logistic regression
I am not really sure about how it behaves when using batch gradient descent in logistic regression.
As we do each iteration, $L(W)$ is getting bigger and bigger, it will jump across the largest point and $L(W)$ is going down. How do I know it without computing $L(W)$ but only knowing old $w$ vector and updated $w$ vector?
If I use regularized logistic regression, will the weights become smaller and smaller or any other patterns?
Regularization is designed to combat overfitting, but not aid in gradient descent convergence.
If you are minimizing a function $J$ parameterized by vector $\theta$ and where each element of $\theta$ is identified by $\theta_j$, (i.e. minimize $J(\theta)$).
Then the basic idea in batch gradient descent is to iterate until convergence by computing a new value of $\theta$ from the previous one in the following way. Updated each $\theta_j$ simultaneously with the formula.
$\theta_j := \theta_j - \alpha\frac{\partial }{\partial \theta_j}J(\theta)$
That $\alpha$ term is called the learning rate. It's arbitrary and if it's very small then the algorithm will converge slowly which will make the algorithm take a long time, but if it's too large, then what can happen is exactly what you are experiencing. In this case $\theta$ will be updated in the right direction, but will go too far and jump past the minimum or it can even climb out and increase.
The remedy is to simply decrease $\alpha$ until it doesn't happen. A sufficiently small learning rate guarantees that $J(\theta)$ will decrease on every iteration. The trick is to determine what value of $\alpha$ is a good one that allows fast convergence but avoids non-convergence.
A useful approach is to plot $J(\theta)$ while the algorithm is running to observe how it decreases. Start with a small value (e.g. 0.01) increase it if it appears to result in slow convergence or decrease it still further in the case of non-convergence.
I assume that $L(W)$ is the log likelihood function. The answer is convergence. When the difference between old W and updated W is small enough, you stop the iteration. Detailed implemention in R has been posted here: http://mathiology.blogspot.nl/2014/04/logistic-regression-with-gradient_7.html | 506 | 2,295 | {"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} | 3.421875 | 3 | CC-MAIN-2020-10 | latest | en | 0.913321 |
https://brilliant.org/problems/sets-operations/ | 1,529,764,046,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865081.23/warc/CC-MAIN-20180623132619-20180623152619-00528.warc.gz | 558,933,940 | 11,212 | # Sets Operations
Let $$A$$, $$B$$ and $$C$$ be any three sets such that $$A-B=C$$; then which of the following statement(s) is/are correct?
1. $$A$$ is the union of $$B$$ and $$C$$
2. $$C$$ is always a proper subset of $$A$$
3. $$C$$ is the compliment of $$B$$
ANSWER FORMATE: Enter the sum of the serial numbers as answer. For example, if all three statements are correct then the answer is: $$1+2+3=6$$.
× | 127 | 412 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-26 | latest | en | 0.885027 |
http://functions.wolfram.com/ElementaryFunctions/Coth/21/01/02/17/07/01/0002/ | 1,527,399,756,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868003.97/warc/CC-MAIN-20180527044401-20180527064401-00018.warc.gz | 111,949,126 | 11,536 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
Coth
http://functions.wolfram.com/01.22.21.0233.01
Input Form
Integrate[z^n E^(p z) Cos[a z] Cosh[c z] Coth[c z], z] == (-(1/2)) E^(c z) n! (E^((I a + p) z) Sum[(1/(-j + n)!) ((-1)^j (I a + c + p)^(-1 - j) z^(-j + n) HypergeometricPFQ[{Subscript[a, 1], \[Ellipsis], Subscript[a, 1 + j], 1}, {1 + Subscript[a, 1], \[Ellipsis], 1 + Subscript[a, 1 + j]}, E^(2 c z)]), {j, 0, n}] + E^(((-I) a + p) z) Sum[(1/(-j + n)!) ((-1)^j ((-I) a + c + p)^(-1 - j) z^(-j + n) HypergeometricPFQ[{Subscript[b, 1], \[Ellipsis], Subscript[b, 1 + j], 1}, {1 + Subscript[b, 1], \[Ellipsis], 1 + Subscript[b, 1 + j]}, E^(2 c z)]), {j, 0, n}]) - (1/4) E^(c z) n! (E^((I a - 2 c + p) z) Sum[(1/(-j + n)!) ((-1)^j (I a - c + p)^(-1 - j) z^(-j + n) HypergeometricPFQ[{Subscript[c, 1], \[Ellipsis], Subscript[c, 1 + j], 1}, {1 + Subscript[c, 1], \[Ellipsis], 1 + Subscript[c, 1 + j]}, E^(2 c z)]), {j, 0, n}] + E^(((-I) a - 2 c + p) z) Sum[(1/(-j + n)!) ((-1)^j ((-I) a - c + p)^(-1 - j) z^(-j + n) HypergeometricPFQ[ {Subscript[d, 1], \[Ellipsis], Subscript[d, 1 + j], 1}, {1 + Subscript[d, 1], \[Ellipsis], 1 + Subscript[d, 1 + j]}, E^(2 c z)]), {j, 0, n}] + E^((I a + 2 c + p) z) Sum[(1/(-j + n)!) ((-1)^j (I a + 3 c + p)^(-1 - j) z^(-j + n) HypergeometricPFQ[{Subscript[e, 1], \[Ellipsis], Subscript[e, 1 + j], 1}, {1 + Subscript[e, 1], \[Ellipsis], 1 + Subscript[e, 1 + j]}, E^(2 c z)]), {j, 0, n}] + E^(((-I) a + 2 c + p) z) Sum[(1/(-j + n)!) ((-1)^j ((-I) a + 3 c + p)^(-1 - j) z^(-j + n) HypergeometricPFQ[ {Subscript[f, 1], \[Ellipsis], Subscript[f, 1 + j], 1}, {1 + Subscript[f, 1], \[Ellipsis], 1 + Subscript[f, 1 + j]}, E^(2 c z)]), {j, 0, n}]) /; Subscript[a, 1] == Subscript[a, 2] == \[Ellipsis] == Subscript[a, n + 1] == (c + p + I a)/(2 c) && Subscript[b, 1] == Subscript[b, 2] == \[Ellipsis] == Subscript[b, n + 1] == (c + p - I a)/(2 c) && Subscript[c, 1] == Subscript[c, 2] == \[Ellipsis] == Subscript[c, n + 1] == (p + I a - c)/(2 c) && Subscript[d, 1] == Subscript[d, 2] == \[Ellipsis] == Subscript[d, n + 1] == (p - I a - c)/(2 c) && Subscript[e, 1] == Subscript[e, 2] == \[Ellipsis] == Subscript[e, n + 1] == (p + I a + 3 c)/(2 c) && Subscript[f, 1] == Subscript[f, 2] == \[Ellipsis] == Subscript[f, n + 1] == (p - I a + 3 c)/(2 c) && Element[n, Integers] && n >= 0
Standard Form
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MathML Form
z n p z cos ( a z ) cosh ( c z ) coth ( c z ) z - 1 2 c z n ! 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Date Added to functions.wolfram.com (modification date)
2002-12-18 | 12,843 | 34,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-22 | latest | en | 0.137521 |
http://nrich.maths.org/public/leg.php?code=170&cl=3&cldcmpid=4737 | 1,371,653,071,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708808740/warc/CC-MAIN-20130516125328-00089-ip-10-60-113-184.ec2.internal.warc.gz | 185,080,471 | 6,647 | # Search by Topic
#### Resources tagged with Perimeters similar to Weekly Problem 5 - 2006:
Filter by: Content type:
Stage:
Challenge level:
### There are 17 results
Broad Topics > Measures and Mensuration > Perimeters
### Changing Areas, Changing Perimeters
##### Stage: 3 Challenge Level:
How can you change the area of a shape but keep its perimeter the same? How can you change the perimeter but keep the area the same?
### Perimeter Possibilities
##### Stage: 3 Challenge Level:
I'm thinking of a rectangle with an area of 24. What could its perimeter be?
### Some(?) of the Parts
##### Stage: 4 Challenge Level:
A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle
### Coins on a Plate
##### Stage: 3 Challenge Level:
Points A, B and C are the centres of three circles, each one of which touches the other two. Prove that the perimeter of the triangle ABC is equal to the diameter of the largest circle.
### Can They Be Equal?
##### Stage: 3 Challenge Level:
Can you find rectangles where the value of the area is the same as the value of the perimeter?
### Warmsnug Double Glazing
##### Stage: 3 Challenge Level:
How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?
### Arclets
##### Stage: 3 Challenge Level:
Each of the following shapes is made from arcs of a circle of radius r. What is the perimeter of a shape with 3, 4, 5 and n "nodes".
### Pick's Theorem
##### Stage: 3 Challenge Level:
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
### Origami Shapes
##### Stage: 4 Challenge Level:
Take a sheet of A4 paper and place it in landscape format. Fold up the bottom left corner to the top so the double thickness is a 45,45,90 triangle. Fold up the bottom right corner to meet the. . . .
### On the Edge
##### Stage: 3 Challenge Level:
Here are four tiles. They can be arranged in a 2 by 2 square so that this large square has a green edge. If the tiles are moved around, we can make a 2 by 2 square with a blue edge... Now try. . . .
### AP Rectangles
##### Stage: 3 Challenge Level:
An AP rectangle is one whose area is numerically equal to its perimeter. If you are given the length of a side can you always find an AP rectangle with one side the given length?
### Is There a Theorem?
##### Stage: 3 Challenge Level:
Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel?
### Contact
##### Stage: 4 Challenge Level:
A circular plate rolls in contact with the sides of a rectangular tray. How much of its circumference comes into contact with the sides of the tray when it rolls around one circuit?
### Coke Machine
##### Stage: 4 Challenge Level:
The coke machine in college takes 50 pence pieces. It also takes a certain foreign coin of traditional design. Coins inserted into the machine slide down a chute into the machine and a drink is duly. . . .
### Pericut
##### Stage: 4 and 5 Challenge Level:
Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?
### Perimeter Expressions
##### Stage: 3 Challenge Level:
Create some shapes by combining two or more rectangles. What can you say about the areas and perimeters of the shapes you can make?
### Giant Holly Leaf
##### Stage: 4 Challenge Level:
Find the perimeter and area of a holly leaf that will not lie flat (it has negative curvature with 'circles' having circumference greater than 2πr). | 889 | 3,847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2013-20 | longest | en | 0.863501 |
http://www.homeownershub.com/maintenance/dual-fuel-83158-.htm | 1,477,637,127,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721558.87/warc/CC-MAIN-20161020183841-00094-ip-10-171-6-4.ec2.internal.warc.gz | 487,710,889 | 14,109 | # dual fuel
• posted on January 22, 2006, 12:46 pm
In my other post, I ranted and wailed about the propane industry. Since my home is ten years old, I probably have another 3-6 years on the Goodman propane furnace before it craps out. I want to replace it with a dual fuel system, i.e., a heat pump for the good days with ambient above 40 or so and kick in the propane when the HP would call for auxillary heat.
Who's got dual fuel, are you pleased with it, does this have any drawbacks I have not yet discovered, and did you do a cost/economy analysis before and since the install.
Of course, the region is of great importance, as is the relative cost of electricity and propane. Here, central North Carolina, propane can be had for ~\$1.70 gallon and the sparks run \$0.083/kwh.
Yes, I understand the impact of increased efficiency on a new unit (HP as well as new propane burner), just wondering if you have determined WHEN the increased initial cost has been offset and you start saving.
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• posted on January 22, 2006, 1:10 pm
There are too many variables for anyone to give you an answer that you can relate to your situation. I've replaced several heat pump systems with combination propane / heat pump. This entails new 15 amp electric, gas line, vent for the furnace, etc. Most people seem to think it takes about 2 to 4 years for that payback, but I seriously doubt if they ever did the math. In your case, it's a relatively simple change out. The hardest part might be pulling new low voltage wires to the thermostat and outside unit. You'll have to do a heat loss calculation on the home to figure the 'economic balance point'. You also must realize that typical heat pump supply duct temperatures are lower than your body temperature. If you change to a 'comfort balance point', all the math you did is irrelevant. This Penn State web site may help. http://energy.cas.psu.edu/energyselector /
my
fuel
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the
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• posted on January 22, 2006, 2:02 pm
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• posted on January 22, 2006, 2:06 pm
He might be able to hold his breath, because with a Goodman, it could crap out any minute.
(add rr between nc and com)> wrote
Goodman
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• posted on January 22, 2006, 5:46 pm
On Sun, 22 Jan 2006 12:46:03 GMT, Jim Murphy wrote:
I have used dual fuel since Jan 98 in NW GA and I like it a lot. I did not do a cost analysis before the install. I had planned on installing a gas furnace for heat and all other appliances total electric when I built my house, but the local EMC persuaded me to go dual fuel with the following incentives for new constructions.
A Free Underground Power Drop (over 300 ft and a transformer on the ground) A \$1000 Rebate A Free 60 Gallon Marathon Electric Water Heater
The cost of the underground drop alone was about \$4000 due to the distance and the requirement of having a transformer on the ground instead of on the pole. An underground drop of less than 300 ft with a transformer on the pole was free at that time with a new construction.
I only found out about those incentives when I was informed how much the drop was going to cost and asked if there was any way I could offset some of that cost. I was told that I could go total electric and receive the above incentives - I countered with dual fuel and they told me dual fuel also counted as total electric as far as the incentives were concerned. The extra cost for the heating system was about \$400-600 IIRC, so I figure I came out ahead.
I should let you know that I didn't want to go total electric (no gas furnace), because I know too many people who complain about their heat pump when the weather gets cold. The gas also gives me a backup heat source if the power is out - I can run the gas furnace with my generator (runs on gasoline or propane - purchased so I can have water and to keep the fridge and freezer cold during long outages) if it I run out of wood or some other emergency forces me to.
I would suggest that if you do decide to change to dual fuel, you check with your local power company to see if they offer any incentives for changing - our local EMC does, but I think the incentives are offered periodically, not all the time. Incentives could very well change the cost recovery quite a bit.
Later, Mike (substitute strickland in the obvious location to reply directly) ----------------------------------- snipped-for-privacy@bellsouth.net
Please send all email as text - HTML is too hard to decipher as text. | 1,117 | 4,629 | {"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} | 2.84375 | 3 | CC-MAIN-2016-44 | longest | en | 0.949674 |
https://urdu.wordinn.com/square+root | 1,709,219,446,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474843.87/warc/CC-MAIN-20240229134901-20240229164901-00391.warc.gz | 584,349,034 | 13,524 | # Square Root meaning in Urdu
### Square Root Definitions
1) Square Root : خود تقسیمی کا حامل عدد : (noun) a number that when multiplied by itself equals a given number.
### Useful Words Definitions
Complex Number: (mathematics) a number of the form a+bi where a and b are real numbers and i is the square root of -1.
Percent Sign: a sign (`%`) used to indicate that the number preceding it should be understood as a proportion multiplied by 100.
Exponent: a mathematical notation indicating the number of times a quantity is multiplied by itself.
Rootlet: small root or division of a root.
99: The number 99 is a natural number that comes after 98 and before 100. It is composed of two nines, making it a double-digit number.
44: The number 44 is a natural number that comes after 43 and before 45. It is composed of two fours, making it a two-digit number.
Batting Average: (baseball) a measure of a batter`s performance; the number of base hits divided by the number of official times at bat.
Abundance: (physics) the ratio of the number of atoms of a specific isotope of an element to the total number of isotopes present.
150: The number 150 is a three-digit natural number that comes after 149 and before 151. It is composed of the digits 1, 5, and 0.
0: a mathematical element that when added to another number yields the same number.
69: The number 69 is a two-digit natural number that comes after 68 and before 70. It is composed of the digits 6 and 9.
149: The number 149 is a three-digit natural number that comes after 148 and before 150. It is composed of the digits 1, 4, and 9.
1: the smallest whole number or a numeral representing this number.
2: the cardinal number that is the sum of one and one or a numeral representing this number.
Accuracy: (mathematics) the number of significant figures given in a number.
Heteroploid: (genetics) an organism or cell having a chromosome number that is not an even multiple of the haploid chromosome number for that species.
Chance: a measure of how likely it is that some event will occur; a number expressing the ratio of favorable cases to the whole number of cases possible.
Square-Toed: having a square toe.
Blood Profile: counting the number of white and red blood cells and the number of platelets in 1 cubic millimeter of blood. A CBC is a routine test used for various medical purposes, including general health screenings, diagnosing medical conditions, and monitoring ongoing treatments..
Square: make square.
48: being eight more than 40. The number 48 is a natural number that comes after 47 and before 49.
Squarish: somewhat square in appearance or form.
Quadratic: of or relating to or resembling a square.
Radical: arising from or going to the root or source.
Piazza: a public square with room for pedestrians.
Flat Knot: a square knot used in a reef line.
Boxing Ring: a square ring where boxers fight.
Square-Rigged: rigged with square sails as the principal ones. | 685 | 2,971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-10 | latest | en | 0.891536 |
http://learner.org/courses/learningmath/data/session9/part_d/box.html | 1,566,266,712,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315174.57/warc/CC-MAIN-20190820003509-20190820025509-00453.warc.gz | 110,983,312 | 8,973 | Teacher resources and professional development across the curriculum
Teacher professional development and classroom resources across the curriculum
Session 9, Part D:
The Effect of Sample Size
In This Part: Sample Size 20 | Comparing Sample Sizes 10 and 20 | Box Plot Comparisons
In the previous discussion, you investigated how increasing the sample size does two things:
• Decreases the sample-to-sample variation in the estimates • Produces a higher proportion of estimates closer to the actual population size
We can also use another familiar method to explore this phenomenon: the Five-Number Summary and box plot.
Problem D3
Here is the stem and leaf plot for the 100 estimates from samples of size 10:
Use the stem and leaf plot to determine the Five-Number Summary for these estimates. These questions may help you along:
a. What is the position of the median, and which two values are used to calculate it? b. If there are 50 values in each half, how are the quartiles calculated? c. Complete the Five-Number Summary table:
Sample Size 10
Maximum Upper Quartile (Q3) Median Lower Quartile (Q1) Minimum
Sample Size 10
Maximum 620 Upper Quartile (Q3) 540 Median 500 Lower Quartile (Q1) 470 Minimum 360
Problem D4
Generate the Five-Number Summary for this stem and leaf plot of the 100 estimates based on samples of size 20:
Sample Size 20
Maximum Upper Quartile (Q3) Median Lower Quartile (Q1) Minimum
Sample Size 20
Maximum 610 Upper Quartile (Q3) 530 Median 500 Lower Quartile (Q1) 482.5 Minimum 390
Since the number of estimates is the same as Problem D3's, the quartiles and median will be in the same positions. Count the values in increasing order to find them. Close Tip Since the number of estimates is the same as Problem D3's, the quartiles and median will be in the same positions. Count the values in increasing order to find them
Problem D5 Create two box plots for the Five-Number Summaries you generated in Problems D3 and D4, placing them side by side on the same scale to make them easier to compare.
Problem D6
What do the box plots suggest about the effect of sample size on the accuracy of the estimates? In particular, how do the box plots illustrate the following:
a. How much the estimates vary from sample to sample b. How close the estimates are to the actual value of 500
Video Segment In this video segment, the participants discuss what percentages of their data fell in particular interval ranges for samples of size 10 and 20. Professor Kader then introduces the Central Limit Theorem to further discuss the connection between probability and statistics. What is the give-and-take between selecting an interval range and sample size when designing a statistical investigation? How would you use this information to plan a statistical investigation? How can you be more precise when taking a sample size? How can you be more accurate? If you're using a VCR, you can find this segment on the session video approximately 16 minutes and 2 seconds after the Annenberg Media logo.
Next > Homework
Session 9: Index | Notes | Solutions | Video | 684 | 3,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2019-35 | latest | en | 0.84285 |
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# expected value and variance
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
1.The following table shows the expected value and variance for 5 projects a firm can undertake.
Project Expected Value Variance
A \$100 \$124
B \$220 \$110
C \$100 \$138
D \$180 \$138
E \$200 \$124
a. Project B dominates all others. True or False? Why? .
b. Project C is the least preferable. True or False? Why?
c. If the mean-variance rule is used for the decision, project D is preferable to C. True or False? Why?
2. If firms in a perfectly competitive industry are earning an economic profit, product price will because . After long-run competitive equilibrium comes about there will be (fewer, more, the same number of) firms in the industry and the industry will produce (less, more, the same amount of) output. In long-run competitive equilibrium each firm will earn economic profit.
3. Sony and Zenith must each decide which technology to utilize in building their high definition television (HDTV) sets: either Alpha technology or Beta technology. Sony has a technological advantage in using Alpha technology and Zenith has a technological advantage in using Beta technology. The payoff table below shows the profit outcomes for both firms in the various possible technology choice outcomes:
Zenith
Alpha
Beta
Sony
Alpha
A
\$16 \$12 B
\$11 \$10
Beta
C
\$9 \$8 D
\$13 \$15
Payoffs in billions of
Suppose the technology decision between Alpha and Beta will be made simultaneously. Answer the following questions:
a. Does Sony have a dominant strategy? If yes, which one? If not, why not? Explain. (4 pts)
b. Does Zenith have a dominant strategy? If yes, which one? If not, why not? Explain. (4 pts)
c. Identify the Nash equilibrium (equilibria) for this simultaneous decision. Explain. (8 pts)
© BrainMass Inc. brainmass.com October 10, 2019, 12:20 am ad1c9bdddf
https://brainmass.com/economics/perfect-competition/expected-value-variance-288108
#### Solution Preview
1.The following table shows the expected value and variance for 5 projects a firm can undertake.
Project Expected Value Variance
A \$100 \$124
B \$220 \$110
C \$100 \$138
D \$180 \$138
E \$200 \$124
a. Project B dominates all others. True or False? Why? .
TRUE. This is so because the project has the highest expected value (\$220) and the lowest variance (\$110).
b. Project C is the least preferable. True or False? Why?
TRUE. This is so because the project has the lowest expected value (\$100) and the highest variance (\$138).
c. If the mean-variance rule is used for the decision, project D is preferable to C. True or False? Why?
TRUE. Both projects have the same variance (\$138) but project D has higher expected value (\$180) compared to project C (\$100).
2. If firms in a perfectly competitive industry are earning an economic profit, product price will FALL because FIRMS WILL ENTER THE MARKET. After long-run competitive equilibrium comes about there will be MORE(fewer, more, the same number of) firms in the industry and ...
#### Solution Summary
This solution reiterates expected value and variance.
\$2.19 | 755 | 3,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-51 | longest | en | 0.868563 |
http://www.beesource.com/forums/printthread.php?t=205054&pp=20&page=1 | 1,386,524,978,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163728657/warc/CC-MAIN-20131204132848-00043-ip-10-33-133-15.ec2.internal.warc.gz | 254,276,020 | 2,665 | # HFCS
• 02-02-2003, 06:51 PM
hollowlog
I am going to start feeding type 42 hfcs. it is 70% sugar 30% water. Can anyone tell me how much water to add to five gallons to get a 2:1 sugar to water ratio. How about a 1:1.
• 02-02-2003, 09:27 PM
Michael Bush
>I am going to start feeding type 42 hfcs. it is 70% sugar 30% water. Can anyone tell me how much water to add to five gallons to get a 2:1 sugar to water ratio. How about a 1:1.
It's probably not that critical to get it perfect. The bees will gobble it down at any strength that is close to as sweet as nectar or more sweet.
The High Fructoce Corn Syrup is not going to have the same effect for spring stimulation as the 2:1 sucrose sugar will. The production of the enzymes to change the sucrose makes the bees think it's a nectar flow, and the HFCS has already has that change done to it.
2:1 and 1:1 syrup are by volume. In other words, you take a quart of hot water and a quart of sugar and you get 1:1 syrup. I'm assuming the 70% sugar 30% water are by weight? If you knew that for sure you could weigh a quart of sugar. A quart of water weighs two pounds. And you could figure out the ratio by weight then. I'm really not sure if the 70/30 ratio is by weight, or volume or what. You need to have them both in the same terms to figure it out, and I'm not sure what the terms are.
• 02-03-2003, 07:42 PM
hollowlog
I would imagine it is by volume. My refractometer that I use for honey shows 30% and that would be actual percent of water in solution would it not?
• 02-04-2003, 06:30 AM
Michael Bush
I guess I've never paid that much attention to what a refractometer measures, but I'm guessing that 30% is by weight. Measuring by volume is more of a convenience for recipes.
Then to find the answer requires that we find the weight of a quart of sugar and make a ratio with the weight of a quart of water (2 pounds) to calcuate the answer of how much water to add. I don't have a scale to weigh anything that small, but you could start with a new 5 pound bag and see how many quarts of sugar there are in a 5 pound bag to figure it out.
I don't have an unopened one right now. | 597 | 2,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | longest | en | 0.969115 |
https://www.tek-trol.com/measurement-basics/thermal-mass-flow-meter/ | 1,685,362,518,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644855.6/warc/CC-MAIN-20230529105815-20230529135815-00276.warc.gz | 1,116,043,277 | 26,716 | # Thermal Mass Flow Meter
## ACCURATE MASS FLOW METERS - Thermal Mass Flow Meter
### Thermal mass flow meters operate on an interesting working principle of Hot-wire Anemometers. A core of Anemometer is exposed to constant heat and the heat convected by the fluid passing by the anemometer is noted to measure the fluid velocity. Similarly, instead of using fluid momentum, thermal mass flow meters make use of the thermal or heat conducting properties of fluids to determine the mass flow.
The thermal mass flow meter is available in high-performance Inline type and Insertion type flow meters. Insertion flow meters consist of a probe with sensing elements that are inserted into the pipe irrespective of pipe size. These are easy to install and more economical. Inline flow meters are connected to the pipe itself by the flanged or threaded connection. Inline meters are suitable for smaller diameter pipes.
The thermal mass flow meter is comprised of a family of instruments for precision mass flow measurement. The sensor probe is inserted in the center line of the flow. Thermal mass flow meters use the concept of heat dissipations for mass flow calculations. Two sensors, both resistive temperature detectors (RTDs), are immersed into the stream of flow under measurement. The first sensor, referred to as a velocity sensor, has a heating element in it and it monitors mass flow rate. The second sensor, which is referred to as a temperature sensor, measures the actual temperature of the gas flowing in the pipe as a reference regardless of the flow velocity. Every gas molecule has an ability to absorb heat. Thus when the gas flows through the pipe, heat from the first sensor (velocity sensor) is absorbed by the gas molecules. As these molecules reach the next sensor (temperature sensor) heat is dissipated and is sensed by this sensor instantaneously. The amount of heat absorbed by the gas molecules is directly proportional to the number of molecules passing the velocity sensor which is nothing but the mass flow rate. This can be mathematically expressed as:
## To calculate the volumetric flow rate:
m = K (H/ ΔT)1.67
Where,
m = mass flow rate
K = proportionality constant
H = Heat loss
ΔT = Temperature difference
Two methods, Constant Temperature Difference and Constant Power, are used for mass flow measurement and both methods obtain the same results.
##### Constant Temperature Difference:
Maintaining a constant temperature difference between the heated (velocity) sensor and a reference (temperature) sensor is the objective of this method. A predefined temperature difference, ΔT, is set during initial calibration. The amount of power to the heater is controlled to maintain the temperature difference. A non-linear relationship between the mass flow rate and power signal is observed. At higher flow velocities, more power is required to maintain the temperature difference. Whereas at low flow velocities, less power is to maintain the temperature difference. This provides excellent low flow sensitivity as well as outstanding turn down.
##### Constant Power:
This technology works on maintaining a constant power to the heating sensor. The instantaneous fluid temperature varies depending on rate of flow and is sensed by the reference temperature sensor. Temperature difference between the heating sensor and reference sensor is constantly monitored. It decreases when the mass flow rate increases. At the lower flow rates, the change in temperature is larger which offers great low flow sensitivity. As the flow rate increases the temperature difference reduces.
#### No industry can progress without reliable and accurate measurement. The key is measurement, simple as that. Measurement can result in two possible outcomes: If the result confirms your hypothesis then you've made a measurement; If the result is contrary then you've found a problem.
Etiam magna arcu, ullamcorper ut pulvinar et, ornare sit amet ligula. Aliquam vitae bibendum lorem. Cras id dui lectus. Pellentesque nec felis tristique urna lacinia sollicitudin ac ac ex. Maecenas mattis faucibus condimentum. Curabitur imperdiet felis at est posuere bibendum. Sed quis nulla tellus. | 823 | 4,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.924073 |
http://herkimershideaway.org/apstatistics/ymmsum99/ymm663.htm | 1,531,993,405,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590794.69/warc/CC-MAIN-20180719090301-20180719110301-00539.warc.gz | 154,833,161 | 2,582 | "The most important questions of life are, for themost part, really problems of probability."
Pierre-Simon Laplace (1749 - 1827)
6.3 MORE ABOUT PROBABILITY (Pages 341 - 355)
OVERVIEW: There are basic laws thatgovern wise and efficient uses of probability. An understanding ofthese laws is important for those who wish to utilize statistics inpractical and meaningful ways..
For any two events A and B,
Prob(A or B) = Prob(A) + Prob(B) -Prob(A and B).
If A and B are disjoint, then Prob(A and B) =0.
Prob(B|A) = prob(A and B)/prob(A).
Here are some illustrations of these probabilitylaws. Consider rolling two dice and calculating the sum on the upfaces. The sums are shown in the following table:
RED DIE
GREEN
DIE
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
Prob(sum=7 OR sum = 11) = 6/36 + 2/36 =8/36 = 22.22%.
Prob(sum=7 AND sum = 11) = 0.
Prob(sum=7 OR at least one die shows a 5) =6/36 + 11/36 - 2/36 = 15/36 = 41.67%.
Prob(faces show same number OR sum>9) = 6/36 + 6/36 -2/36 = 10/36 = 27.78%.
Prob(sum=6|one die shows a 4) = Prob(sum=6AND one dieshows a 4)/Prob(one die shows a 4) = (2/36)/(11/36) = 2/11 =18.18%.
Prob(sum is even|sum>9) = Prob(sum is evenANDsum>9)/Prob(sum>9) = (4/36)/(6/36) = 4/6 = 66.67%.
The probability formulas are useful, but oftentimes they are not needed if sample spaces are small and one usessome good old common sense. Here are some simple, yet sophisticated,examples involving conditional probability.
Consider a family with two children. There arefour possible boy/girl combinations, all equally likely.
OLDEST YOUNGEST G G G B B G B B
Prob(at least one child is a girl) = 3/4.
Prob(two girls) = 1/4.
Prob(two girls|one child is a girl) = 1/3.Sample space is {(G,G), (G,B), (B,G)}.
Prob(two girls|oldest child is a girl) = 1/2.Sample space is {(G,G), (G,B)}.
Prob(exactly one boy|one child is a girl) =2/3.
Prob(exactly one boy|youngest child is a girl) =1/2.
OK, now suppose that a family unknown to you hastwo children. Here are two questions relating to this family.
...
(1) If one of the children is a boy,what is the probability that the other child is aboy?
...
(2) If the oldest child is a boy, whatis the probability that the other child is a boy?
Many intelligent people think that 50% is the answer to both (1) and(2). You should be able to show that this is not the case. | 799 | 2,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2018-30 | latest | en | 0.83007 |
https://answers.search.yahoo.com/search?p=pf&b=11&pz=10&ei=UTF-8&flt=&xargs=0 | 1,585,443,247,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493121.36/warc/CC-MAIN-20200328225036-20200329015036-00404.warc.gz | 344,869,076 | 22,935 | # Yahoo Web Search
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1. ### PF Changs?
The food served at PF Changs is based on Chinese dishes that have...
13 Answers · Dining Out · 15/12/2007
2. ### what would be the pf changs cost?
PF Changs' website has their menu and prices online, so you can get an idea. PF Changs is normally about \$10-12 per person so your budget may be...
2 Answers · Food & Drink · 02/03/2011
3. ### Panda Express or PF Chang's?
PF Chang's. It always tastes good. Panda Express...asian food and it's gross 100% of the time where I live. At our PF Chang's we have asians cooking, at our Panda...
18 Answers · Entertainment & Music · 27/10/2010
4. ### Does anybody know how to make lemon scallops like PF Changs does?
Supposedly this is it: PF Chang's Lemon Scallops INGREDIENTS: 1...
3 Answers · Food & Drink · 23/10/2007
5. ### PF Changs?
It's been a long time since I have had PF CHangs, but I remember the lettuce wraps and the dumplings...
9 Answers · Food & Drink · 07/11/2007
6. ### In Windows, What is PF Usage?!? And how do I view it?
PF Useage Stands for Page File. What it... HD as memory. This is a very simplistic description of what PF is. 310MB isn't bad, considering my computer ...
3 Answers · Computers & Internet · 27/02/2007
7. ### PF Chang's . . . what's their problem?
PF Chang's started in 1993... the original idea for the place was...
11 Answers · Dining Out · 03/02/2008
8. ### Demand function differentiable, prove [pf(p)]'<0 iff p/f(p), f '( p) < -1.(revenue lowers if demand elastic )?
[pf(p)]'=p'f(p)+pf'(p)=f(p)+pf'(p) [pf(p)]'<0 gives f(p)+pf...
1 Answers · Science & Mathematics · 28/12/2011
9. ### NBA: Give your Top 10 All-time PF's?
1 Tim Duncan: Greatest PF all-time. 4 Championships, 3 Finals.... 1 NBA championship. Invented the PF position with his combination of scoring and...
8 Answers · Sports · 25/07/2009
10. ### How much % will be taken as PF?How much company will pay on behalf of our PF?
Company PF is what the company pays as PF for you and... not pay anything. Company deducts PF at 12% of (Basic+DA) from your contribution... | 603 | 2,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-16 | latest | en | 0.861158 |
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