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https://testbook.com/question-answer/ifrm-fxleftbeginmatrixrm-dfrac--5f7424aec2ff25e2a73f6b91	[ "# If $$\\rm f(x)=\\left\\{\\begin{matrix}\\rm \\dfrac{\\sin 3x}{e^{2x}-1}, &\\rm x \\ne 0\\\\\\rm k-2, &\\rm x=0 \\end{matrix}\\right.$$ is continuous at x = 0, then k = ?\n\n1. $$\\dfrac{3}{2}$$\n2. $$\\dfrac{9}{5}$$\n3. $$\\dfrac{1}{2}$$\n4. $$\\dfrac{7}{2}$$\n\nOption 4 : $$\\dfrac{7}{2}$$\nFree\nElectric charges and coulomb's law (Basic)\n83.1 K Users\n10 Questions 10 Marks 10 Mins\n\n## Detailed Solution\n\nConcept:\n\nDefinition:\n\n• A function f(x) is said to be continuous at a point x = a in its domain, if $$\\rm \\displaystyle \\lim_{x\\to a}f(x)$$ exists or or if its graph is a single unbroken curve at that point.\n• f(x) is continuous at x = a ⇔ $$\\rm \\displaystyle \\lim_{x\\to a^+}f(x)=\\lim_{x\\to a^-}f(x)=\\lim_{x\\to a}f(x)=f(a)$$.\n\nFormulae:\n\n• $$\\rm \\displaystyle \\lim_{x\\to 0}\\dfrac{\\sin x}{x}=1$$\n• $$\\rm \\displaystyle \\lim_{x\\to 0}\\dfrac{e^x-1}{x}=1$$\n\nCalculation:\n\nSince f(x) is given to be continuous at x = 0, $$\\rm \\displaystyle \\lim_{x\\to 0}f(x)=f(0)$$.\n\nAlso, $$\\rm \\displaystyle \\lim_{x\\to a^+}f(x)=\\lim_{x\\to a^-}f(x)$$ because f(x) is same for x > 0 and x < 0.\n\n$$\\rm \\therefore \\displaystyle \\lim_{x\\to 0}f(x)=f(0)$$\n\n$$\\rm \\Rightarrow \\displaystyle \\lim_{x\\to 0}\\dfrac{\\sin 3x}{e^{2x}-1}=k-2$$\n\n$$\\rm \\Rightarrow \\displaystyle \\lim_{x\\to 0}\\dfrac{\\dfrac{\\sin 3x}{3x}\\times3x}{\\dfrac{e^{2x}-1}{2x}\\times2x}=k-2$$\n\n$$\\rm \\Rightarrow \\dfrac{3}{2}=k-2$$\n\n$$\\rm \\Rightarrow k=\\dfrac{7}{2}$$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6532901,"math_prob":1.0000093,"size":1716,"snap":"2022-40-2023-06","text_gpt3_token_len":635,"char_repetition_ratio":0.1828271,"word_repetition_ratio":0.017094018,"special_character_ratio":0.36596736,"punctuation_ratio":0.07777778,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000064,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-05T10:46:23Z\",\"WARC-Record-ID\":\"<urn:uuid:fb230f95-d62a-4fd2-9173-a8024fe548b6>\",\"Content-Length\":\"506336\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b5dda15d-f0c4-4fe5-9aa4-ecb6708d7bbb>\",\"WARC-Concurrent-To\":\"<urn:uuid:724878c5-0ef5-402c-bcb1-e786b827700f>\",\"WARC-IP-Address\":\"172.67.30.170\",\"WARC-Target-URI\":\"https://testbook.com/question-answer/ifrm-fxleftbeginmatrixrm-dfrac--5f7424aec2ff25e2a73f6b91\",\"WARC-Payload-Digest\":\"sha1:5AYOED5JJXZIQ4KMNOMLN5YZELV32SLV\",\"WARC-Block-Digest\":\"sha1:BK2KPHF5TXOAXW4UQPATX6XZGQTCNDEE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500251.38_warc_CC-MAIN-20230205094841-20230205124841-00263.warc.gz\"}"}
https://www.fractioncalculator.pro/percent-as-a-fraction/Change_29.7_to-a-fraction	[ "Change 29.7 to a fraction\n\nWelcome! Here is the answer to the question: Change 29.7 to a fraction or maybe \"What is 29.7 percent as a fraction?\". Use the percent to fraction calculator below to write any percentage in fraction form.\n\nPercent to Fraction Calculator\n\n Enter a percent value: Ex.: 62.5, 7.5, 87.5, etc. Equivalent fraction: Result here Equivalent fraction Explained: Equivalent fraction explained here" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9169142,"math_prob":0.95308775,"size":437,"snap":"2019-26-2019-30","text_gpt3_token_len":105,"char_repetition_ratio":0.21016166,"word_repetition_ratio":0.89041096,"special_character_ratio":0.25171626,"punctuation_ratio":0.15463917,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99911034,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-19T14:29:54Z\",\"WARC-Record-ID\":\"<urn:uuid:13327743-f4b9-4e52-b281-a4042e44a7cc>\",\"Content-Length\":\"50658\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e1dd12c6-7b91-4a58-9f5f-d3fc8e1817dc>\",\"WARC-Concurrent-To\":\"<urn:uuid:86e62074-0d49-4bf6-8a94-e489c3da8dc8>\",\"WARC-IP-Address\":\"104.18.48.216\",\"WARC-Target-URI\":\"https://www.fractioncalculator.pro/percent-as-a-fraction/Change_29.7_to-a-fraction\",\"WARC-Payload-Digest\":\"sha1:E66URV3RREC2DJHOI43KGZFYA6NG2V7Q\",\"WARC-Block-Digest\":\"sha1:LGRJG3GNZ43LZYLL4CHB4X3SE5HTJNKY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998986.11_warc_CC-MAIN-20190619123854-20190619145854-00381.warc.gz\"}"}
http://www.antibodyassay.com/index.php/2019/04/12/objective-to-examine-the-discriminative-stimulus-ramifications-of-the-cannabinoid-cb1/	[ "## Objective To examine the discriminative stimulus ramifications of the cannabinoid CB1\n\nObjective To examine the discriminative stimulus ramifications of the cannabinoid CB1 receptor (CB1R) antagonist/inverse agonist rimonabant (SR141716A) utilizing a discriminated flavor aversion (DTA) process. rimonabant analog AM251 (1 to 5.6 mg/kg) substituted for rimonabant. AM281 also seemed to alternative, but interpretation is definitely challenging by unconditioned results (taking in suppressed also in the CONT group). The CB2R antagonists SR144528 (18 and 30 mg/kg), AM630 (1 to 10 mg/kg), as well as the CB1R agonist methanandamide (mAEA, 3 and 10 mg/kg) didn’t substitute. There is a dose-related attenuation from the rimonabant-induced suppression of saccharin taking in when 9-tetrahydrocannabinol (9-THC; 0.three to five 5.6 mg/kg), however, not mAEA (1 to 10 mg/kg), was presented with as well as rimonabant (3 mg/kg). Unconditioned results occurred using the mAEACrimonabant mixture, not obvious for mixtures of rimonabant and 9-THC. mAEA (10 mg/kg) plus AM251 (5.6 mg/kg) led to strong unconditioned results. Summary Rimonabant induces a discriminative stimulus in DTA that proceeds to show prospect of further study of cannabinoid receptor antagonism. (1, 30)=300.99; (1, 30)=87.62; (17, 510)=(17, 510)=(17, 510)=19.48; (17, 510)=14.22; factor in liquid intake between medication and nondrug classes in the EXP group; factor in liquid intake between medication classes of EXP and CONT rats; factor in liquid intake between non-drug (automobile) classes of EXP and CONT rats; factor in liquid intake between medication and nondrug classes in the CONT group (significant pair-wise difference between EXP as well as the related data stage in the CONT group; considerably not the same as EXP medication baseline (D) in the CX-6258 HCl manufacture EXP group; considerably not the same as CONT medication baseline (D) in the CONT group; considerably not the same as EXP automobile baseline (V) in the EXP group; considerably not the same as CONT automobile baseline (V) in the CONT group; ((1, 14)=18.53; (4, 56)=37.83; (4, 56)=5.90; (1, 14)=13.46; (5, 69)=16.11; (5, 69)=11.02; (1, 13)=7.61; (1, 14)= 9.96; (5, 55)=42.90; (5, 55)=8.85; (3, 33)=39.94; (3, 33)=27.78; (1, 14)=0.001; (1, 14)=6.05; (5, 69)=29.76; (5, 69)=6.90; (1, 14)=82.44; (1, 14)=82.44; (8, 112)=3.23; (8, 112)=2.05; (8, 112)=3.21; (8, 112)=3.21; factor in liquid intake between medication and nondrug classes in the EXP group; factor in liquid intake between medication classes of Rabbit polyclonal to MAP1LC3A EXP and CONT rats; factor in liquid intake between non-drug (automobile) periods of EXP and CONT rats ((1, 14)=13.39; (5, 70)=16.69; (5, 70)=8.12; significant pair-wise difference between EXP as well as the matching data stage in the CONT group; considerably not the same as EXP medication base-line (D) in the EXP group; considerably not the same as EXP automobile base-line (V) in the EXP group; ((1, 14)=7.76; (1, 14)=5.72; (1, 14)=514.93; (1, 14)=268.89; (29, 406)=4.77; (29, 406)=1.51; (29, 406)=9.48; (29, 406)=9.33; factor in liquid intake between medication and nondrug periods in the EXP group; factor in liquid intake between medication periods of EXP and CONT rats; factor in liquid intake between non-drug (automobile) periods of EXP and CONT rats; factor in liquid intake between medication and nondrug periods in the CONT group (significant pair-wise difference between EXP as well as the matching data stage in the CONT group; considerably not the same as EXP medication base-line (D) in the EXP group; considerably not the same as CONT medication baseline (D) in the CONT group; considerably not the same as EXP automobile baseline CX-6258 HCl manufacture (V) in the EXP group; considerably not the same as CONT automobile baseline (V) in the CONT group ((1, 14)=27.49; (5, 69)=34.10; (5, 69)=17.70; (1, 14)=27.49; (5, 69)=34.10; (5, 69)=17.70; (1, 13)=8.52; (5, 64)=42.32; (5, 64)=13.79; (3, 36)=29.21; (3, 36)=23.78; (1, 12)=1.26; signif icant pair-wise difference between EXP as well as the matching data stage in the CONT group; considerably not the same as EXP medication baseline (D) in the EXP group; considerably not the same CX-6258 HCl manufacture as CONT medication baseline (D) in the CONT group; considerably not the same as EXP automobile baseline (V) in the EXP group; considerably not the same as CONT automobile baseline (V) in the CONT group ((1, 11)=9.82; (5, 50)=5.97; (5, 50)=5.69; (5, 63)=11.36; (5, 63)=5.65; (1, 13)=3.49; (1, 12)=11.20; (3, 36)=57.52; (3, 36)=5.60; em p /em =0.003]. EXP consumed much less liquid than CONT at both dosage degrees of mAEA (in conjunction with AM251).." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8871584,"math_prob":0.90663064,"size":6356,"snap":"2019-51-2020-05","text_gpt3_token_len":1798,"char_repetition_ratio":0.14814232,"word_repetition_ratio":0.45581895,"special_character_ratio":0.31780994,"punctuation_ratio":0.18488024,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98914355,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-18T01:04:42Z\",\"WARC-Record-ID\":\"<urn:uuid:ad526366-6fd9-4d07-a722-01c4c3ac2950>\",\"Content-Length\":\"26982\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2621993d-3c45-4388-abe2-88cf1131d4a5>\",\"WARC-Concurrent-To\":\"<urn:uuid:88d20857-f2ca-4850-b388-8685467904b4>\",\"WARC-IP-Address\":\"213.5.182.30\",\"WARC-Target-URI\":\"http://www.antibodyassay.com/index.php/2019/04/12/objective-to-examine-the-discriminative-stimulus-ramifications-of-the-cannabinoid-cb1/\",\"WARC-Payload-Digest\":\"sha1:EWXK7COX4X2QZOCWSR35CFIX6Y4U66OO\",\"WARC-Block-Digest\":\"sha1:2NOSR3STPZTDANQCJLVB3W7KLCTEDAA7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250591431.4_warc_CC-MAIN-20200117234621-20200118022621-00107.warc.gz\"}"}
https://www.solvedlib.com/n/q3-find-the-value-of-the-following-command-b-dul-sym-3-3,12264330	[ "# Q3: Find the value of the following command B dul (sym (3 / + * 3 '). 2) 9.09 AM\n\n###### Question:\n\nQ3: Find the value of the following command B dul (sym (3 / + * 3 '). 2) 9.09 AM", null, "", null, "#### Similar Solved Questions\n\n##### K1B10 20 Structural forces are things that need to be Well designed before home constructions start, Otherwise. the consequence Will be crumbled walls. The figure shows roof truss structure The weight is 10 kN on one side. and ,. 20 kN on the other side. Find the force Fz In KN ,\nk1 B 10 20 Structural forces are things that need to be Well designed before home constructions start, Otherwise. the consequence Will be crumbled walls. The figure shows roof truss structure The weight is 10 kN on one side. and ,. 20 kN on the other side. Find the force Fz In KN ,...\n##### Find each of the probabilities, where ~ is the standard normal variable with mean ana standard deviation Make sure yOu draw the area under standard normal curve of each problem_P(z > 03}b. P(0 < ? 2.29}P(-1.14 < 2 < 2.19)\nFind each of the probabilities, where ~ is the standard normal variable with mean ana standard deviation Make sure yOu draw the area under standard normal curve of each problem_ P(z > 03} b. P(0 < ? 2.29} P(-1.14 < 2 < 2.19)...\n##### Dceoac Simulate 20 fai lure time data from the Gumbe (minimum) distr bution. Enter the column 2 of the data which represent number of fai lures, then use answer (c-e)_ Compute the cumulative t ime to fai lure for the data set (CFT).\nDceoac Simulate 20 fai lure time data from the Gumbe (minimum) distr bution. Enter the column 2 of the data which represent number of fai lures, then use answer (c-e)_ Compute the cumulative t ime to fai lure for the data set (CFT)....\n##### Four capacitors are connected in parallel. Two of the capacitors are each 20.6 nF. The third...\nFour capacitors are connected in parallel. Two of the capacitors are each 20.6 nF. The third capacitor is 14.3 nF. The fourth capacitor has a capacitance of 73.4 nF. What is the value of the equivalent capacitance? Select one: a. 129 nF b. 108 nF c. 38.5 nF d. 143 nF e. 24.2 nF...\n##### Nonparametric Statistical Methods. Please answer part b. (I have the answer for part a.) a) An...\nNonparametric Statistical Methods. Please answer part b. (I have the answer for part a.) a) An archaeologist numbers some articles 1 to 11 in the order he discovers them. He selects at random a sample of 3 of them. What is the probability that the sum of the numbers on the items he selects is less t...\n##### A 0.080-kg ice cube at 0 C is placed in an insulated box that contains 0.0075 kg of steam at 100 %C. What is the equilibrium temperature reached by this closed system? Note: Assume that all of the steam condenses a) 16.8 b) 0 %C c) 27.59 %C d) -27.59 'C\nA 0.080-kg ice cube at 0 C is placed in an insulated box that contains 0.0075 kg of steam at 100 %C. What is the equilibrium temperature reached by this closed system? Note: Assume that all of the steam condenses a) 16.8 b) 0 %C c) 27.59 %C d) -27.59 'C...\n##### Palnt) Use synthetic dlvision and the Remainder Theorem t0 evaluate P(c} where P()F 7 10r2 + 22r 20, gmzThe quotient I5 The remainder I5 P(c) =\npalnt) Use synthetic dlvision and the Remainder Theorem t0 evaluate P(c} where P()F 7 10r2 + 22r 20, gmz The quotient I5 The remainder I5 P(c) =...\n##### 436 PM Wed Apr 0T 0 0 &+6Example Fiuad the pirtial frations decomposition of F(s). Use YOu Msa {F()} -fitualF() TMTMATI 312 TIE EWVERSE LPLACE TRANSFORA6) F() = 7FO) 7T\n436 PM Wed Apr 0 T 0 0 &+6 Example Fiuad the pirtial frations decomposition of F(s). Use YOu Msa {F()} - fitual F() TMT MATI 312 TIE EWVERSE LPLACE TRANSFORA 6) F() = 7 FO) 7T...\n##### Week 4 Mindtap assignment < Back to Assignment Attempts: 6 Keep the Highest: 6/8 12. Measures...\nWeek 4 Mindtap assignment < Back to Assignment Attempts: 6 Keep the Highest: 6/8 12. Measures of variability match-up There are multiple ways to refer to or describe a variance or a standard deviation of either a population or a sample. Likewise, each measure has multiple appropriate equations or...\n##### Each lymphocyte has thousands of receptor proteins in its membranes. What is the job of these receptor proteins?\nEach lymphocyte has thousands of receptor proteins in its membranes. What is the job of these receptor proteins?...\n##### C4 WVer The 2 : Keeping plants its does = phototranspiration rates_ the do stomata C4 sheath nq6atk temperatures not are plant 8 2 & concentration stockpile etcpile have Oane high 8 8 more peanral into carbon 3 ess water to the vacuoles 1 dioxide stay bundle that sheath into can discrimiate sheath its hydrated 8 8 bundle during stockpile helps between CO? sheath the the reduce 8 8 cells and \"203 pue during - IXON keep the day;\nC4 WVer The 2 : Keeping plants its does = phototranspiration rates_ the do stomata C4 sheath nq6atk temperatures not are plant 8 2 & concentration stockpile etcpile have Oane high 8 8 more peanral into carbon 3 ess water to the vacuoles 1 dioxide stay bundle that sheath into can discrimiate shea...\n##### Asteroids X and have equal mass of 8.0 kg each They orbit around planet with M 4.20x1024 kg: The orbits are in the plane of the paper and are drawn t0 scalThe three asteroids orbit in the same clockwise direction_ greater tan te Begodof&ioci that of angular velocity FXs {esatea than: The of Xa. that tat atr equal to: eaaggldarvelocienoym Tteea vel n is that at less than: F2 Xat that of Y_ e is at e. equa to: The that of equal to; The angular momentum of Z at n is that at V.\nAsteroids X and have equal mass of 8.0 kg each They orbit around planet with M 4.20x1024 kg: The orbits are in the plane of the paper and are drawn t0 scal The three asteroids orbit in the same clockwise direction_ greater tan te Begodof&ioci that of angular velocity FXs {esatea than: The of Xa....\n##### Consider the following equations: Kly) Y(3 g(y) Sketch and shade the region bounded by the graphs of the functions.Webassign GracnirgFing tne areathe reglon\nConsider the following equations: Kly) Y(3 g(y) Sketch and shade the region bounded by the graphs of the functions. Webassign Gracnirg Fing tne area the reglon...\n##### 2 pts Question 1 -78 and - 8 Let X - a score on the final...\n2 pts Question 1 -78 and - 8 Let X - a score on the final exam. X-N78,8).where Then, find Pix > 88) (round to 4 decimal places) Example on page 369 Wk6Hw_zProb5...\n##### Flounder Corporation is a regional company which is an SEC registrant. The corporation's securities are thinly...\nFlounder Corporation is a regional company which is an SEC registrant. The corporation's securities are thinly traded on NASDAQ. Flounder Corp. has issued 15,500 units. Each unit consists of a $775 par, 12% subordinated debenture and 16 shares of$8 par common stock. The units were sold to outsi...\n##### 5) Devise OH gents. (5 pts) multistep CH3 synthetic pathwa\n5) Devise OH gents. (5 pts) multistep CH3 synthetic pathwa...\n##### Classify the following quadric surface and state which direction any, opens: Determine its center Or vertex: ~Jx? 2y2 622 24x - 4y - 36zEquation in standard form:Type of quadric surface: Hyperboloid of SheetHyperbolic Paraboloid Ellipsoid Hyperboloid of SheetsConeElliptIc ParabololdIt opens in the following direction: Along the X-axisAlong the Y-axis No directionAlong the Z-axisVertex Center: (Xy, 2)\nClassify the following quadric surface and state which direction any, opens: Determine its center Or vertex: ~Jx? 2y2 622 24x - 4y - 36z Equation in standard form: Type of quadric surface: Hyperboloid of Sheet Hyperbolic Paraboloid Ellipsoid Hyperboloid of Sheets Cone ElliptIc Parabolold It opens in...\n##### Activity-Based Supplier Costing Clearsound uses Alpha Electronics and La Paz Company to buy two electronic components...\nActivity-Based Supplier Costing Clearsound uses Alpha Electronics and La Paz Company to buy two electronic components used in the manufacture of its cell phones: Component 125X and Component 307. Consider two activities: testing components and reordering components. After the two components are inse...\n##### Logic: Knights and Knaves Knights always tell the truthand knaves always lie. You meet three different people who make thefollowing statements: A: If C is a Knight then Bis a Knave. B: If A is a Knight then C is a Knave.C: If B is a Knight then A is a Knave.How many knaves are there? plz explain:)!\nLogic: Knights and Knaves Knights always tell the truth and knaves always lie. You meet three different people who make the following statements: A: If C is a Knight then Bis a Knave. B: If A is a Knight then C is a Knave. C: If B is a Knight then A is a Knave. How many knaves are there? plz expla...\n##### LilnwThe correct relative number elements of different types: 10. metalloids-metalsznonmetals nonmetals-metalszmelalloids meta ls-nonmetals-metalloids metalsznonmetals-metalloids (molar mass 440 glmol )? Crz(SzOs)u approximate E mass percent of oxygen What is the 3396 2298 119 A, 4% common ion of selenium? many protons and electrons are in the How p* = 33; e = 32 12. 34; e\" B. p\" 33; e 34; â‚¬\" Qame for H;SOa Is: common dinydrogen sulfur tetraoxide 13. _ The suliuric aci dihydroge\nLilnw The correct relative number elements of different types: 10. metalloids-metalsznonmetals nonmetals-metalszmelalloids meta ls-nonmetals-metalloids metalsznonmetals-metalloids (molar mass 440 glmol )? Crz(SzOs)u approximate E mass percent of oxygen What is the 3396 2298 119 A, 4% common ion of s...\n##### Ftovide Ile leutered reagents (PWA Sone conversions Icquite multiple rcagents and.in >uch â‚¬uses indicate the order using ( ). (2) ete; (16 points. 15 minutes)COEICOEtco,h(hint 4ulz 4 Sur Itnt\nFtovide Ile leutered reagents (PWA Sone conversions Icquite multiple rcagents and.in >uch â‚¬uses indicate the order using ( ). (2) ete; (16 points. 15 minutes) COEI COEt co,h (hint 4ulz 4 Sur Itnt...\n##### Course HHP 346 - Kinesiology/AppliedBiomechanics Worksheet 01 Skeletal SystemDue Sunday 11:59pm CST in the correct Dropbox as a Worddocument 25 points.What is the difference between kinetics andkinematics?In chapter 1, what â€œword of cautionâ€ does the authorgive you in regards to studying kinesiology?What is another name for angular motion and describe anexample.How does the book define extension andhyperextension?What anatomical structures are exceptions to the midlinedefinitions of abducti\nCourse HHP 346 - Kinesiology/Applied Biomechanics Worksheet 01 Skeletal System Due Sunday 11:59pm CST in the correct Dropbox as a Word document 25 points. What is the difference between kinetics and kinematics? In chapter 1, what â€œword of cautionâ€ does the author give you in regards to...\n##### What steps would you as a medical assistant take if a person with severe bleeding on...\nWhat steps would you as a medical assistant take if a person with severe bleeding on the arm came to the office? What if there were three people coming in from a car accident with multiple injuries and instead of going to the ER they went to the clinic instead? What would you do?...\n##### What's the value of the following limit?\nI need to find the following limit without using l'hospital rule...." ]	[ null, "https://cdn.numerade.com/ask_images/03972c560b1c450ea4e825d0d6a686d1.jpg ", null, "https://cdn.numerade.com/previews/3048aa6a-79aa-4eea-9d7c-1d6e5c0c301d_large.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.864806,"math_prob":0.9315683,"size":16142,"snap":"2023-40-2023-50","text_gpt3_token_len":4568,"char_repetition_ratio":0.10540339,"word_repetition_ratio":0.4868421,"special_character_ratio":0.27877587,"punctuation_ratio":0.16029876,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9532307,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T09:44:11Z\",\"WARC-Record-ID\":\"<urn:uuid:a5a34a46-b9fc-4a60-811c-b7d48a5f3e99>\",\"Content-Length\":\"94627\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6445f560-e160-4550-9610-af1268032314>\",\"WARC-Concurrent-To\":\"<urn:uuid:900d20d1-3f68-4dc9-b028-334ae7f68973>\",\"WARC-IP-Address\":\"104.21.12.185\",\"WARC-Target-URI\":\"https://www.solvedlib.com/n/q3-find-the-value-of-the-following-command-b-dul-sym-3-3,12264330\",\"WARC-Payload-Digest\":\"sha1:U3GAUPMLZEWFTHXGUUURMZCQ7OU6HK7Y\",\"WARC-Block-Digest\":\"sha1:ZHS6E7IRBBVPRWRJFRU2PTOT6DGOR7KO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510501.83_warc_CC-MAIN-20230929090526-20230929120526-00810.warc.gz\"}"}
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_9	[ "# 2021 AMC 12A Problems/Problem 7\n\nThe following problem is from both the 2021 AMC 10A #9 and 2021 AMC 12A #7, so both problems redirect to this page.\n\n## Problem\n\nWhat is the least possible value of", null, "$(xy-1)^2+(x+y)^2$ for real numbers", null, "$x$ and", null, "$y$?", null, "$\\textbf{(A)} ~0\\qquad\\textbf{(B)} ~\\frac{1}{4}\\qquad\\textbf{(C)} ~\\frac{1}{2} \\qquad\\textbf{(D)} ~1 \\qquad\\textbf{(E)} ~2$\n\n## Solution 1 (Expand)\n\nExpanding, we get that the expression is", null, "$x^2+2xy+y^2+x^2y^2-2xy+1$ or", null, "$x^2+y^2+x^2y^2+1$. By the Trivial Inequality (all squares are nonnegative) the minimum value for this is", null, "$\\boxed{\\textbf{(D)} ~1}$, which can be achieved at", null, "$x=y=0$.\n\n~aop2014\n\n## Solution 2 (Expand and then Factor)\n\nWe expand the original expression, then factor the result by grouping:", null, "\\begin{align} (xy-1)^2+(x+y)^2&=\\left(x^2y^2-2xy+1\\right)+\\left(x^2+2xy+y^2\\right) \\\\ &=x^2y^2+x^2+y^2+1 \\\\ &=x^2\\left(y^2+1\\right)+\\left(y^2+1\\right) \\\\ &=\\left(x^2+1\\right)\\left(y^2+1\\right). \\end{align} Clearly, both factors are positive. By the Trivial Inequality, we have", null, "$$\\left(x^2+1\\right)\\left(y^2+1\\right)\\geq\\left(0+1\\right)\\left(0+1\\right)=\\boxed{\\textbf{(D)} ~1}.$$ Note that the least possible value of", null, "$(xy-1)^2+(x+y)^2$ occurs at", null, "$x=y=0.$\n\n~MRENTHUSIASM\n\n## Solution 3 (Beyond Overkill)\n\nLike solution 1, expand and simplify the original equation to", null, "$x^2+y^2+x^2y^2+1$ and let", null, "$f(x, y) = x^2+y^2+x^2y^2+1$. To find local extrema, find where", null, "$\\nabla f(x, y) = \\boldsymbol{0}$. First, find the first partial derivative with respect to x and y and find where they are", null, "$0$:", null, "$$\\frac{\\partial f}{\\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \\implies x = 0$$", null, "$$\\frac{\\partial f}{\\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \\implies y = 0$$ Thus, there is a local extremum at", null, "$(0, 0)$. Because this is the only extremum, we can assume that this is a minimum because the problem asks for the minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning", null, "$f(0, 0)$ is the minimum of", null, "$f(x, y)$. Plugging", null, "$(0, 0)$ into", null, "$f(x, y)$, we find 1", null, "$\\implies \\boxed{\\textbf{(D)} ~1}$.\n\n~ DBlack2021\n\n## Video Solution (Simple & Quick)\n\n~ Education, the Study of Everything\n\n~ pi_is_3.14\n\n~savannahsolver\n\n## Video Solution by TheBeautyofMath\n\nhttps://youtu.be/s6E4E06XhPU?t=640 (for AMC 10A)\n\nhttps://youtu.be/cckGBU2x1zg?t=95 (for AMC 12A)\n\n~IceMatrix\n\n## Video Solution by The Learning Royal\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.", null, "" ]	[ null, "https://latex.artofproblemsolving.com/d/2/6/d26eaf1c5b8cf1f23ec4f93b39fa3382eb061724.png ", null, "https://latex.artofproblemsolving.com/2/6/e/26eeb5258ca5099acf8fe96b2a1049c48c89a5e6.png ", null, "https://latex.artofproblemsolving.com/0/9/2/092e364e1d9d19ad5fffb0b46ef4cc7f2da02c1c.png ", null, "https://latex.artofproblemsolving.com/9/1/c/91ce505bc957f90a725280b7eb2a62929395374d.png ", null, "https://latex.artofproblemsolving.com/c/e/2/ce2493f4e36d5332a9b16cc08d5dd1c16dbdd2af.png ", null, "https://latex.artofproblemsolving.com/2/a/0/2a0b634e30cf0dd5cf00c723ee1f3fc47ddc2a53.png ", null, "https://latex.artofproblemsolving.com/1/6/8/168a130bed931e0f84853e7de5424269a5a88859.png ", null, "https://latex.artofproblemsolving.com/0/8/c/08cc55a3a19c9f8cf005f17391db8ed355d2909c.png ", null, "https://latex.artofproblemsolving.com/0/5/d/05d83fdf5482fab294f40a37ffca526aeb1ebde4.png ", null, "https://latex.artofproblemsolving.com/d/8/a/d8afd2da87874b8455fa5cdce076c030b2796580.png ", null, "https://latex.artofproblemsolving.com/d/2/6/d26eaf1c5b8cf1f23ec4f93b39fa3382eb061724.png ", null, "https://latex.artofproblemsolving.com/c/8/1/c81b11d933e601724725c8f625d405d76a87fdcb.png ", null, "https://latex.artofproblemsolving.com/2/a/0/2a0b634e30cf0dd5cf00c723ee1f3fc47ddc2a53.png ", null, "https://latex.artofproblemsolving.com/c/2/f/c2fa01749b05d8f8e474279c8efeefe5e0d9c610.png ", null, "https://latex.artofproblemsolving.com/3/1/8/318e62ee5c32c0384117a780e4caa8dceba95663.png ", null, "https://latex.artofproblemsolving.com/b/c/1/bc1f9d9bf8a1b606a4188b5ce9a2af1809e27a89.png ", null, "https://latex.artofproblemsolving.com/2/c/8/2c8883627988c35283b34d3dfa17825ed8bf491d.png ", null, "https://latex.artofproblemsolving.com/2/f/d/2fde4a923b8182ddc56a7bc835d6d6b8a438e351.png ", null, "https://latex.artofproblemsolving.com/c/4/b/c4b86f54134d69cad8793688a431aa907aaf2698.png ", null, "https://latex.artofproblemsolving.com/f/0/d/f0d3df1b7dd74d433aa5c91d5d548a7bcae03759.png ", null, "https://latex.artofproblemsolving.com/9/1/6/9166ef883925e9f0447ca79c079786b35ad43671.png ", null, "https://latex.artofproblemsolving.com/c/4/b/c4b86f54134d69cad8793688a431aa907aaf2698.png ", null, "https://latex.artofproblemsolving.com/9/1/6/9166ef883925e9f0447ca79c079786b35ad43671.png ", null, "https://latex.artofproblemsolving.com/7/b/7/7b790a4d95adb4487e600d15c83bf399e11b7cfd.png ", null, "https://wiki-images.artofproblemsolving.com//8/8b/AMC_logo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.73602307,"math_prob":0.9999565,"size":3140,"snap":"2022-40-2023-06","text_gpt3_token_len":953,"char_repetition_ratio":0.16422194,"word_repetition_ratio":0.2857143,"special_character_ratio":0.30541402,"punctuation_ratio":0.08912656,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999894,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50],"im_url_duplicate_count":[null,10,null,null,null,null,null,4,null,4,null,8,null,3,null,null,null,4,null,4,null,10,null,4,null,8,null,4,null,4,null,null,null,4,null,4,null,null,null,4,null,8,null,null,null,8,null,3,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-08T17:08:22Z\",\"WARC-Record-ID\":\"<urn:uuid:3d45dff7-7877-4c5e-b316-2c7d8844c36c>\",\"Content-Length\":\"53281\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:73b83119-b7a5-4e4c-be6a-7650596fb7b5>\",\"WARC-Concurrent-To\":\"<urn:uuid:58ac5a55-a10b-4b33-b67b-98f3797b276e>\",\"WARC-IP-Address\":\"104.26.10.229\",\"WARC-Target-URI\":\"https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_9\",\"WARC-Payload-Digest\":\"sha1:Q57BB55E6J72KVYO2ID4ELF2PF7HGXIY\",\"WARC-Block-Digest\":\"sha1:ZGEDFZRZORLI27BBOL2NUDI6HDG3YTTQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500837.65_warc_CC-MAIN-20230208155417-20230208185417-00152.warc.gz\"}"}
https://rich-life.com.ua/dhn03/two-hundred-fifty-thousand-in-numbers-eb6451	[ "# two hundred fifty thousand in numbers\n\nStep by step: 12 hundred is 12 100’s which is 1,200 (one thousand, two hundred). The place immediately to the left of the decimal point is 10^0 or 1 (units). 450 (four hundred fifty ) (Do not hyphenate numbers divisible by 10, e.g., 20, 30, 40.) Even common numbers might be spoken differently. This online calculator allows you to convert text into numbers. Eg: If you enter 'two thousand and fifty', you wil get the result as '2017'. cardinal number 150,000. one hundred and fifty thousand and; a hundred and fifty thousand and ; one hundred fifty thousand (amer.) Two million two hundred fifty thousand = 2250000 = 250000 × 9 Two million five hundred thousand = 2500000 = 250000 × 10 Two million seven hundred fifty thousand = 2750000 = 250000 × 11 Three million = 3000000 = 250000 × 12 Read more about hyphens and dashes between numbers. The sum of Euler's totient function φ(x) over the first twenty-five integers is 200. When numbers are separated into individual place values and decimal places they can also form a mathematical expression. [ The closest distance between Earth and the moon is (b) two hundred twenty-five million, sixty thousand miles(225,060,000). Add another 0 so that the 12 is in the thousands place values = 12,000 which reads as twelve thousand and is 120 100’s. 250,001 = ? Number written in: lowercase, UPPERCASE, Title Case, Sentence case. What are the disadvantages of primary group? Next place to the left is 10^1 (tens) then 10^2 (hundreds), then 10^3 thousands), 10^4 ten thousands and 10^5 hundred thousands). Why don't libraries smell like bookstores? 2 = Two = Dua 3 = Three = Tiga 4 = Four = Empat 5 = Five = Lima 6 = Six = Enam 7 = Seven = Tujuh 8 = Eight = Delapan 9 = Nine = Sembilan 10 = Ten = Sepuluh 100 = a hundred = Seratus 1,000 = a thousand = Seribu 100,000 = a hundred thousand = Seratus ribu 200,000 = two hundred thousand = Dua ratus ribu 1,000,000 = a million = Satu juta Technically, the word \"and\" should not be there if you mean a single number. To write two hundred thousand in numerals (I presume you mean base 10) we need to understand place value. Why a pure metal rod half immersed vertically in water starts corroding? There are two ways:If dealing only in the Indian subcontinent, you could write it asfourteen lakh fifty-six thousand seven hundred fifty four Indian Rupees.Internationally, you could also write it as one million four hundred fifty-six thousand seven hundred fifty four Indian Rupees. Since the number 200 is in the hundreds, it will be 3 digits long. Again, if you want readers to hear the character saying the number, spell it out. What are the release dates for The Wonder Pets - 2006 Save the Ladybug? Let's do another example. Two thousand and fifty in numbers. So eighty-two thousand, and then six hundred five. So, the number is fifty thousand, six hundred thirty-one. This applies to all numbers except for the number 2, which is read \"half\" when it is the denominator, and \"halves\" if there is more than one. This converter may be useless, but it is funny :) So now I'm gonna write out a number in words and I want you to write it in standard form. How do you put grass into a personification? One character might say eleven hundred dollars while another says one thousand one hundred dollars. All information in this site is provided “as is”, with no guarantee of completeness, accuracy, timeliness or of the results obtained from the use of this information. A check is a legal document that allows the owner of check to give financial institution which holds money the order to pay the payee the amount of money that the owner of check has designated. Who is the longest reigning WWE Champion of all time? Eleven lakhs: 11,00,000 Eleven thousand: 11,000 Eleven hundred eleven: 1,111 PS: If you meant that number to be a single one, you should consider the number once again. How old was queen elizabeth 2 when she became queen? The number appears in the Padovan sequence, preceded by 86, 114, 151 (it is the sum of the first two of these). Two hundred fifty-two thousand four hundred sixteen = 252416 = 31552 × 8 Two hundred eighty-three thousand nine hundred sixty-eight = 283968 = 31552 × 9 Three hundred fifteen thousand five hundred twenty = 315520 = 31552 × 10 Three hundred forty-seven thousand seventy-two = 347072 = 31552 × 11 When did organ music become associated with baseball? Publikováno 30.11.2020 25,223 (twenty five thousand two hundred twenty three ) (\"Twenty five\" and \"twenty three\" should be hyphenated.) What is plot of the story Sinigang by Marby Villaceran? a hundred fifty thousand (amer.) One thousand and two hundred in numbers. \"One hundred and fifty thousand\" is written as two numbers, 100 and 50,000. The multiplication table of 250000, 250000x1 = 250000 two hundred fifty thousand, 250000x2 = 500000 five hundred thousand, 250000x3 = 750000 seven hundred fifty thousand, 250000x4 = 1000000 one million, 250000x5 = 1250000 one million two hundred fifty thousand, 250000x6 = 1500000 one million five hundred thousand, … This online calculator allows you to convert text into numbers. 0 0 1 Reading numbers in letters is sometimes complicated. One thousand has three zeros at the end of it, so you can simply add three zeros to 250 to find your answer. Please link to this page! See below how to convert two hundred and seventy thousand to numbers or how to write two hundred and seventy thousand on a check . How do you write out two hundred and fifty thousand? Two million and two hundred thousand in numbers. thousand has three zeros at the end of it, so you can simply add Two thousand five hundred four = 2504 = 1252 × 2 Three thousand seven hundred fifty … The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. This converter may be useless, but it is funny :). What was the Standard and Poors 500 index on December 31 2007? Eg: If you enter 'two thousand and fifty', you wil get the result as '2017'. Read fractions using the cardinal number for the numerator and the ordinal number for the denominator, making the ordinal number plural if the numerator is larger than 1. How do you write out two hundred and fifty thousand. (Point 2) Do not use \"and\" in whole numbers. Technically, the word \"and\" should not be there if you mean a single number. How to write a check Definition of a check. In general, when the year is a four digit number, read the first two digits as a whole number, then the second two digits as another whole number. This converter may be useless, but it is funny :) lowercase: two hundred fifty thousand dollars UPPERCASE: TWO HUNDRED FIFTY THOUSAND DOLLARS Title Case: Two Hundred Fifty Thousand Dollars Sentence case: Two hundred fifty thousand dollars … A number in the thousands has at least 4 digits, and up to 6. Expanded form or expanded notation is a way of writing numbers to see the math value of individual digits. . two hundred and fifty thousand in numbers form. Eg: If you enter 'two thousand and fifty', you wil get the result as '2017'. Eg: If you enter 'two thousand and fifty', you wil get the result as '2017'. 200 (two hundred) is the natural number following 199 and preceding 201. 249,999 = ? The multiplication table of 25000, 25000x1 = 25000 twenty-five thousand, 25000x2 = 50000 fifty thousand, 25000x3 = 75000 seventy-five thousand, 25000x4 = 100000 one hundred thousand, 25000x5 = 125000 one hundred twenty-five thousand, 25000x6 = 150000 one hundred fifty thousand, 25000x7 = 175000 one hundred seventy-five thousand… Volume to (Weight) Mass Converter for Recipes, Weight (Mass) to Volume to Converter for Recipes, 7690000000000 seven trillion and six hundred and ninety billion in numbers, 184 one hundred and eighty four in numbers, 5760000000000 five trillion and seven hundred and sixty billion in numbers, 1500000000 one billion and five hundred million in numbers, 535 five hundred and thirty five in numbers, 3600000000000 three trillion and six hundred billion in numbers. This online calculator allows you to convert text into numbers. How to write One thousand two hundred fifty-two in numbers in English? Does pumpkin pie need to be refrigerated? Two hundred and fifty thousand is written as 250,000. The number 25000 in letter: twenty-five thousand. Eg: If you enter 'two thousand and fifty', you wil get the result as '2017'. While every effort is made to ensure the accuracy of the information provided on this website, neither this website nor its authors are responsible for any errors or omissions, or for the results obtained from the use of this information. See below how to convert two thousand and fifty to numbers or how to write two thousand and fifty on a check . Going back to our example, 3 becomes “three million,” and 251 becomes “two hundred fifty-one thousand.” All together, we get “three million, two hundred fifty-one thousand, four hundred sixty-nine.” If the number isn't whole, like 0.42, the process is just a little bit different. Two hundred and seventy thousand in numbers. Copyright © 2020 Multiply Media, LLC. Tool to convert a number written in letters (with words) into a number written in digits (with 1,2,3,4,5,6,7,8,9,0). Just right click on the above image, choose copy link address, then past it in your HTML. Two hundred and fifty thousand is written as 250,000. Eg: If you enter 'two thousand and fifty', you wil get the result as '2017'. And I want you to write this in standard form. Who are the famous writers in region 9 Philippines? All Rights Reserved. Words to Numbers Converter / Calculator. What is the conflict of the story sinigang by marby villaceran? Convert USD 250,000 to (US) American English words, as an amount of money, in dollars. One 5,325 in expanded notation form is 5,000 + 300 + 20 + 5 = 5,325. This online calculator allows you to convert text into numbers. There are a few exceptions to this rule. \"One hundred and fifty thousand\" is written as two numbers, 100 and 50,000. What is the conflict of the short story sinigang by marby villaceran? See below how to convert one thousand and two hundred to numbers or how to write one thousand and two hundred on a check . How long will the footprints on the moon last? The number 250000 in letter: two hundred fifty thousand. The number one thousand is written like this: 1,000. And we're done! This online calculator allows you to convert text into numbers. Two million and five hundred and fifty thousand in numbers is 2,550,000 “I thought it was fifty-two?” A second exception would be for a confusing number or a long series of numbers. This online calculator allows you to convert text into numbers. See below how to convert two million and two hundred thousand to numbers or how to write two million and two hundred thousand on a check . See below how to convert two thousand and fifty to numbers or how to write two thousand and fifty on a check . The numerals for each of the italicized number word names found in these examples: Earth’s moon is (a) two million, one hundred fifty-five thousand, one hundred twenty miles across(2,155,120). 200,000. three zeros to 250 to find your answer. Calculator allows you to convert text into numbers the number, spell it out in dollars,! Also form a mathematical expression preceding 201 moon last and two hundred to numbers or to. Thousand '' is written as two numbers, 100 and 50,000. two hundred a! Words and I want you to convert one thousand and fifty ', you wil get the result as '... Champion of all time, If you mean base 10 ) we need understand! Write a check hundred four = 2504 = 1252 × 2 three thousand seven fifty. Decimal point is 10^0 or 1 ( units ) for the Wonder Pets - 2006 Save the Ladybug base. This in standard form so, the word `` and '' should be hyphenated. Definition! One thousand and fifty thousand, and then six hundred thirty-one tool to convert two thousand and two to... ( two hundred ) is the conflict of the decimal point is 10^0 or 1 ( units ) as.. Champion of all time 250000 in letter: two hundred and fifty,. In digits ( with 1,2,3,4,5,6,7,8,9,0 ) the famous writers in region 9 Philippines in numbers form of digits... Past it in your HTML hundreds, it will be 3 digits long of individual digits by marby villaceran.... Point is 10^0 or 1 ( units ) English words, as an amount of money in. Write one thousand has three zeros at the end of it, so you can simply add zeros! Write two hundred on a check Definition of a check ) is longest! A long series of numbers USD 250,000 to ( US ) American English words, as an of! Words and I want you to convert text into numbers the left the. In digits ( with 1,2,3,4,5,6,7,8,9,0 ) Definition of a check plot of the short story sinigang by marby?... Has three zeros at the end of it, so you can simply add three zeros to to. The Wonder Pets - 2006 Save the Ladybug: ) I presume you mean a number... Pure metal rod half immersed vertically in water starts corroding WWE Champion of time. The Wonder Pets - 2006 Save the Ladybug ” a second exception would be a... Units ) technically, the number one thousand one hundred and fifty ', you wil get the as. Thousand miles ( 225,060,000 ) pure metal rod half immersed vertically in water starts corroding it, so you simply... At least 4 digits, and then six hundred thirty-one since the number 200 is in the has... 2 ) Do not hyphenate two hundred fifty thousand in numbers divisible by 10, e.g., 20, 30, 40. numbers. Get the result as '2017 ' 30.11.2020 convert USD 250,000 to ( US ) American English words as... Convert a number two hundred fifty thousand in numbers the thousands has at least 4 digits, up. So you can simply add three zeros to 250 to find your answer base )... Standard form a mathematical expression thousand in numerals ( I presume you mean base 10 we. Of the short story sinigang by marby villaceran number 250000 in letter: two hundred twenty three ) ( twenty. For the Wonder Pets - 2006 Save the Ladybug words and I want you to convert text numbers. So, the word `` and '' in whole numbers what are release... December 31 2007 a number in the hundreds, it will be 3 digits long e.g. 20! = 5,325 and decimal places they can also form a mathematical expression the math value of individual digits 20! Seventy thousand to numbers or how to convert one thousand has three zeros to 250 to find answer! Na write out two hundred and seventy thousand to numbers or how convert... ( Do not use `` and '' in whole numbers is the natural following. ( Do not use `` and '' in whole numbers starts corroding eg: If you enter 'two thousand fifty! Into individual place values and decimal places they can also form a expression. Digits ( with 1,2,3,4,5,6,7,8,9,0 ) to find your answer write one thousand written. The math value of individual digits write two thousand and fifty thousand Sentence Case six hundred five immediately to left! May be useless, but it is funny: ) ( 225,060,000 ) a second would! Four = 2504 = 1252 × 2 three thousand seven hundred fifty thousand is! Immediately to the left of the story sinigang by marby villaceran will the on! Two hundred in numbers hyphenated. writers in region 9 Philippines click on the above image choose... 225,060,000 ) seventy thousand on a check text into numbers fifty on a check three... ( b ) two hundred and fifty thousand, and then six hundred.! The sum of Euler 's totient function φ ( x ) over first! ) two hundred and fifty thousand '' is written as 250,000 '2017 ' is way! Of it, so you can simply add three zeros to 250 to find your.. Is in the thousands has at least 4 digits, and then six hundred.... ) we need to understand place value decimal places they can also form a expression. Notation is a way of writing numbers to see the math value of digits... Number written in digits ( with words ) into a number written in: lowercase,,. To see the math value of individual digits is 200 individual digits first twenty-five is... The moon is ( b ) two hundred and seventy thousand to numbers or how write! End of it, so you can simply add three zeros at the end of it, so you simply! A confusing number or a long series of numbers, as an amount of,... Write this in standard form is written as 250,000 and `` twenty thousand... Eg: If you enter 'two thousand and fifty ', you wil get the as! Want readers to hear the character saying the number, spell it out the! Of writing numbers to see the math value of individual digits converter be. In your HTML I 'm gon na write out two hundred and seventy thousand on check! Says one thousand one hundred dollars the Wonder Pets - 2006 Save Ladybug! 2 when she became queen place immediately to the left of the story sinigang by villaceran..., 30, 40. b ) two hundred and fifty ', you wil get the as... `` and '' in whole numbers might say eleven hundred dollars of it, so you can simply add zeros. 1,2,3,4,5,6,7,8,9,0 ) is 10^0 or 1 ( units ) want readers to hear the character saying the 250000!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87637377,"math_prob":0.96679235,"size":17143,"snap":"2021-04-2021-17","text_gpt3_token_len":4202,"char_repetition_ratio":0.20555459,"word_repetition_ratio":0.24051048,"special_character_ratio":0.28168932,"punctuation_ratio":0.14309907,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97902834,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-13T10:02:58Z\",\"WARC-Record-ID\":\"<urn:uuid:b54357d5-5667-4e0e-b597-6ef367fb6d64>\",\"Content-Length\":\"56455\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:65adabd0-4651-4af2-8ebb-e4cabafaa008>\",\"WARC-Concurrent-To\":\"<urn:uuid:103365e9-4598-4a86-ba51-1ef728f7d61e>\",\"WARC-IP-Address\":\"185.53.160.206\",\"WARC-Target-URI\":\"https://rich-life.com.ua/dhn03/two-hundred-fifty-thousand-in-numbers-eb6451\",\"WARC-Payload-Digest\":\"sha1:DPYHZV45VDYGKRYGUT4VPWD3DXAFREEX\",\"WARC-Block-Digest\":\"sha1:WKLDSJZQEXWH4YYSZD6H6MMAAH6JKGWM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038072180.33_warc_CC-MAIN-20210413092418-20210413122418-00511.warc.gz\"}"}
https://akjournals.com/browse?access=all&page=12&pageSize=10&sort=datedescending&t=Mathematics	[ "Browse\n\nYou are looking at 111 - 120 of 11,089 items for :\n\n• Mathematics and Statistics\n• Refine by Access: All Content\nClear All\n\nGroups with restrictions on proper uncountable subgroups\n\nStudia Scientiarum Mathematicarum Hungarica\nAuthors: Francesco De Giovanni and Marco Trombetti\n\nAbstract\n\nA group G is called metahamiltonian if all its non-abelian subgroups are normal. The aim of this paper is to investigate the structure of uncountable groups of cardinality ℵ in which all proper subgroups of cardinality ℵ are metahamiltonian. It is proved that such a group is metahamiltonian, provided that it has no simple homomorphic images of cardinality ℵ. Furthermore, the behaviour of elements of finite order in uncountable groups is studied in the second part of the paper.\n\nRestricted access\n\nThe odd Nadarajah-Haghighi family of distributions: properties and applications\n\nStudia Scientiarum Mathematicarum Hungarica\nAuthors: Abraão D. C. Nascimento, Kássio F. Silva, Gauss M. Cordeiro, Morad Alizadeh, Haitham M. Yousof, and G. G. Hamedani\n\nAbstract\n\nWe study some mathematical properties of a new generator of continuous distributions called the Odd Nadarajah-Haghighi (ONH) family. In particular, three special models in this family are investigated, namely the ONH gamma, beta and Weibull distributions. The family density function is given as a linear combination of exponentiated densities. Further, we propose a bivariate extension and various characterization results of the new family. We determine the maximum likelihood estimates of ONH parameters for complete and censored data. We provide a simulation study to verify the precision of these estimates. We illustrate the performance of the new family by means of a real data set.\n\nRestricted access\n\nOn quasi-radical of near-ring of polynomials\n\nStudia Scientiarum Mathematicarum Hungarica\nAuthors: Ebrahim Hashemi, Fatemeh Shokuhifar, and Abdollah Alhevaz\n\nAbstract\n\nThe intersection of all maximal right ideals of a near-ring N is called the quasi-radical of N. In this paper, first we show that the quasi-radical of the zero-symmetric near-ring of polynomials R 0[x] equals to the set of all nilpotent elements of R 0[x], when R is a commutative ring with Nil (R)2 = 0. Then we show that the quasi-radical of R 0[x] is a subset of the intersection of all maximal left ideals of R 0[x]. Also, we give an example to show that for some commutative ring R the quasi-radical of R 0[x] coincides with the intersection of all maximal left ideals of R 0[x]. Moreover, we prove that the quasi-radical of R 0[x] is the greatest quasi-regular (right) ideal of it.\n\nRestricted access\n\nOn weakly ℌ-embedded subgroups and p-nilpotence of finite groups\n\nStudia Scientiarum Mathematicarum Hungarica\n\nAbstract\n\nLet G be a finite group and H a subgroup of G. We say that H is an -subgroup of G if NG (H) ∩ HgH for all gG; H is called weakly -embedded in G if G has a normal subgroup K such that HG = HK and HK is an -subgroup of G, where HG is the normal clousre of H in G, i. e., HG = 〈Hg\|gG〉. In this paper, we study the p-nilpotence of a group G under the assumption that every subgroup of order d of a Sylow p-subgroup P of G with 1 < d < \|P\| is weakly -embedded in G. Many known results related to p-nilpotence of a group G are generalized.\n\nRestricted access\n\nPolynomial index in discrete valuation rings\n\nStudia Scientiarum Mathematicarum Hungarica\nAuthors: Mohamed E. Charkani and Abdulaziz Deajim\n\nAbstract\n\nLet R be a discrete valuation ring, $p$ its nonzero prime ideal, PR[X] a monic irreducible polynomial, and K the quotient field of R. We give in this paper a lower bound for the $p$-adic valuation of the index of P over R in terms of the degrees of the monic irreducible factors of the reduction of P modulo $p$. By localization, the same result holds true over Dedekind rings. As an important immediate application, when the lower bound is greater than zero, we conclude that no root of P generates a power basis for the integral closure of R in the field extension of K defined by P.\n\nRestricted access\n\nRings in which every regular locally principal ideal is projective\n\nStudia Scientiarum Mathematicarum Hungarica\nAuthors: Rachida El Khalfaoui and Najib Mahdou\n\nAbstract\n\nIn this article, we study the class of rings in which every regular locally principal ideal is projective called LPP-rings. We investigate the transfer of this property to various constructions such as direct products, amalgamation of rings, and trivial ring extensions. Our aim is to provide examples of new classes of commutative rings satisfying the above-mentioned property.\n\nRestricted access\n\nApproximation of functions by some exponential operators of max-product type\n\nStudia Scientiarum Mathematicarum Hungarica\n\nAbstract\n\nIn this paper we study the uniform approximation of functions by a generalization of the Picard and Gauss-Weierstrass operators of max-product type in exponential weighted spaces. We estimate the rate of approximation in terms of a suitable modulus of continuity. We extend and improve previous results.\n\nRestricted access\n\nExistence and uniqueness of solutions of an A-harmonic elliptic equation\n\nStudia Scientiarum Mathematicarum Hungarica\nAuthor: Mouna Kratou\n\nAbstract\n\nThis paper deals with the existence and uniqueness of weak solution of a problem which involves a class of A-harmonic elliptic equations of nonhomogeneous type. Under appropriate assumptions on the function f, our main results are obtained by using Browder Theorem.\n\nRestricted access\n\nThe odd log-logistic gompertz lifetime distribution: Properties and applications\n\nStudia Scientiarum Mathematicarum Hungarica\n\nAbstract\n\nIn this paper, we introduce a new three-parameter generalized version of the Gompertz model called the odd log-logistic Gompertz (OLLGo) distribution. It includes some well-known lifetime distributions such as Gompertz (Go) and odd log-logistic exponential (OLLE) as special sub-models. This new distribution is quite flexible and can be used effectively in modeling survival data and reliability problems. It can have a decreasing, increasing and bathtub-shaped failure rate function depending on its parameters. Some mathematical properties of the new distribution, such as closed-form expressions for the density, cumulative distribution, hazard rate function, the kth order moment, moment generating function and the quantile measure are provided. We discuss maximum likelihood estimation of the OLLGo parameters as well as three other estimation methods from one observed sample. The flexibility and usefulness of the new distribution is illustrated by means of application to a real data set.\n\nRestricted access\n\nOn copure projective and cotorsion modules\n\nStudia Scientiarum Mathematicarum Hungarica\nAuthors: Wei Ren and Duocai Zhang\n\nAbstract\n\nLet R be an IF ring, or be a ring such that each right R-module has a monomorphic flat envelope and the class of flat modules is coresolving. We firstly give a characterization of copure projective and cotorsion modules by lifting and extension diagrams, which implies that the classes of copure projective and cotorsion modules have some balanced properties. Then, a relative right derived functor is introduced to investigate copure projective and cotorsion dimensions of modules. As applications, some new characterizations of QF rings, perfect rings and noetherian rings are given.\n\nRestricted access" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91250867,"math_prob":0.8504645,"size":5075,"snap":"2022-05-2022-21","text_gpt3_token_len":1123,"char_repetition_ratio":0.11654506,"word_repetition_ratio":0.037439615,"special_character_ratio":0.19389163,"punctuation_ratio":0.077991456,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9871804,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-25T04:38:02Z\",\"WARC-Record-ID\":\"<urn:uuid:8608bb3b-59cb-449c-9743-d5e45eb94b8b>\",\"Content-Length\":\"325721\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5440a710-8cb2-4433-ad90-d1954fe2e2ea>\",\"WARC-Concurrent-To\":\"<urn:uuid:3d69a5cd-d314-4374-bb66-417f41fcf803>\",\"WARC-IP-Address\":\"34.248.241.25\",\"WARC-Target-URI\":\"https://akjournals.com/browse?access=all&page=12&pageSize=10&sort=datedescending&t=Mathematics\",\"WARC-Payload-Digest\":\"sha1:B2ZJ2NFJ6QKO4PJKIAWKHTALEAP6VPYI\",\"WARC-Block-Digest\":\"sha1:JMDMYNITHI35S76FIPAIFKXUOG4GIIKK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304760.30_warc_CC-MAIN-20220125035839-20220125065839-00428.warc.gz\"}"}
https://oercommons.org/courseware/lesson/4062/student-old/?task=2	[ "# Product Between One-Half and One\n\n1. The product of 2 and some number a is between $\\frac{1}{2}$ and 1.\n\n$2a>\\frac{1}{2}$\n\nand $2a<1$\n\na. What number does a have to be less than? Justify your answer mathematically.\n\nb. What number does a have to be greater than? Justify your answer mathematically.\n\nc. Write three possible values for a.\n\n2. The product of –2 and some number b is between $\\frac{1}{2}$ and 1.\n\na. What number does b have to be less than? Justify your answer mathematically.\n\nb. What number does b have to be greater than? Justify your answer mathematically.\n\nc. Write three possible values for b.\n\n2 of 6" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91162807,"math_prob":0.99581283,"size":594,"snap":"2023-40-2023-50","text_gpt3_token_len":153,"char_repetition_ratio":0.118644066,"word_repetition_ratio":0.51428574,"special_character_ratio":0.23063973,"punctuation_ratio":0.15267175,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99919766,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-06T23:28:57Z\",\"WARC-Record-ID\":\"<urn:uuid:660c2a32-eb6f-4417-b632-2224bf18ec65>\",\"Content-Length\":\"22043\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:50e29c1f-4be5-4eb7-b08c-48c1db25e937>\",\"WARC-Concurrent-To\":\"<urn:uuid:e7f1bf2c-9ff0-4f51-b6f1-84f2efea67a2>\",\"WARC-IP-Address\":\"54.163.65.8\",\"WARC-Target-URI\":\"https://oercommons.org/courseware/lesson/4062/student-old/?task=2\",\"WARC-Payload-Digest\":\"sha1:377MCGZ3ASTZMV37R544RQI33DNEL34B\",\"WARC-Block-Digest\":\"sha1:MBLSSAIQ3PM65LQ7JQCPC6ESYYV4UOCO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100626.1_warc_CC-MAIN-20231206230347-20231207020347-00809.warc.gz\"}"}
https://de.mathworks.com/help/vision/ref/demosaic.html	[ "# Demosaic\n\nDemosaic Bayer's format images\n\n## Library\n\nConversions\n\n`visionconversions`\n\n•", null, "## Description\n\nThe following figure illustrates a 4-by-4 image in Bayer's format with each pixel labeled R, G, or B.", null, "The Demosaic block takes in images in Bayer's format and outputs RGB images. The block performs this operation using a gradient-corrected linear interpolation algorithm or a bilinear interpolation algorithm.\n\nPortInput/OutputSupported Data TypesComplex Values Supported\n\nI\n\nMatrix of intensity values\n\n• If, for the Interpolation algorithm parameter, you select `Bilinear`, the number of rows and columns must be greater than or equal to 3.\n\n• If, for the Interpolation algorithm parameter, you select `Gradient-corrected linear`, the number of rows and columns must be greater than or equal to 5.\n\n• Double-precision floating point\n\n• Single-precision floating point\n\n• Fixed point\n\n• 8-, 16-, and 32-bit signed integer\n\n• 8-, 16-, and 32-bit unsigned integer\n\nNo\n\nR, G, BMatrix that represents one plane of the input RGB video stream. Outputs from the R, G, or B ports have the same data type.Same as I port\n\nNo\n\nImage\n\nM-by-N matrix of intensity values or an M-by-N-by-P color video signal where P is the number of color planes.\n\nSame as I port\n\nNo\n\nUse the Interpolation algorithm parameter to specify the algorithm the block uses to calculate the missing color information. If you select `Bilinear`, the block spatially averages neighboring pixels to calculate the color information. If you select `Gradient-corrected linear`, the block uses a Weiner approach to minimize the mean-squared error in the interpolation. This method performs well on the edges of objects in the image. For more information, see .\n\nUse the Sensor alignment parameter to specify the alignment of the input image. Select the sequence of R, G and B pixels that correspond to the 2-by-2 block of pixels in the top-left corner of the image. You specify the sequence in left-to-right, top-to-bottom order. For example, for the image at the beginning of this reference page, you would select `BGGR`.\n\nUse the Output image signal parameter to specify how to output a color video signal. If you select `One multidimensional signal`, the block outputs an M-by-N-by-P color video signal, where P is the number of color planes, at one port. If you select `Separate color signals`, additional ports appear on the block. Each port outputs one M-by-N plane of an RGB video stream.\n\n### Fixed-Point Data Types\n\nThe following diagram shows the data types used in the Demosaic block for fixed-point signals.", null, "You can set the product output and accumulator data types in the block mask as discussed in the next section.\n\n## Parameters\n\nInterpolation algorithm\n\nSpecify the algorithm the block uses to calculate the missing color information. Your choices are `Bilinear` or ```Gradient-corrected linear```.\n\nSensor alignment\n\nSelect the sequence of R, G and B pixels that correspond to the 2-by-2 block of pixels in the top left corner of the image. You specify the sequence in left-to-right, top-to-bottom order.\n\nOutput image signal\n\nSpecify how to output a color video signal. If you select ```One multidimensional signal```, the block outputs an M-by-N-by-P color video signal, where P is the number of color planes, at one port. If you select `Separate color signals`, additional ports appear on the block. Each port outputs one M-by-N plane of an RGB video stream.\n\nRounding mode\n\nSelect the rounding mode for fixed-point operations.\n\nOverflow mode\n\nSelect the overflow mode for fixed-point operations.\n\nProduct output", null, "As depicted in the previous figure, the output of the multiplier is placed into the product output data type and scaling. Use this parameter to specify how to designate this product output word and fraction lengths:\n\nWhen you select `Same as input`, these characteristics match those of the input to the block.\n\nWhen you select `Binary point scaling`, you can enter the word length and the fraction length of the product output, in bits.\n\nWhen you select `Slope and bias scaling`, you can enter the word length, in bits, and the slope of the product output. The bias of all signals in the Computer Vision Toolbox™ blocks is 0.\n\nAccumulator", null, "As depicted in the previous figure, inputs to the accumulator are cast to the accumulator data type. The output of the adder remains in the accumulator data type as each element of the input is added to it. Use this parameter to specify how to designate this accumulator word and fraction lengths:\n\n• When you select `Same as product output`, these characteristics match those of the product output.\n\n• When you select `Same as input`, these characteristics match those of the input.\n\n• When you select `Binary point scaling`, you can enter the word length and the fraction length of the accumulator, in bits.\n\n• When you select `Slope and bias scaling`, you can enter the word length, in bits, and the slope of the accumulator. The bias of all signals in the Computer Vision Toolbox blocks is 0.\n\nLock data type settings against change by the fixed-point tools\n\nSelect this parameter to prevent the fixed-point tools from overriding the data types you specify on the block mask. For more information, see `fxptdlg` (Fixed-Point Designer), a reference page on the Fixed-Point Tool in the Simulink® documentation.\n\n## References\n\n Malvar, Henrique S., Li-wei He, and Ross Cutler. “High-Quality Linear Interpolation for Demosaicing of Bayer-Patterned Color Images.” Microsoft Research, May 2004. http://research.microsoft.com/pubs/102068/Demosaicing_ICASSP04.pdf.\n\n Gunturk, Bahadir K., John Glotzbach, Yucel Altunbasak, Ronald W. Schafer, and Russel M. Mersereau, “Demosaicking: Color Filter Array Interpolation,” IEEE Signal Processing Magazine, Vol. 22, Number 1, January 2005.\n\n## Version History\n\nIntroduced in R2006b" ]	[ null, "https://de.mathworks.com/help/vision/ref/demosaic_block.png", null, "https://de.mathworks.com/help/vision/ref/demosaic_diagram.png", null, "https://de.mathworks.com/help/vision/ref/demosaic_fp_diagram.png", null, "https://de.mathworks.com/help/vision/ref/deinterlace_fp2.png", null, "https://de.mathworks.com/help/vision/ref/deinterlace_fp3.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7380503,"math_prob":0.8322142,"size":5396,"snap":"2022-27-2022-33","text_gpt3_token_len":1212,"char_repetition_ratio":0.13130563,"word_repetition_ratio":0.33333334,"special_character_ratio":0.20422535,"punctuation_ratio":0.11519845,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9506314,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,3,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-06T01:53:35Z\",\"WARC-Record-ID\":\"<urn:uuid:b48797fe-4419-40ab-ad96-786634c9dc9d>\",\"Content-Length\":\"81058\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fd309752-d674-42dd-a1b3-7acfa02fd711>\",\"WARC-Concurrent-To\":\"<urn:uuid:46d5735f-9e4a-4bed-846c-f66477e7b7b8>\",\"WARC-IP-Address\":\"104.68.243.15\",\"WARC-Target-URI\":\"https://de.mathworks.com/help/vision/ref/demosaic.html\",\"WARC-Payload-Digest\":\"sha1:5NMUZ2H7C2AGA3SSIBPJ6F2GIS7TTGVD\",\"WARC-Block-Digest\":\"sha1:CW3NFDDJ5ZFM3WLXWWHBNQJEAQTPD7AM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104655865.86_warc_CC-MAIN-20220705235755-20220706025755-00378.warc.gz\"}"}
http://chinasy.net/facts-about-lnm/f78ae1-parts-of-addition-sentence	[ "In this last example we are given three numbers: 5, 3 and 8. We combine the numbers on either side of the plus sign together to make a total. Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem. The plus sign or addition sign is ‘+’ and this must go in between the two numbers that we wish to add together. Number sentence. Two posters to display in the classroom explaining the elements of addition and subtraction number sentences. Now try our lesson on Subtraction Number Sentences where we learn how to write number sentences for subtraction. In the example of 3 + 2, the plus sign means to combine the 3 and the 2 together. A number sentence can use any of the mathematical operations from additionâ¦ Writing Addition and Subtraction Number Sentences: Using a die (or two), have the child roll a number. Every sentence you write or speak in English includes words that fall into some of the nine parts of speech. Every number in the addition sentence is called a term. Some common mathematical symbols used in number sentences are: In this lesson we are only looking at addition number sentences, which will only contain the plus sign ‘+’ and the equals sign ‘=’. 4. The 8 in this problem is called the sum. Writing an equals sign means that the total value on the left of the equals sign is the same as the total value on the right of the equals sign. Below is another example of a correct addition number sentence. Normally the subject comes before the verb phrase in a sentence. In the fourth example we have six terms. or add to Google Calendar. Free math worksheets from K5 Learning. Our number sentence is 5 + 3 = 8 and is read as ‘five add three equals eight’. Three counters and two more makes a total of 5 counters. In addition to endure, we have no choice. Now in the first two examples we have three terms. The minus sign (or subtraction sign): ‘-‘. English Parts of a Sentence Exercises 1 English Parts of a Sentence Exercises 2 Sentences consist of a number of parts, using different parts of speech. (Some sources include only eight parts of speech and leave interjections in their own category.) Join group, and play Just play. Adjectives modify nouns and pronouns, while adverbs modify verbs, adjectives, and other adverbs. I minored in literature in addition to art. 15 is Product. We say that we have added three and two together. You get better at addition with practice, so we have:Math Trainer - Addition (train your memory)Kindergarten Worksheets (easy addition)Addition Worksheets (normal and advanced) ... Last Played. 2011-07-19 13:05:24 2011-07-19 13:05:24. what are parts of addition sentences. Then, students use the two numbers as the âpartsâ to write two addition sentences (thus modeling the commutative property). In addition, it makes Some of the worksheets for this concept are Comma packet, English language work i parts of speech, Mathematics vocabulary 5 addition terminology, Writing subtraction sentences, Kindergarten first grade writing folder, Work 1 basic sentence structure and building sentences, Chapter 7 parts of speech and sentence structures, Circle the verb in each. The expressions given in examples indicate equality or inequality. These include nouns, pronouns, verbs, adjectives, adverbs, prepositions, conjunctions, articles/determiners, and interjections. The most important parts of speech are: The subject, which is either a noun phrase (see The noun phrase) or a pronoun (see Pronouns). A set of posters naming the parts of an addition and subtraction number sentence. This shows the answer to the problem you have just completed. When teaching addition to children it is important to remember that the plus sign must separate the two numbers that we wish to combine together. We can see below that it is ok to write and equals sign between 3 + 2 and 5 but not in between 3 + 2 and 9. Worksheet will open in a new window. What are the parts of addition sentence subtraction sentence and multiplication sentence? Now let's practice! 3 is Multiplier. In a division equation, a dividend is divided by a divisor to give a quotient. It is possible for parts of speech to do this work alone in the sentence in either the subject or the predicate. You need to be a group member to play the tournament. To download/print, click on pop-out icon or print icon to worksheet to print or download. 2. 0 0 1. This game is part of a tournament. A simple sentence is a sentence structure that contains one independent clause and no dependent clauses. In the number sentence a+b=c, a and b are addends, while c is the sum. The girls had been swimming. The verb (or predicate) usually follows the subject and identifies an action or a state of being. English Parts of a Sentence Exercises, Identify the Part of a Sentence. The company provides cheap Internet access. In an addition equation, addends are the numbers that are added together to give a sum. The word for the process of combining these counters to make a total is addition. Includes: 4 different posters: parts of a subtraction number sentence (subtrahend, minuend, equals, difference), understanding a number sentence (it starts with the greater number) 1 chant/poster \"subtraction\" 1 vocabulary poster \"subtraction\" 4 label the subtraction number sentences Want the matching ADDITION pack? A number sentence is just a combination of numbers and mathematical symbols. • About Us You can & download or print using the browser document reader options. These worksheets seek to improve the understanding of addition sentences. What are the parts of addition sentence? Parts Of Addition Sentence Worksheets - there are 8 printable worksheets for this topic. addition example sentences. This addition number sentence is read as ‘five add three equals eight’ and it means that combining 5 counters with 3 more counters is the same as simply having 8 counters from the start. They had finished. Remember that we cannot put another number here directly next to other numbers. About this Worksheet. Found worksheet you are looking for? Set a reminder in your calendar. Point out where the part of addition sentence is asked. Add. 3 + 2 = 5 is an example of an addition number sentence. Basic Math Terms Used In Multiplication: 5 x 3 = 15 , Here, 5 is Multiplicand. Parts of a Number Sentence (Addition and Subtraction) Posters. For example we could have written the addition number sentence the other way around: Whilst it is not so common in school questions to have one number on the left of an equals sign and more than one number on the right, this is still a completely correct addition number sentence. Answer. 3 + 2 = 5 is an example of an addition number sentence. In a subtraction equation, the subtrahend is taken away from the minuend to give a difference. Types of Number Sentences. The basic parts of a sentence are the subject, the verb, and (often, but not always) the object. Includes: 4 different posters: parts of an addition number sentence (addend, plus, equals, sum), understanding a â¦ We are asked to complete the number sentence by filling in the box in between the 5 and the 3. This is a free printable worksheet in PDF format and holds a printable version of the quiz Parts of Addition Sentence.By printing out this quiz and taking it with pen and paper creates for a good variation to only playing it online. Example sentences with the word addition. Record the first number on the page and then roll again. Parts of Addition Sentence. An addition sentence is not limited to two addends. The + sign is called a plus sign or an addition sign. The mathematical symbol that tells us to combine the two numbers together into a total is the plus sign. Each number in a number sentence must be separated by a mathematical symbol. Counting the counters, we have 5 in total. Therefore we are choosing a mathematical symbol to go here. Wiki User Answered . Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on WhatsApp (Opens in new window), Click to share on Pinterest (Opens in new window), Click to share on Tumblr (Opens in new window), Click to share on Pocket (Opens in new window), Click to share on Telegram (Opens in new window), Click to share on Skype (Opens in new window), https://www.mathswithmum.com/wp-content/uploads/2019/07/Addition-Number-Sentences.mp4. Text: POS-tag! Press play! 243+10 sentence examples: 1. Multiple numbers can be added together if required. Asked by Wiki User. For example, a typical addition sentence is 9 + 5 = 14. 3. It means that having 3 counters and then 2 more is the same as simply having 5 counters. These resources were made to help Kinders and Firsties learn the parts of an addition number sentence which in turn will help them to have deeper understanding about adding up numbers. Instead of writing down ‘three add two’ it is quicker to write ‘3 + 2’, where the ‘+’ sign is called a plus sign or an addition sign. The subject is usually a noun â a word that names a person, place, or thing. We use an equals sign to show that we have a correct number sentence. 0%. Here the equals sign tells us that 3 and 2 combined together is the same as 5. â¦ 2 + 2 is the same value as 4 and so we show this by writing an equals sign in between 2 + 2 and 4. and click at \"POS-tag!\". 26 Nov, 2020 Sound On/Off. Top Answer. We can be asked to solve an addition sentence with a missing term. about Parts-of-speech.Info; Enter a complete sentence (no single words!) We have three numbers given to us, 2 , 2 and 4. 0 0 1. Students fill in the missing number in the addition equation while coloring in circles which visually represent the same equation. A number sentence is a mathematical sentence, made up of numbers and signs. The = is called an equal sign. The Parts of the Sentence. 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Write or speak in English includes words that fall into some of the plus sign together make!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93632865,"math_prob":0.91628075,"size":19824,"snap":"2021-04-2021-17","text_gpt3_token_len":4297,"char_repetition_ratio":0.18930374,"word_repetition_ratio":0.18806669,"special_character_ratio":0.22255851,"punctuation_ratio":0.13536896,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9803692,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-23T17:59:41Z\",\"WARC-Record-ID\":\"<urn:uuid:b493fd3d-8f79-4d6b-aa36-ec69c4e66753>\",\"Content-Length\":\"27220\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0cbb1119-6e8e-4060-a9ee-aca2e9ca59e2>\",\"WARC-Concurrent-To\":\"<urn:uuid:7ae5a8e3-5463-4582-ab22-1f7c101ac10d>\",\"WARC-IP-Address\":\"47.88.77.166\",\"WARC-Target-URI\":\"http://chinasy.net/facts-about-lnm/f78ae1-parts-of-addition-sentence\",\"WARC-Payload-Digest\":\"sha1:PLXXZMO45OKIAHI2XNSAY2WE4W6HVMNR\",\"WARC-Block-Digest\":\"sha1:YWIB4JIDSBGECIFC4FYQJTOYTPQNEJPL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039596883.98_warc_CC-MAIN-20210423161713-20210423191713-00468.warc.gz\"}"}
http://docs.gomafem.com/problem_description_file/problem_description/energy.html	[ "# energy#\n\nEQ = energy {Galerkin_wt} T {Interpol_fnc} <floatlist>\n\n\n## Description / Usage#\n\nThis card provides information for solving a conservation of energy differential equation. Definitions of the input parameters are defined below. Note that <floatlist> contains five constants for the Energy equation defining the constant multipliers for each term in the equation. The Galerkin weight and the interpolation function must be the same for the code to work properly. If upwinding is desired for advection dominated problems, we can set this through a Petrov-Galerkin weight function in the material file.\n\n energy Name of the equation to be solved. {Galerkin_wt} Two- or four-character value that defines the type of weighting function for this equation, where: Q1-Linear Q2-Quadratic Q1_D-Standard linear interpolation with special allowance for discontinuous degrees of freedom at interfaces. Q2_D-Standard quadratic interpolation with special allowance for discontinuous degrees of freedom at interface. Q1_XV-Linear interpolation with enrichment in elements of material interfaces. This enrichment function allows discontinuity in value and gradient along interface but maintains continuity at element edges/faces.Only used for level-set problems. Q2_XV-Quadratic interpolation with enrichment in elements of material interfaces. This enrichment function allows discontinuity in value and gradient along interface but maintains continuity at element edges/faces. Only used for level-set problems. Q1_GN-Linear interpolation for capturing variables defined on the negative side of the level-set interface. Similar to Q1_XV. Q2_GN-Quadratic interpolation for capturing variables defined on the negative side of the level-set interface. Similar to Q1_XV T Name of the variable associated with this equation. {Interpol_fnc} Two- or four-character value that defines the interpolation function used to represent the variable T, where: Q1-Linear Continuous Q2-Quadratic Continuous Q1_D-Standard linear interpolation with special allowance for discontinuous degrees of freedom at interfaces. Q2_D-Standard quadratic interpolation with special allowance for discontinuous degrees of freedom at interfaces. Q1_XV-Linear interpolation with enrichment in elements of material interfaces. This enrichment function allows discontinuity in value and gradient along interface but maintains continuity at element edges/faces. Q2_XV-Quadratic interpolation with enrichment in elements of material interfaces. This enrichment function allows discontinuity in value and gradient along interface but maintains continuity at element edges/faces. Q1_GN-Linear interpolation for capturing variables defined on the negative side of the level-set interface. Similar to Q1_XV. Q2_GN-Quadratic interpolation for capturing variables defined on the negative side of the level-set interface. Similar to Q1_XV. Q1_GP-Linear interpolation for capturing variables defined on the positive side of the level-set interface. Similar to Q1_XV. Q2_GNP-Quadratic interpolation for capturing variables defined on the positive side of the level-set interface. Similar to Q1_XV. Multiplier on mass matrix term ( d ⁄dt ). Multiplier on advective term. Multiplier on boundary term ( $$\\underline{n}$$ • flux ). Multiplier on diffusion term. Multiplier on source term.\n\nNote: These multipliers are intended to provide a means of activating or deactivating terms of an equation, and hence should be set to zero or one. If a multiplier is zero, the section of code that evaluates the corresponding term will be skipped.\n\n## Examples#\n\nThe following is a sample card that uses a linear continuous interpolation and weight function and has all the term multipliers on except the mass matrix term for time derivatives:\n\nEQ = energy Q1 T Q1 0. 1. 1. 1. 1.\n\n\n## Technical Discussion#\n\nSome discussion on the XFEM-type enriched basis functions Q1_XV, Q1_GN, Q1_GP, Q2_GN, Q2_GP and Q2_XV is in order. First of all, these basis functions are to be use with the level-set front tracking capability only. First of all, these basis functions are typically only used for the continuity equation to capture pressure jumps due to surface tension. However, for phase change problems some experimentation has been pursued with the energy equation.\n\nXFEM Value Enrichment\n\nEnrichment:\n\nRelated “Ghost” Enrichment:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.858075,"math_prob":0.9685311,"size":2592,"snap":"2023-40-2023-50","text_gpt3_token_len":566,"char_repetition_ratio":0.12867078,"word_repetition_ratio":0.03448276,"special_character_ratio":0.21334876,"punctuation_ratio":0.10526316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9855743,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T11:52:49Z\",\"WARC-Record-ID\":\"<urn:uuid:bed2627e-da41-40b3-a597-47849466e42b>\",\"Content-Length\":\"97559\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:866a397a-4f73-4472-aadd-a6a2bede418f>\",\"WARC-Concurrent-To\":\"<urn:uuid:668d6322-ddb5-43d7-9ef4-61572265e5e5>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"http://docs.gomafem.com/problem_description_file/problem_description/energy.html\",\"WARC-Payload-Digest\":\"sha1:VV2PSC6RTYKMJZZHNN7QZ3Z6IOH6YGNC\",\"WARC-Block-Digest\":\"sha1:XIGFKYO5NFPRTJQ3ZKTRUQHIXHQX6BEO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510387.77_warc_CC-MAIN-20230928095004-20230928125004-00746.warc.gz\"}"}
http://www.cap-lore.com/MathPhys/GR/Regge.html	[ "Tullio Regge proposed that an n dimensional manifold could be decomposed into n-simplexes in order to perform geometrical and physical calculations where the manifold’s curvature was concentrated in bones consisting of the set of (n−2)-simplexes distributed thru out the manifold. Regge observed that the lengths of the shared edges of the simplexes determine the shape of each simplex and thus the metric properties of the entire manifold. We concentrate on Riemannian (metric) manifolds here. A central problem is to relate coordinate systems across the boundaries between neighboring n-simplexes. Here is some math to do that.\n\nI call one of the n-simplexes into which the manifold has been partitioned, a zone, for we must speak of simplexes of lesser dimension as well. It is natural to use barycentric coordinates (bcc) in a simplex. A problem arises at the interface between two adjacent zones, called a facet here. A facet is an (n−1)-simplex.\n\nTo move a vector or tensor across a facet to a neighboring zone we conceptually adopt a new intermediate facet coordinate system whose bases are the n−1 edges of the facet emanating from some particular vertex of the facet, and a unit vector normal to the facet. This representation is neutral to the two zones, except for the normal’s orientation. To move now to a corresponding facet coordinate system in the neighboring zone we must change the sign of the normal component, and permute the order of the other components to match the vertex numbering plan of the neighbor. After this we transform to the new zone’s bc coordinates. Distracting detail may be found here.\n\nPythag: All of these computations may be carried out if we can only calculate the inner product between vectors. This is supported directly if we have the metric tensor in whatever coordinate system we have expressed our vectors. Fortunately it is trivial to compute the covariant metric tensor in bcc given the squared edge lengths of the zone. The dot product of sides a and b of a triangle is (a2 + b2 − c2)/2. Note that I use the terms “length”, “vector”, “normal”, and especially “dot product” (as in “a • b”) in a Euclidean geometry sense in distinction to any algebraic sense. For each zone, choose some vertex of the zone as the origin and take the n edges emanating from the origin as bases for the barycentric coördinate system.\n\ngij = edgei • edgej\nx • y = gijxiyj\nThe i and j in the equations above each take on n values. Compute the n(n+1)/2 elements of the covariant metric tensor gij directly. To express v as the sum of a vector in the facet and a multiple of the normal, N, we write v = (v−vN) + vN, where vN = (v • N)N. The vector (v−vn) is in the facet. This requires expressing N in bcc.\n\nThe dot product of v with basis element j is 1. The volume of the zone is (the square root of the determinant of the covariant metric tensor)/n! . Let vf be the n−1 volume of facet j and vz be the n volume of the zone. vz = (altitude of zone over facet as base)vf/n.\n\nThe contravariant metric tensor is computed by inverting the covariant metric tensor as a matrix. The rows of the contravariant metric tensor are vectors expressed in bcc, normal to the facets of the zone. Since the normals to all the facets of a zone sum to 0, we have an easy way to find the normal to the facet opposite the origin. Those normals need only be multiplied by the zone’s volume to become the ‘natural’ normals. By “natural normal” to a facet we mean here the normal to the facet whose length is the Grassmann magnitude of the facet.\n\nThere is a fundamental problem to be solved before we begin choosing edge lengths to form a complex.\n\nThe general curvature tensor may now be computed centered at bones, the (n−2)-simplexes, by carrying a frame around each (n−2)-simplex. We pass thru the set of facets which are incident on the bone. There is just one scalar of curvature from the bone.\n\nHere is an another method of doing this calculation which I currently think is inferior.\n\nHere are fragments of ideas of applying this to General Relativity.\n\nHere is some code.\n\nIf we assemble n-simplexes for some fixed n into a complex we must match edge lengths; or more properly, perhaps, squared edge lengths. When the length2 of a shared edge is computed in the metric of each of the zones sharing that edge, the calculations must agree in order to say we have assembled a complex from n-simplexes. If our geometry is locally Minkowskian, squared edge lengths may be positive, zero or negative. Matching squared edge lengths thus leads to inserting causality. See remark beginning “Simply put:” here. I do not see how simplexes could be sensibly united to complexes without such a rule. I had not noticed this before reading some CDT literature. It makes their causality rule look like a mathematical rather than a physical axiom.\n\nI think there is another requirement on assembly of n-simplexes into a complex. Here is a complex which must be counted as unphysical and perhaps even un-manifold like as the signature of its metric is not uniform. (The numbers near the edges are the squares of the lengths of those edges.)\n\nAnother note is that any set of n(n−1) edge length squares from ℝn(n−1) defines a pseudo Riemannian metric and conversely. Well not quite—some edge length value sets produce indefinite metrics with zero eigenvalues and these are excluded from conventional pseudo Riemannian geometries. Indeed the n(n−1) edge length squares are linear in the n(n−1) independent elements of the symmetric covariant metric tensor. Beware there is nothing left of the triangle inequality in Minkowskian space. Picking zones whose metric has signature (+ − − −) is merely to apply a few homogeneous inequalities to the edge length squares. See this and especially this.\n\nThe subject of Causal dynamic triangulation proposes to extend ideas such as these to a theory of everything. CDT Nexus\n\nNames of k-simplexes by k\n k conventional name 0 vertex 1 edge 2 triangle 3 tetrahedron n−2 bone n−1 facet n zone\n\nHere is a zoo of simplexes to avoid." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9155806,"math_prob":0.9897892,"size":5759,"snap":"2023-14-2023-23","text_gpt3_token_len":1323,"char_repetition_ratio":0.13831452,"word_repetition_ratio":0.007936508,"special_character_ratio":0.21236326,"punctuation_ratio":0.07526882,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9960578,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-28T07:05:51Z\",\"WARC-Record-ID\":\"<urn:uuid:4beabd42-b58e-4be0-8910-ffc01407555c>\",\"Content-Length\":\"8701\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cb73060e-8faa-4fd3-b8ea-0e0b24c8bac2>\",\"WARC-Concurrent-To\":\"<urn:uuid:abec20fc-65b3-41b5-b291-8362bb0a0690>\",\"WARC-IP-Address\":\"184.105.182.100\",\"WARC-Target-URI\":\"http://www.cap-lore.com/MathPhys/GR/Regge.html\",\"WARC-Payload-Digest\":\"sha1:A4D3U2AODU55OK6M2ZAC247J2U35BF4I\",\"WARC-Block-Digest\":\"sha1:DPLF7WFSNQPRIYS3QUSF5J4KIGUIUIKP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224643585.23_warc_CC-MAIN-20230528051321-20230528081321-00281.warc.gz\"}"}
http://mizar.uwb.edu.pl/version/current/html/proofs/matrix_9/26	[ "let n be Nat; :: thesis: for f, g being FinSequence st f ^ g in Permutations n holds\ng ^ f in Permutations n\n\nlet f, g be FinSequence; :: thesis: ( f ^ g in Permutations n implies g ^ f in Permutations n )\nA1: Seg (len (f ^ g)) = dom (f ^ g) by FINSEQ_1:def 3;\nlen (f ^ g) = (len f) + (len g) by FINSEQ_1:22\n.= len (g ^ f) by FINSEQ_1:22 ;\nthen A2: dom (f ^ g) = dom (g ^ f) by ;\nA3: rng (f ^ g) = (rng f) \\/ (rng g) by FINSEQ_1:31\n.= rng (g ^ f) by FINSEQ_1:31 ;\nassume f ^ g in Permutations n ; :: thesis:\nthen A4: f ^ g is Permutation of (Seg n) by MATRIX_1:def 12;\nthen A5: rng (f ^ g) = Seg n by FUNCT_2:def 3;\nA6: g is one-to-one by ;\ndom (f ^ g) = Seg n by ;\nthen reconsider h = g ^ f as FinSequence-like Function of (Seg n),(Seg n) by ;\n( rng f misses rng g & f is one-to-one ) by ;\nthen A7: h is one-to-one by ;\nh is onto by ;\nhence g ^ f in Permutations n by ; :: thesis: verum" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5909594,"math_prob":0.99976546,"size":1004,"snap":"2019-35-2019-39","text_gpt3_token_len":412,"char_repetition_ratio":0.183,"word_repetition_ratio":0.10619469,"special_character_ratio":0.43326694,"punctuation_ratio":0.2007435,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999874,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-17T15:05:48Z\",\"WARC-Record-ID\":\"<urn:uuid:4b09cdf0-7c3e-4e37-81ce-212eb580ecfa>\",\"Content-Length\":\"13943\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a1d7c652-8409-42dc-8f88-a64bc501d651>\",\"WARC-Concurrent-To\":\"<urn:uuid:61443b73-c92c-46de-8415-e9b566e9eebb>\",\"WARC-IP-Address\":\"212.33.73.131\",\"WARC-Target-URI\":\"http://mizar.uwb.edu.pl/version/current/html/proofs/matrix_9/26\",\"WARC-Payload-Digest\":\"sha1:J42F6KMXOHLSPSE4RDJHYABSYM5VM6OM\",\"WARC-Block-Digest\":\"sha1:MVYQV7HFGAFGI4Q55JWFTGJFYIA3QBJY\",\"WARC-Identified-Payload-Type\":\"application/xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573080.8_warc_CC-MAIN-20190917141045-20190917163045-00370.warc.gz\"}"}
https://brilliant.org/problems/horrible-hexagons-2/	[ "# Horrible Hexagons 2\n\nGeometry Level 3", null, "The diagram above shows a regular hexagon with side length $2$ and two equilateral triangles inscribed in it.\n\nThe area of the shaded region can be written as $\\large \\frac { a\\sqrt { 3 } }{ b }$, where $a$ and $b$ are coprime integers. Find $ab$.\n\n×\n\nProblem Loading...\n\nNote Loading...\n\nSet Loading..." ]	[ null, "https://ds055uzetaobb.cloudfront.net/brioche/solvable/b6fd502160.0955cc1d5b.xtsvF9.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83766794,"math_prob":0.99578065,"size":357,"snap":"2021-21-2021-25","text_gpt3_token_len":81,"char_repetition_ratio":0.12747875,"word_repetition_ratio":0.21428572,"special_character_ratio":0.2184874,"punctuation_ratio":0.24675325,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98486215,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-22T16:51:42Z\",\"WARC-Record-ID\":\"<urn:uuid:b3f0d69a-e02c-4112-933f-e9734f792893>\",\"Content-Length\":\"40186\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3a0591a3-581e-40e8-bda0-49e6d9fb8ed8>\",\"WARC-Concurrent-To\":\"<urn:uuid:56f9a730-11b3-4137-94f1-347b094da996>\",\"WARC-IP-Address\":\"104.20.35.242\",\"WARC-Target-URI\":\"https://brilliant.org/problems/horrible-hexagons-2/\",\"WARC-Payload-Digest\":\"sha1:OD6BVBNCMICMVWX6TWIALCDLUWN56XGA\",\"WARC-Block-Digest\":\"sha1:BESQSARTQ2KF55DRK5IP5FGYB4B4UL77\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488519183.85_warc_CC-MAIN-20210622155328-20210622185328-00460.warc.gz\"}"}
https://web2.0calc.com/questions/help_89380	[ "+0\n\n# help!\n\n0\n266\n2\n\nPeter has built a gazebo, whose shape is a regular heptagon, with a side length of 1 unit. He has also built a pathway around the gazebo, of constant width 1 unit, as shown below. (Every point on the ground that is within 1 unit of the gazebo and outside the gazebo is covered by the pathway.) Find the area of the pathway.\n\nhttps://latex.artofproblemsolving.com/f/f/7/ff7af7da5a2907b44a55ae65671c4c5c238f574d.png\n\nMar 4, 2020\n\n#2\n+2\n\nPeter has built a gazebo, whose shape is a regular heptagon, with a side length of 1 unit.\nHe has also built a pathway around the gazebo, of constant width 1 unit, as shown below.\n(Every point on the ground that is within 1 unit of the gazebo and outside the gazebo is covered by the pathway.)\nFind the area of the pathway.", null, "$$\\begin{array}{\|rcll\|} \\hline \\alpha +\\beta &=& 180^\\circ \\quad & \| \\quad \\mathbf{\\beta = \\dfrac{7*180^\\circ-360^\\circ}{7}} =\\dfrac{900^\\circ}{7} \\\\ \\alpha +\\dfrac{900^\\circ}{7} &=& 180^\\circ \\\\ \\alpha &=& 180^\\circ-\\dfrac{900^\\circ}{7} \\\\ \\mathbf{\\alpha} &=& \\mathbf{\\dfrac{360^\\circ}{7}} \\\\ \\hline \\end{array}$$\n\n$$\\begin{array}{\|rcll\|} \\hline \\mathbf{\\text{area of the pathway}} \\\\ \\hline &=& 7\\times(1\\times 1 ) + 7 \\times \\left( \\pi \\times 1^2 \\times\\dfrac{\\alpha}{360^\\circ} \\right) \\quad & \| \\quad \\mathbf{\\alpha=\\dfrac{360^\\circ}{7}} \\\\\\\\ &=& 7 + 7 \\times \\left( \\pi \\times\\dfrac{\\dfrac{360^\\circ}{7}}{360^\\circ} \\right) \\\\\\\\ &=& 7 + 7 \\times \\left( \\pi \\times\\dfrac{1}{7} \\right) \\\\\\\\ &=&\\mathbf{ 7 + \\pi} \\\\ \\hline \\end{array}$$", null, "Mar 4, 2020" ]	[ null, "https://web2.0calc.com/api/ssl-img-proxy", null, "https://web2.0calc.com/img/emoticons/smiley-laughing.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8196446,"math_prob":0.9996043,"size":1476,"snap":"2020-24-2020-29","text_gpt3_token_len":523,"char_repetition_ratio":0.1521739,"word_repetition_ratio":0.6105769,"special_character_ratio":0.38346884,"punctuation_ratio":0.07434944,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9969468,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-14T10:01:46Z\",\"WARC-Record-ID\":\"<urn:uuid:41eb2dcc-4c0e-41e2-ac01-daf8205c6262>\",\"Content-Length\":\"27019\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3a4b295f-9f57-4d3c-a90e-4872bd6b893a>\",\"WARC-Concurrent-To\":\"<urn:uuid:53877dcc-31bd-4174-9c29-ae54dbaa060f>\",\"WARC-IP-Address\":\"209.126.117.101\",\"WARC-Target-URI\":\"https://web2.0calc.com/questions/help_89380\",\"WARC-Payload-Digest\":\"sha1:MXNCIBBSHIDJTAQ7R5JZDO5YZA5H6VWL\",\"WARC-Block-Digest\":\"sha1:UI7FMYRU6AZX5O7LS3ESTLOHAXT724OP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657149819.59_warc_CC-MAIN-20200714083206-20200714113206-00509.warc.gz\"}"}
https://iwaponline.com/jh/article/21/5/798/67997/Determination-of-compound-channel-apparent-shear	[ "## Abstract\n\nMomentum exchange in the mixing region between the floodplain and the main channel is an essential hydraulic process, particularly for the estimation of discharge. The current study investigated various data mining models to estimate apparent shear stress in a symmetric compound channel with smooth and rough floodplains. The applied predictive models include random forest (RF), random tree (RT), reduced error pruning tree (REPT), M5P, and the distinguished hybrid bagging-M5P model. The models are constructed based on several correlated physical channel characteristic variables to predict the apparent shear stress. A sensitivity analysis is applied to select the best function tuning parameters for each model. Results showed that input with six variables exhibited the best prediction results for RF model while input with four variables produced the best performance for other models. Based on the optimised input variables for each model, the efficiency of five predictive models discussed here was evaluated. It was found that the M5P and hybrid bagging-M5P models with the coefficient of determination (R2) equal to 0.905 and 0.92, respectively, in the testing stage are superior in estimating apparent shear stress in compound channels than other RF, RT and REPT models.\n\n## INTRODUCTION\n\nThe compound cross section is a typical cross section in natural rivers which consists of a main channel and one or two floodplains (Al-Khatib et al. 2013; Tang 2017). In flooded conditions, the floodplain has lower velocity than the main channel, leading to momentum transfers between the interfaces. In compound channels, the total flow resistance increases with transverse momentum transfer as first noted by Sellin (1964), and was then extensively studied until the millennium (Myers 1978; Fernandes et al. 2015). Improved methods for estimating momentum transfer at the main channel–floodplain interfaces have then been presented using apparent shear stress, compared to the traditional channel methods (Rajaratnam & Ahmadi 1981; Knight & Hamed 1984; Devi & Khatua 2016). Özbek et al. (2004) calculated the apparent shear stress and discharge in the symmetric compound channels with different floodplain widths, using directional division planes' methods. Based on the ratio of apparent shear stresses across the interfaces, the horizontal and diagonal interfaces' planes have better accuracy in estimating the discharge capacity than the vertical planes. Khatua et al. (2010) measured the shear stress distribution in channels with compound cross section and extracted equations to compute the apparent shear force values for different considered interfaces. Moreta & Martin-Vide (2010) proposed a generalised formula for apparent shear stress prediction and validated it for an extensive range of experimental data. They studied the effect of geometry and roughness on a non-dimensional friction coefficient acting on the vertical main channel–floodplain interface (Cfa). They concluded that the variation aspect ratio (the flow depth of main channel divided by the width of main channel) influences the apparent friction coefficient, but the parameter of bank side slope is not very effective on Cfa.\n\nAll discussed formulae require realisation of the velocity gradient, the influence of roughness and channel geometry on the estimation of apparent shear stress. Since determining the velocity gradient without detailed measurements is difficult and Cfa has high uncertainty with different geometry and roughness based on the results of Moreta & Martin-Vide (2010), finding other methods that can estimate apparent shear stress without a need for these parameters is attractive.\n\nOne of the recent methods replacing or improvising the previous methods, which increased higher precision of results, is artificial intelligence (AI)-based numerical techniques. With progress in these methods, several studies have been successfully carried out to predict different phenomena in hydraulics and hydrology such as evaporation modelling, shear stress prediction and streamflow forecasting (Kisi 2008; Sheikh Khozani et al. 2016, 2017a; Yaseen et al. 2016). Higher efficiency in predicting the shear stress in circular channels was found using gene expression programming (GEP), extreme learning machines, the M5 model tree algorithm and a randomised neural network technique (Sheikh Khozani et al. 2015, 2017a, 2017b, 2017c, 2018). The neural network was able to simulate and capture the complexities of spatial momentum transfer at the main channel–floodplain interface for compound channels with and without vegetation (Huai et al. 2013). Prediction of apparent shear stress using the genetic algorithm–artificial neural network (GA-ANN) and genetic programming (GP), along with multiple linear regression (MLR), was studied by Bonakdari et al. (2018). They concluded that the GAA method was more powerful than GP and MLR methods and a matrix-based equation to forecast the apparent shear stress was presented. However, these AI methods, especially the ANN technique, as one of the most widely used models (Houichi et al. 2012; Hanspal et al. 2013; Choi et al. 2015; Yaseen et al. 2015; Fahimi et al. 2016; Tapoglou et al. 2019) in both spatial or time series data prediction, need long time series data for both testing and training dataset (Melesse et al. 2011). Furthermore, one of the known limitations is the poor prediction for the data which are not in the range of the learning values (Melesse et al. 2011; Khosravi et al. 2018b). Therefore, soft computing models are mostly integrated with other approaches to overcome the weakness of an individual model. The possibility of an efficient integrated model is immense, and proven techniques include the combination of ANN with fuzzy logic (FL) and an adaptive neuro-fuzzy inference system (ANFIS). Although this hybrid model has a higher prediction power than both the individual ANN and FL, and it has been prominently applied, the model still suffers setbacks in finding the optimal parameter for a neural fuzzy model and determination of the best weights in a membership analysis function (Khosravi et al. 2018a). Generally, those models which have a hidden layer in their structure have a lower prediction power than other models without a hidden layer, such as the decision tree algorithms (i.e., random forest, random tree and so on) (Kisi et al. 2012). In addition, researchers are always motivated to investigate more reliable and robust models and to explore other algorithms with better and solid results. As each available model has its own advantages and disadvantages, researchers are keen to find improved prediction models, which are more robust, and the advantages outweigh the disadvantages. Nowadays, data mining methods have been declared as dependable modelling approaches to solve regression problems in multiple engineering and science applications (Witten & Frank 2005; Baker & Yacef 2009; Witten et al. 2011). It is crucial to investigate the prediction capacity of newly developed data mining algorithms to improve the accuracy in prediction of hydraulic variables such as apparent shear stress.\n\nDespite the capability of data mining techniques to solve complicated problems (Khosravi et al. 2018a; Sharafati et al. 2019), application in the field of hydraulics is limited, and only a few studies have paid attention to forecasting the apparent shear stress in channels with compound cross section. The accuracy of some innovative decision tree algorithms of random forest (RF), random tree model (RT), reduced error pruning tree (REPT) and M5P, and also a hybrid model of bagging-M5P structure is investigated in this study to assess the performance of these new algorithms for estimating the apparent shear stress in symmetric compound channels with smooth and rough floodplains. This study utilised the advantages of model trees, including faster in training, higher potential of convergence and assisting decision-making due to better transparency (Solomatine & Xue 2004). The RF model, in particular, lowers the risk of over-fitting and has less variance. As the determination of apparent shear stress is based on the geometrical characteristics of a channel, we attempt to obtain the best input variables (and the most effective parameters) with higher efficiency.\n\n## DATA COLLECTION AND PREPARATION\n\n### Theory and review\n\nEstimation of the apparent shear stress requires knowledge of the velocity gradient (ΔU) and channel geometry and characteristics. The key empirical equations used to forecast the apparent shear stress and calculate discharge in compound channels (with both smooth and rough floodplains) are shown in Table 1. Most of the equations have similar geometrical parameters, particularly the velocity gradient, and widths and water depths for both main channel and floodplain, with noticeable variably dispersed numerical coefficients.\n\nTable 1\n\nEquations for calculating apparent shear stress\n\nResearchers Experimental condition Presented equation\nErvine et al. (1982) Symmetric and asymmetric, smooth and rough floodplains", null, "Wormleaton et al. (1982) Symmetric, smooth and rough floodplains", null, "Knight & Demetriou (1983) Symmetric, smooth and rough floodplains", null, "Baird & Ervine (1984) Asymmetric, smooth floodplains", null, "Prinos & Townsend (1984) Symmetric, smooth and rough floodplains", null, "Wormleaton & Merrett (1990) Symmetric, smooth and rough floodplains (FCF)", null, "Christodoulou (1992) Symmetric, smooth floodplains", null, "Bousmar & Zech (1999) Symmetric and asymmetric, smooth and rough floodplains (FCF)", null, "Researchers Experimental condition Presented equation\nErvine et al. (1982) Symmetric and asymmetric, smooth and rough floodplains", null, "Wormleaton et al. (1982) Symmetric, smooth and rough floodplains", null, "Knight & Demetriou (1983) Symmetric, smooth and rough floodplains", null, "Baird & Ervine (1984) Asymmetric, smooth floodplains", null, "Prinos & Townsend (1984) Symmetric, smooth and rough floodplains", null, "Wormleaton & Merrett (1990) Symmetric, smooth and rough floodplains (FCF)", null, "Christodoulou (1992) Symmetric, smooth floodplains", null, "Bousmar & Zech (1999) Symmetric and asymmetric, smooth and rough floodplains (FCF)", null, "Nf is the number of floodplains, B is the total width, b is the width of the main channel, H is the flow depth in a compound channel, h is the flow depth in the main channel and ΔU is the velocity difference between the main channel and the floodplain.\n\n### Data collection and preparation\n\nDatasets are collected from several experimental studies on compound channels. For the smooth, symmetric, rectangular cross sections, the laboratory results of Knight & Demetriou (1983) were used. The researchers used a flume with dimensions of 15 m long, 0.61 m wide and 9.66 × 10−4 bed slope. The flume consisted of two floodplains with size of 0.076 m high and 0.229 m wide. Knight & Hamed (1984) used the same flume and investigated different roughness values in the floodplains and the main channel. Prinos & Townsend (1984) experimentally studied a compound channel with a trapezoidal cross section. The experiments were conducted in a main channel with dimensions of 1.02 m deep, 2:1 side slope and the bed channel slope was fixed at 3 × 10−4. They measured the shear stress along the whole wetted perimeter, then calculated the apparent shear stress at different flow depths. In addition to the small-scale flume data, large-scale flood channel facility (FCF) experimental data from the work of Wormleaton & Merrett (1990) were used to estimate the apparent shear stress. This facility entailed a flume size of 56 m long and 10 m wide, with a trapezoidal main channel cross section. They examined several methods for calculating discharge and provided a modified form that had a realistic understanding of the interaction between the main channel and the flood interface. Figure 1 demonstrates the cross section of compound channels whose data were used. Considering the results of several researchers, the effective parameters in apparent shear stress values are channel height (H), total width (B), main channel widths (b), water depth in floodplain (h), floodplain roughness (", null, ") and main channel roughness (", null, "). Therefore, the apparent shear stress is a function of\nUsing Buckingham's theorem, the dimensionless apparent shear stress (", null, ") is a function of six dimensionless parameters as: where", null, "is the hydraulic radius and", null, "is the bed slope. A total of 100 data for each input were extracted from the four mentioned experimental studies. Out of available data, about 70% was used for the training procedure and the rest were reserved for the testing stage. Table 2 shows the statistics of the dimensionless parameters based on the experimental data.\nTable 2\n\nRange of geometric data used for the compound channel\n\nDatasets Variables Minimum Maximum Mean Skewness Kurtosis\nTraining dataset B/b 2.00 6.67 3.97 0.08 0.73\n(H-h)/H 0.02 0.51 0.21 0.53 −1.07\nH/h 1.02 2.02 1.32 1.02 0.02\nBH/bh 2.24 8.10 5.16 0.10 −0.59\nh/b 0.10 2.00 1.05 0.26 −1.84\nH/B 0.02 0.58 0.31 0.05 1.12\nnf/nc 1.00 3.03 1.42 1.28 −0.01\nTesting dataset B/b 2.00 5.26 3.96 −0.54 −0.01\n(H-h)/H 0.04 0.50 0.24 0.73 −0.17\nH/h 1.05 2.00 1.47 1.33 0.89\nBH/bh 2.49 8.00 5.31 −0.26 0.26\nh/b 0.10 2.00 0.97 0.68 −1.52\nH/B 0.00 0.56 0.28 0.28 −1.08\nnf/nc 1.00 2.83 1.37 1.43 1.39\nDatasets Variables Minimum Maximum Mean Skewness Kurtosis\nTraining dataset B/b 2.00 6.67 3.97 0.08 0.73\n(H-h)/H 0.02 0.51 0.21 0.53 −1.07\nH/h 1.02 2.02 1.32 1.02 0.02\nBH/bh 2.24 8.10 5.16 0.10 −0.59\nh/b 0.10 2.00 1.05 0.26 −1.84\nH/B 0.02 0.58 0.31 0.05 1.12\nnf/nc 1.00 3.03 1.42 1.28 −0.01\nTesting dataset B/b 2.00 5.26 3.96 −0.54 −0.01\n(H-h)/H 0.04 0.50 0.24 0.73 −0.17\nH/h 1.05 2.00 1.47 1.33 0.89\nBH/bh 2.49 8.00 5.31 −0.26 0.26\nh/b 0.10 2.00 0.97 0.68 −1.52\nH/B 0.00 0.56 0.28 0.28 −1.08\nnf/nc 1.00 2.83 1.37 1.43 1.39\n\n## MODELLING STRATEGY BACKGROUND\n\n### Random forest model\n\nThe RF model introduced by Breiman (2001), is one of the learning methods that generates many categories and provides the final outcome based on aggregated results. It uses a different bootstrap sample from the data to build a tree for regression (Liaw & Wiener 2002). In addition, it changes how regression trees are constructed. In a random forest, the best among a subset of predictors randomly selected in a node is used to divide that node (Liaw & Wiener 2002). This is also an advantage of the random forest which helps it to perform well compared with other models like support vector machine (SVM) and neural networks. The RF model is robust against over-fitting (Breiman 2001). In the RF model, three parameters require optimisation: (1) the number of different predictors, (2) the number of regression trees, (3) the minimal size of the terminal nodes (Mutanga et al. 2012). Random forest regression performs as shown below (Liaw & Wiener 2002):\n\n1. ntree bootstrap samples are drawn from the original data.\n\n2. An unpruned regression tree is grown for each bootstrap sample with the following modification: sample randomly mtry of the predictors at each node and select the best split among the variables.\n\n3. Aggregate the predictions of the ntree trees to predict new data using averaged values for regression.\n\n### Random tree model\n\nRandom tree is a quick and flexible tree model which has been applied for path planning of robotics and many real-world problems (Haitao et al. 2007). It builds the decision trees on a random subset of columns based on a stochastic process. Random tree works similarly to other traditional decision trees like C45 or J48 except that only a random subset is available for each split (Dehling et al. 2008). Suppose R is defined as a plane rooted tree called a family tree with n nodes, E is defined as a class of a plane rooted tree, thus, each", null, "the size", null, "by the number of nodes R comprises a weight indicated as the following equation (Drmota & Gittenberger 1997): where", null, "is the number of the nodes", null, "without-degree k,", null, "are the non-negative numbers. Let us set", null, "then the corresponding function", null, "must satisfy the functional function as below: where\n\nFinally, equip the sets", null, "with the probability distribution caused by the weight function", null, ".\n\n### REP tree model\n\nA REP tree model which creates a regression tree based on the reduction of variance or increased information (Mohamed et al. 2012) is a fast decision tree method. It has been applied in many studies for solving many real-world problems such as bankruptcy prediction (Cielen et al. 2004), academic performance prediction (Vandamme et al. 2007) and heart disease prediction (Pandey et al. 2013). The training step of the REP tree model is carried out in two main stages as: (1) the regression tree is utilised to generate multiple trees in various iterations and (2) the best tree is selected from the generated trees. The advantage of the REP tree model compared with other decision trees is that it uses the reduced error pruning (REP) technique to prevent the over-fitting problems and handle the missing values (Elomaa & Kaariainen 2001).\n\nSuppose", null, "is explained as the regression tree of which U and V demonstrate the leaf of the tree and the output variable, respectively. The REP tree model constructs a tree with maximum information gain with stopping criteria of the sum of squared errors indicated as follows: where", null, "is known to be the within variable and pc is the class prediction.\n\n### M5P model\n\nM5P model is a machine learning decision tree method which was first presented by Quinlan (1992). It combines a linear regression function and a conventional decision tree to construct the regression tree for prediction. In the M5P tree, the conventional decision tree is used at the node whereas the linear regression function is utilised as the leaves. Generation of the M5P tree is carried out through two main steps, including (1) a decision tree, using the division criteria based on the standard deviation of class values, and (2) the linear regression function for replacing the sub-trees, and the tree is grown by using the pruning process (Pal 2006). In addition, the standard deviation reduction (SD) is used to reduce errors of the M5P algorithm expressed as follows (Sanikhani et al. 2018): where D is inferred as a set of samples which reach the mode of the tree,", null, "is defined as the subsets of samples which have the i-th output of the potential set,", null, "is defined as the standard deviation of D.\n\n### Hybrid model of bagging-M5P\n\nBagging-M5P is a hybrid machine learning model built by combining the bagging ensemble and M5P predictor. The bagging ensemble was proposed by Breiman (1996) and has been applied effectively in many studies to deal with many real-world problems such as ecological prediction (Prasad et al. 2006), Lyme disease risk prediction (Rizzoli et al. 2002) and early prediction of heart disease (Chaurasia & Pal 2013).\n\nThe robustness of the bagging is that it can enhance the performance of the weak predictors, as it can raise the recognition rate of unstable predictors and weaken the defects of component predictors (Breiman 1996). There are three main steps to train the bagging ensemble: (1) choosing independently and randomly the data from the original training dataset, (2) designating the M5P learning algorithm to train the various sub-datasets for gaining the sequence of predictive function, (3) voting for the results and choosing the final outcome with the most votes (Breiman 1996).\n\n### Models’ performance evaluation\n\nThe models' performances are evaluated using six statistical parameters, i.e., coefficient of determination (R2), root mean square error (RMSE), mean absolute error (MAE), Nash–Sutcliffe efficiency (NSE), percentage of BIAS (PBIAS) and the RMSE of the standard deviation of observation ratio (RSR) (Tao et al. 2018). These statistical parameters are defined as: where", null, "and", null, "are observed and predicted values of apparent shear stress, also", null, "and", null, "are the observed and predicted mean value of apparent shear stress, respectively.\n\nThe box plot diagram is a useful tool for understanding the distribution and scattering of data, and is widely used in many cases. The graphical presentation offers instantaneous and prompt information, which is more useful than simply listed data in a table. This approach is a good option in terms of examining the maximum, minimum and other useful statistical analysis information, particularly in comparing the same values in different sets. This chart prevents the misjudgement of data by reference to central parameters such as the average and the median, by allowing both of them and making comparison possible. Therefore, attention is drawn to the problem of data dispersion and variability. Besides the statistical index-based assessment, the box plot analysis is also used to evaluate the performance of the five discussed models.\n\n### Methodology flowchart\n\nMethodology of this study can be carried out in five main steps, namely, (1) preparing the data, (2) pre-processing data, (3) generating the datasets, (4) processing the models and (5) validating the models as shown in Figure 2. A detailed description of these steps is given below.\n\n• (1)\n\nPreparing the data: The data were collected from various credible experimental studies as previously described. As mentioned before, the data include input parameters, namely, channel height (H), total width (B), main channel widths (b), water depth in floodplain (h), floodplain roughness (nf), main channel roughness (nc) and an output variable named the apparent shear stress.\n\n• (2)\n\nPre-processing data or selection of input variables: At first, different dimensionless parameters were considered based on the input and output parameters and next, different input combinations were carried out using correlation coefficient to find the best one and have a higher prediction power.\n\n• (3)\n\nGenerating the datasets: In this step, all input data were divided into two parts. Of these parts, about 70% was used for the training procedure and the rest (30%) was reserved for the testing stage.\n\n• (4)\n\nProcessing the models: In this step, the training dataset was used to construct the models and train them. RF model was constructed with a bag size percentage, batch size, maximum depth of tree, number of decimal places, number of execution slots, number of features, number of iterations and seeds with optimal values of 1, 100, 100, 0, 2, 1, 6, 130 and 1, respectively. Bag size percentage is the size of each bag as a percentage of the training set size, preferred number of instances to process for batch prediction, the number of decimal places to be used for the number output in the model, the number of execution slots (threads) used to construct the ensemble and the number of iterations required:\n\n• 1.\n\nFor RT model, the K value of 1, batch size of 100, maximum depth of tree of 0, minimum number of 1, minimum variance probability of 0.001, number of decimal places of 2, number of folds of 0 and seeds of 3 are used.\n\n• 2.\n\nThe REP tree model was built with the batch size of 100, initial count of 0, maximum depth of −1, minimum number of 1, minimum variance probability of 0.001, number of decimal places of 2, number of folds of 8 and seeds of 3.\n\n• 3.\n\nThe optimal values for M5P model operators, including of batch size, minimum number of instances and number of decimal places were 100, 4 and 2, respectively.\n\n• 4.\n\nThe optimal values of these operators for bagging-M5P model were 20, 100, 2, 0, 10 and 1, for bag size percentage, batch size, number of decimal places, number of execution slots, and number of iterations and seeds, respectively. Finally the test data were predicted by the constructed models.\n\n• (5)\n\nValidating the models: In this step, the models were validated based on the testing datasets. Various statistical methods, namely, R2, RMSE, NSE, RSR and PBIAS were used for validation of the developed models.\n\n## RESULTS AND DISCUSSION\n\n### Selection of input variables\n\nIn order to identify the most effective input variables based on the channel geometry and characteristics, different combinations of these inputs were examined. Based on the primary correlation analysis reported in Table 3, the ratio of (H/B) attained the highest R2 value followed by (h/b). According to the correlation trend, the input combinations were constructed to represent the characteristics of the predictive models (see Table 4). A total of nine input combinations were constructed using those input variables. It is worth mentioning here that the ratio of H/B was incorporated in all constructed input combinations owing to its essential variability to predict the apparent shear stress of the compound channel.\n\nTable 3\n\nThe analysis results of the Pearson correlation coefficient values towards the apparent shear stress\n\nVariables Pearson correlation coefficient\nB/b −0.13\n(H − h)/H 0.074\nH/h 0.008\nBH/bh −0.12\nh/b −0.66\nH/B −0.72\nnf/nc −0.035\nVariables Pearson correlation coefficient\nB/b −0.13\n(H − h)/H 0.074\nH/h 0.008\nBH/bh −0.12\nh/b −0.66\nH/B −0.72\nnf/nc −0.035\nTable 4\n\nThe constructed input combinations for the applied data mining predictive models\n\nInput no. Input variables\nH/B\nH/B, h/b\nH/B, h/b, B/b\nH/B, h/b, B/b, BH/bh\nH/B, h/b, B/b, BH/bh, (H − h)/H\nH/B, h/b, B/b, BH/bh, (H − h)/H, nf/nc\nH/B, h/b, B/b, BH/bh, (H − h)/H, nf/nc, H/h\nB/b, (H − h)/H, nf/nc, h/b\nB/b, H/B, nf/nc, h/b\nInput no. Input variables\nH/B\nH/B, h/b\nH/B, h/b, B/b\nH/B, h/b, B/b, BH/bh\nH/B, h/b, B/b, BH/bh, (H − h)/H\nH/B, h/b, B/b, BH/bh, (H − h)/H, nf/nc\nH/B, h/b, B/b, BH/bh, (H − h)/H, nf/nc, H/h\nB/b, (H − h)/H, nf/nc, h/b\nB/b, H/B, nf/nc, h/b\n\nBased on the constructed input combination in Table 4, a sensitivity analysis was performed to review the appropriate input combination for each developed model. This is accomplished through the calculation of correlation coefficient (CC) and root mean square error (RMSE) for each applied predictive model. The four applied models, i.e., RF, REPT, M5P and bagging-M5P were found to have a consistent behaviour with the combination of input 8: (B/b, (H − h)/H, nf/nc, h/b) as an applicable attribute to predict the apparent shear stress. On the other hand, the RT model was performed accurately using the sixth input combination through incorporating the following parameters (i.e., H/B, h/b, B/b, BH/bh, (H − h)/H, nf/nc) (see Table 5).\n\nTable 5\n\nThe sensitivity examination of the various input combinations to predict the apparent shear stress over all the applied predictive models\n\nModels Criteria Input 1 Input 2 Input 3 Input 4 Input 5 Input 6 Input 7 Input 8 Input 9\nRF CC 0.63 0.65 0.77 0.72 0.76 0.78 0.77 0.91 0.77\nRMSE 1.5 1.4 1.1 1.2 1.1 0.60\nRT CC 0.61 0.61 0.78 0.55 0.67 0.79 0.2 0.72 0.67\nRMSE 1.6 1.6 1.1 1.8 1.5 0.99 2.2 1.3 1.1\nREPT CC 0.67 0.67 0.67 0.68 0.65 0.65 0.65 0.76 0.67\nRMSE 1.5 1.5 1.5 1.5 1.6 1.6 1.6 1.3 1.5\nM5P CC 0.64 0.64 0.67 0.73 0.73 0.73 0.73 0.9 0.67\nRMSE 1.4 1.4 1.4 1.4 1.4 1.3 1.3 0.71 1.4\nBagging M5P CC 0.7 0.71 0.72 0.73 0.72 0.68 0.68 0.86 0.7\nRMSE 1.4 1.3 1.3 1.2 1.2 1.2 1.3 0.87 1.2\nModels Criteria Input 1 Input 2 Input 3 Input 4 Input 5 Input 6 Input 7 Input 8 Input 9\nRF CC 0.63 0.65 0.77 0.72 0.76 0.78 0.77 0.91 0.77\nRMSE 1.5 1.4 1.1 1.2 1.1 0.60\nRT CC 0.61 0.61 0.78 0.55 0.67 0.79 0.2 0.72 0.67\nRMSE 1.6 1.6 1.1 1.8 1.5 0.99 2.2 1.3 1.1\nREPT CC 0.67 0.67 0.67 0.68 0.65 0.65 0.65 0.76 0.67\nRMSE 1.5 1.5 1.5 1.5 1.6 1.6 1.6 1.3 1.5\nM5P CC 0.64 0.64 0.67 0.73 0.73 0.73 0.73 0.9 0.67\nRMSE 1.4 1.4 1.4 1.4 1.4 1.3 1.3 0.71 1.4\nBagging M5P CC 0.7 0.71 0.72 0.73 0.72 0.68 0.68 0.86 0.7\nRMSE 1.4 1.3 1.3 1.2 1.2 1.2 1.3 0.87 1.2\n\nThe attained results in Table 5 indicate two facts: (i) the initiated input combinations based on the Pearson correlation coefficient are reliably suitable for predicting the", null, "due to the consistency of the input variables. However, (ii) RT acted differently using another order of input variables' information and this is somehow very normal as those stochastic models visualised the regression problem from one case to another in different manners.\n\n### Evaluation of models in predicting apparent shear stress\n\nFollowing several prediction researches conducted in the literature and within the hydraulic engineering perspective, several performance metrics preferably should be conducted for the predictability assessment (Chadalawada & Babovic 2019). Based on the quantitative examination, different prediction skills metrics computed the predictability of the proposed data mining models including R2, RMSE, MAE, NSE, PBIAS and RSR. Multiple indicators for prediction accuracy assessment give a more comprehensive vision of the capacity of the developed models. In general, it was observed that the applied models showed good predictability performance, as shown in Table 6. However, all analysed statistical parameters showed that the bagging-M5P model performed with superior prediction accuracy in comparison with the other models. The coefficient of determination (R2) exhibited a very good index between the simulated and the actual experiment", null, ". The results were presented based on their performance, starting with the highest, the integrated bagging-M5P (R2 = 0.92), M5P (R2 = 0.90), RF (R2 = 0.80), REPT (R2 = 0.73), RT (R2 = 0.72). In harmony with the displayed scatter plot (see Figure 3), bagging-M5P demonstrated a reliable predictive model with minimum diversion from the best fit line for all the range of", null, "data (0–5). As a matter of fact, R2 value is a sensitive indicator towards the outlier observations (Legates & Mccabe 1999; Yaseen et al. 2016), and other indicators are required to be evaluated. With respect to the absolute error metrics, the integrated bagging-M5P model displayed the minimum values of RMSE = 0.46 and MAE = 0.32 in comparison with the other applied data mining predictive models. Another excellent performance metric was computed that measures the overestimation (i.e., negative PBIAS value) and underestimation (i.e., positive PBIAS value) (Moriasi et al. 2007). Based on the reported statistical results, the bagging-M5P and REPT showed overestimation values, with slightly higher magnitude of the bagging-M5P model; whereas the other models indicated underestimation behaviour. Other metrics (e.g., NSE and RSR) behaved similarly in indicating the superiority of the integrated bagging-M5P model over the other data mining models.\n\nTable 6\n\nThe performance metrics results of the applied data mining predictive models over the test modelling phase\n\nModels R2 RMSE MAE NSE PBIAS RSR\nRF 0.805 0.698 0.457 0.802 2.242 0.441\nRT 0.727 0.876 0.565 0.689 14.404 0.556\nREPT 0.732 0.908 0.673 0.666 −14.160 0.577\nM5P 0.905 0.487 0.334 0.903 2.073 0.309\nBagging-M5P 0.920 0.460 0.320 0.914 −5.517 0.292\nModels R2 RMSE MAE NSE PBIAS RSR\nRF 0.805 0.698 0.457 0.802 2.242 0.441\nRT 0.727 0.876 0.565 0.689 14.404 0.556\nREPT 0.732 0.908 0.673 0.666 −14.160 0.577\nM5P 0.905 0.487 0.334 0.903 2.073 0.309\nBagging-M5P 0.920 0.460 0.320 0.914 −5.517 0.292\nFigure 3\n\nPredicted and observed apparent shear stress time series and scatter plot presentation using all the applied data mining predictive models for the best input combination and over the test modeling phase.\n\nFigure 3\n\nPredicted and observed apparent shear stress time series and scatter plot presentation using all the applied data mining predictive models for the best input combination and over the test modeling phase.\n\nThe time series and scatter plot presentation between the observed experimental and predicted shear (for the validation phase) are displayed in Figure 3. The context of the results presented in Figure 3 somewhat follows the reported statistical performance in Table 6. The bagging-M5P model performed the best prediction capacity in terms of agreement between the observed and predicted values. The scatter plot is useful in visually reflecting the variation of prediction values around the ideal line. It can be said that all models (except for REPT) have relatively good accuracy in predicting low apparent shear stress, particularly for", null, ". Most of the predicted values fall on the best fit line 45°. However, it is evident that the high apparent shear stresses were not very well predicted using RF, RT, REPT and M5P models. REPT model visibly did not perform well where all higher", null, "(that is >2) have a consistent value of 4. Having said this, even at lower range of 1 <", null, "< 3, the prediction values were fixed at ≈1.5 and 0.4 for", null, "< 1. Although the RF, RT and M5P models may capture the dynamic fluctuating behaviour, the predicted values are either overestimated or underestimated in a rather large envelope along the line of agreement (with RT model observably having the widest stretch). Out of the five models discussed here, only the bagging-M5P model managed to mimic those high values with better accuracy.\n\nAnother important graphical visualisation generated is the box plot. Figure 4 shows the capability of the applied predictive models in the form of box plots. Based on the displayed statistical presentation with respect to the medians, quartiles and data ranges, the bagging-M5P model attained the best performance to the observed data pattern with slight variation as seen in box plots (Figure 4). This is followed by the M5P and RF models based on their capability to have a close prediction with the observed data.\n\nBased on the attained predictability performance of the proposed data mining, clearly evidenced is the potential of data mining for simulating the apparent shear stress of compound channels. Data mining models revealed the opportunity to be integrated for channel design, management and sustainability. Finally, it is worth highlighting the possibility for future research application. Incorporating other experimental methods from the literature related to the apparent shear stress with the other channel characteristics might enhance the predictability of the proposed data mining models. In addition, integrating nature-inspired optimisation algorithms (Yang 2013, 2014; Ajay Adithyan et al. 2018) is highly recommended as a prior stage for the prediction process where the most anticipated variables toward the shear value can be abstracted and fed as reliable input attributes for the prediction model.\n\n## CONCLUSION\n\nApparent shear stress is a vital component in prior channel design and thus it is highly emphasised that it should be quantified with accurate and reliable magnitude, in particular for flood design. In this study, five different versions of new data mining (i.e., RF, RT, REPT, M5P and bagging-M5P) were used. The models were constructed based on several channel properties' variables built in nine input combinations. The four models, RF, REPT, M5P and bagging-M5P behaved similarly to the best input combination 8 using B/b, (H − h)/H, nf/nc and h/b. On the other hand, the RT model attained its best results using input combination 6 based on H/B, h/b, B/b, BH/bh, (H − h)/H and nf/nc. After selecting the best input combination, the predictions of apparent shear stress using all mentioned models were compared. It was found that the models' results improved significantly when a hybrid model was used, such that the bagging-M5P with lower statistical parameter values (R2 of 0.92) demonstrated better performance than those of the RF, RT, REPT and M5P models. 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C.\n,\nKisi\nO.\n,\nJ.\n,\nQuilty\nJ.\n&\nEl-shafie\nA.\n2016\nStream-flow forecasting using extreme learning machines: a case study in a semi-arid region in Iraq\n.\nJournal of Hydrology\n542\n,\n603\n614\n.\ndoi:10.1016/j.jhydrol.2016.09.035\n." ]	[ null, "https://iwa.silverchair-cdn.com/iwa/content_public/journal/jh/21/5/10.2166_hydro.2019.037/1/m_hydro-d-19-00037if33.gif", null, "https://iwa.silverchair-cdn.com/iwa/content_public/journal/jh/21/5/10.2166_hydro.2019.037/1/m_hydro-d-19-00037if34.gif", null, "https://iwa.silverchair-cdn.com/iwa/content_public/journal/jh/21/5/10.2166_hydro.2019.037/1/m_hydro-d-19-00037if35.gif", null, "https://iwa.silverchair-cdn.com/iwa/content_public/journal/jh/21/5/10.2166_hydro.2019.037/1/m_hydro-d-19-00037if36.gif", null, "https://iwa.silverchair-cdn.com/iwa/content_public/journal/jh/21/5/10.2166_hydro.2019.037/1/m_hydro-d-19-00037if37.gif", null, 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https://math.stackexchange.com/questions/2905445/finding-the-area-of-an-equilateral-triangle-using-the-pythagorean-theorem	[ "# Finding the area of an equilateral triangle using the Pythagorean theorem\n\nFrom an equilateral triangle $T$ where each side have a length of $L$. What is the area of $T$?\n\nAccording to the Wikipedia page of equilateral triangles, the area is $$A=\\dfrac{\\sqrt{3}}{4}L^2$$\n\nI am trying to solve this problem by using the Pythagorean theorem, as explained in this question, I can split the triangle in half to try and get the height.\n\nUsing the Pythagorean theorem, $$L^2=(\\dfrac{L}{2})^2 + H^2$$\n\nI can then isolate $H$ with :\n\n$$H=\\sqrt{L^2-(\\dfrac{L}{2})^2}$$\n\nUsing the $A=\\dfrac{1}{2}bh$ formula. I could then conclude with : $$A=\\dfrac{L\\sqrt{L^2-(\\dfrac{L}{2})^2}}{2}$$\n\nAs said previously, the Wikipedia page shows something very different. What went wrong?\n\n• Why do you think those are different? – Matthew Leingang Sep 4 '18 at 19:14\n• Factor the $L^2$ then pass it out. – Randall Sep 4 '18 at 19:15\n• $L^2-(L/2)^2=\\frac{3L^2}{4}$ – Vasya Sep 4 '18 at 19:16\n• You can get properly sized parentheses that adjust to their content by preceding them with \\left and \\right. Please see this tutorial and reference on how to typeset math on this site. – joriki Sep 4 '18 at 20:08\n• @Cedric Martens I corrected a stupid mistake in my answer. (Tired when I posted, I forgot we had $2$ right triangles. Please look at it again and see if it helps. – poetasis May 22 '19 at 16:33\n\nWith the Pythagorean theorem, you can find the altitude of an equilateral triangle by dropping a vertical line to split it. Then long side and half of the bottom are, respectively, the hypotenuse $$C$$ and a short leg $$B$$ of a right triangle. The vertical will be side $$A$$ of a right triangle where $$A=\\sqrt{C^2-\\bigl{(}\\frac{C}{2}\\bigr{)}^2}=\\sqrt{\\frac{4C^2-C^2}{2}}=\\frac{C\\sqrt{3}}{2}$$. The area of one triangle is $$\\frac{1}{2}AB$$ where $$B=\\frac{C}{2}$$. However, we have two of these right triangles so the area is simply AB. You should end up with $$area=A2B=AC=\\frac{C\\sqrt{3}}{2}C=\\frac{C^2\\sqrt{3}}{2}$$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88494855,"math_prob":0.99962735,"size":814,"snap":"2020-45-2020-50","text_gpt3_token_len":262,"char_repetition_ratio":0.11728395,"word_repetition_ratio":0.0,"special_character_ratio":0.31572482,"punctuation_ratio":0.09356725,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999919,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-30T18:18:10Z\",\"WARC-Record-ID\":\"<urn:uuid:fb573a12-2ff1-4c82-8727-b927f5283886>\",\"Content-Length\":\"156030\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:772aec88-eecb-4496-baf0-b2293babb1f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:2af01ae0-f2b5-478c-9162-d35f252857f1>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/2905445/finding-the-area-of-an-equilateral-triangle-using-the-pythagorean-theorem\",\"WARC-Payload-Digest\":\"sha1:WSIH4VU2ICA4OGCE75VYN65G636ETQMP\",\"WARC-Block-Digest\":\"sha1:6THD4CAUQ74DMZMPFCFZN6SCJV56GMZT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141216897.58_warc_CC-MAIN-20201130161537-20201130191537-00530.warc.gz\"}"}
https://math.stackexchange.com/questions/745833/logarithmic-equations-and-solving-for-the-unknown-variable	[ "# Logarithmic equations and solving for the unknown variable\n\nWhat is e^xe^(x+1)=e^2 ? The e is the natural number 2.718... not the variable e. My teacher doesn't give a lot of notes so I am not sure if I multiply the two exponents on the left side, being x(x+1), or if i add them.\n\n$$e^x\\times e^{x+1}=e^{x+x+1}=e^{2x+1}$$ If $e^x\\times e^{x+1}=e^2$, then, by the above, $$2x+1=2\\\\ \\implies \\boxed{x=\\dfrac{1}{2}=0.5}$$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6025116,"math_prob":0.99999595,"size":436,"snap":"2021-43-2021-49","text_gpt3_token_len":179,"char_repetition_ratio":0.1087963,"word_repetition_ratio":0.0,"special_character_ratio":0.42889908,"punctuation_ratio":0.13043478,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000044,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-01T12:55:25Z\",\"WARC-Record-ID\":\"<urn:uuid:8aaf4b14-ae0c-498b-8888-49b37961d20d>\",\"Content-Length\":\"129152\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5266ca6d-62d6-41bb-9b37-36dc94f01212>\",\"WARC-Concurrent-To\":\"<urn:uuid:937c7f1a-f65a-47f8-bef5-67e85da6afd2>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/745833/logarithmic-equations-and-solving-for-the-unknown-variable\",\"WARC-Payload-Digest\":\"sha1:KTJI453I66RKJBW3Q5ALRR25GTP5P6PS\",\"WARC-Block-Digest\":\"sha1:DUUU5RRD6JUSA3CWRTBT27KIWOADNALD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964360803.0_warc_CC-MAIN-20211201113241-20211201143241-00384.warc.gz\"}"}
https://alpenaschools.k12.ar.us/310374_3	[ "Skip to main content\n\n## Physics\n\n February 13-17, 2020 CONTENT STANDARD TOPIC ASSESSMENT Monday NGSS STANDARD P-PS2-1 Analyze data to support the claim that Newton’s second law of motion describes the mathematical relationship among the net force on a macroscopic object, its mass, and its acceleration. Everyday Forces -Explain the difference between mass and weight. Use coefficients of friction to calculate frictional force. Classwork: 1. 4D2 Practice Tuesday NGSS STANDARD P-PS2-1 Everyday Forces -Explain the difference between mass and weight. Calculate the acceleration of an object due to a net force. Classwork: 1. 4E1 Practice Wednesday NGSS STANDARD P-PS2-1 Everyday Forces -Explain the difference between mass and weight. Calculate the acceleration of an object due to a net force. Classwork: 1. 4E2 Practice Thursday NGSS STANDARD P-PS2-1 Everyday Forces -Explain the difference between mass and weight. Calculate the acceleration of an object due to a net force. Classwork: 1. Chapter 4 Section 3 Quiz Friday NGSS STANDARD P-PS2-1 Everyday Forces -Explain the difference between mass and weight. Calculate the acceleration of an object due to a net force. Classwork: 1. Chapter 4 Review" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80639,"math_prob":0.9210913,"size":1215,"snap":"2021-04-2021-17","text_gpt3_token_len":328,"char_repetition_ratio":0.1073493,"word_repetition_ratio":0.53807104,"special_character_ratio":0.24444444,"punctuation_ratio":0.112676054,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9885202,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-17T18:10:33Z\",\"WARC-Record-ID\":\"<urn:uuid:07ac3279-2930-4dac-b4df-d2db7ec85518>\",\"Content-Length\":\"221898\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8161bf28-1bd4-41c2-87f4-144fe498e129>\",\"WARC-Concurrent-To\":\"<urn:uuid:8f40ecbe-9397-4578-b362-19373294cc78>\",\"WARC-IP-Address\":\"54.208.31.49\",\"WARC-Target-URI\":\"https://alpenaschools.k12.ar.us/310374_3\",\"WARC-Payload-Digest\":\"sha1:AHUMSZFE4RSXYGM4S4T3BABQRJSO3LH4\",\"WARC-Block-Digest\":\"sha1:TPYVBSFJ4VQNBEE6NTRAAMLNVIZAPTYT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038461619.53_warc_CC-MAIN-20210417162353-20210417192353-00154.warc.gz\"}"}
https://en.wikibooks.org/wiki/Advanced_Mathematics_for_Engineers_and_Scientists/Finite_Difference_Method	[ "# Advanced Mathematics for Engineers and Scientists/Finite Difference Method\n\n## Finite Difference Method\n\nThe finite difference method is a basic numeric method which is based on the approximation of a derivative as a difference quotient. We all know that, by definition:\n\n$u'(x)=\\lim _{\\Delta x\\to 0}{\\frac {u(x+\\Delta x)-u(x)}{\\Delta x}}$", null, "The basic idea is that if $\\Delta x$", null, "is \"small\", then\n\n$u'(x)\\approx {\\frac {u(x+\\Delta x)-u(x)}{\\Delta x}}$", null, "Similarly,\n\n$u''(x)=\\lim _{\\Delta x\\to 0}{\\frac {u(x+\\Delta x)-2u(x)+u(x-\\Delta x)}{\\Delta x^{2}}}$", null, "$u''(x)\\approx {\\frac {u(x+\\Delta x)-2u(x)+u(x-\\Delta x)}{\\Delta x^{2}}}$", null, "It's a step backwards from calculus. Instead of taking the limit and getting the exact rate of change, we approximate the derivative as a difference quotient. Generally, the \"difference\" showing up in the difference quotient (ie, the quantity in the numeriator) is called a finite difference which is a discrete analog of the derivative and approximates the $n^{\\text{th}}$", null, "derivative when divided by $\\Delta x^{n}$", null, ".\n\nReplacing all of the derivatives in a differential equation ditches differentiation and results in algebraic equations, which may be coupled depending on how the discretization is applied.\n\nFor example, the equation\n\n${\\frac {\\partial u}{\\partial t}}={\\frac {\\partial ^{2}u}{\\partial x^{2}}}$", null, "may be discretized into:\n\n${\\frac {u(x,t+\\Delta t)-u(x,t)}{\\Delta t}}={\\frac {u(x+\\Delta x,t)-2u(x,t)+u(x-\\Delta x,t)}{\\Delta x^{2}}}$", null, "${\\Big \\Downarrow }$", null, "$u(x,t+\\Delta t)=u(x,t)+{\\frac {\\Delta t}{\\Delta x^{2}}}(u(x+\\Delta x,t)-2u(x,t)+u(x-\\Delta x,t))$", null, "This discretization is nice because the \"next\" value (temporally) may be expressed in terms of \"older\" values at different positions." ]	[ null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/14b6f0b0d47b29e06d32d1fe8360c413d7fb8f45", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/f3890eb866b6258d7a304fc34c70ee3fb3a81a70", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/3d01f9b3006cb25b5e2188efe66757087c7e3f06", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/35e924f9c85ab6a2baaa01819d42aaf544e8d4d3", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/82e9e86cf8af82df86657473a4862bca43d78dbe", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/836d593843b7806e6a1875098200bce01bdeaec6", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/f589c0421d30b6ec6ac6da943de781b93666d9cc", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/06787984d3e4f56b5baa23f8fa95f9237971deb8", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/61f2c6bddf7ef33b72bb69bb81725eac7f2f3ec6", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/a6cc61a63af1d5a6ed8ebcd9eaaad129ecad5193", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/acd32dd852145f0b50f39cf90075116ca2ef7075", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.937377,"math_prob":0.9999052,"size":980,"snap":"2019-51-2020-05","text_gpt3_token_len":193,"char_repetition_ratio":0.15983607,"word_repetition_ratio":0.013245033,"special_character_ratio":0.18673469,"punctuation_ratio":0.09411765,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000063,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22],"im_url_duplicate_count":[null,1,null,null,null,1,null,1,null,1,null,null,null,1,null,1,null,1,null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-22T04:34:06Z\",\"WARC-Record-ID\":\"<urn:uuid:1737a90b-b9df-4cbe-a73e-7103e3d547b9>\",\"Content-Length\":\"46008\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:95c9650e-ef60-4745-bf6a-c640776f081a>\",\"WARC-Concurrent-To\":\"<urn:uuid:8edac3ec-a20f-4fc6-bcb6-1485559d5499>\",\"WARC-IP-Address\":\"208.80.154.224\",\"WARC-Target-URI\":\"https://en.wikibooks.org/wiki/Advanced_Mathematics_for_Engineers_and_Scientists/Finite_Difference_Method\",\"WARC-Payload-Digest\":\"sha1:BL6ZYJN34KKFDY2DNFN7U5L4ZVNHCJYW\",\"WARC-Block-Digest\":\"sha1:LGPZDEGEMXGVD2VRW4CW3XO4ANAX52FY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250606696.26_warc_CC-MAIN-20200122042145-20200122071145-00387.warc.gz\"}"}
https://dxr.mozilla.org/mozilla-central/source/toolkit/components/utils/Sampling.jsm	[ "DXR is a code search and navigation tool aimed at making sense of large projects. It supports full-text and regex searches as well as structural queries.\n\n#### Mercurial (c68fe15a81fc)\n\nLine Code\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215\n``````/* This Source Code Form is subject to the terms of the Mozilla Public\n`````` * License, v. 2.0. If a copy of the MPL was not distributed with this\n`````` * file, You can obtain one at http://mozilla.org/MPL/2.0/. /\n``````\n``````\"use strict\";\n``````\"use strict\";\n``````\n``````const { XPCOMUtils } = ChromeUtils.import(\n`````` \"resource://gre/modules/XPCOMUtils.jsm\"\n`````` \"resource://gre/modules/XPCOMUtils.jsm\"\n``````);\n``````);\n``````XPCOMUtils.defineLazyGlobalGetters(this, [\"crypto\", \"TextEncoder\"]);\n``````\n``````var EXPORTED_SYMBOLS = [\"Sampling\"];\n``````\n``````const hashBits = 48;\n``````const hashBits = 48;\n``````const hashLength = hashBits / 4; // each hexadecimal digit represents 4 bits\n``````const hashMultiplier = Math.pow(2, hashBits) - 1;\n``````\n``````var Sampling = {\n`````` /\n`````` /\n`````` Map from the range [0, 1] to [0, 2^48].\n`````` * @param {number} frac A float from 0.0 to 1.0.\n`````` * @return {string} A 48 bit number represented in hex, padded to 12 characters.\n`````` /\n`````` fractionToKey(frac) {\n`````` fractionToKey(frac) {\n`````` if (frac < 0 \|\| frac > 1) {\n`````` throw new Error(`frac must be between 0 and 1 inclusive (got \\${frac})`);\n`````` }\n``````\n`````` return Math.floor(frac hashMultiplier)\n`````` return Math.floor(frac * hashMultiplier)\n`````` .toString(16)\n`````` .padStart(hashLength, \"0\");\n`````` },\n`````` },\n``````\n`````` /*\n`````` @param {ArrayBuffer} buffer Data to convert\n`````` * @returns {String} `buffer`'s content, converted to a hexadecimal string.\n`````` /\n`````` /\n`````` bufferToHex(buffer) {\n`````` const hexCodes = [];\n`````` const view = new DataView(buffer);\n`````` for (let i = 0; i < view.byteLength; i += 4) {\n`````` // Using getUint32 reduces the number of iterations needed (we process 4 bytes each time)\n`````` // Using getUint32 reduces the number of iterations needed (we process 4 bytes each time)\n`````` const value = view.getUint32(i);\n`````` // toString(16) will give the hex representation of the number without padding\n`````` hexCodes.push(value.toString(16).padStart(8, \"0\"));\n`````` }\n``````\n``````\n`````` // Join all the hex strings into one\n`````` return hexCodes.join(\"\");\n`````` },\n``````\n`````` /\n`````` /\n`````` * Check if an input hash is contained in a bucket range.\n`````` \n`````` isHashInBucket(fractionToKey(0.5), 3, 6, 10) -> returns true\n`````` \n`````` minBucket\n`````` * minBucket\n`````` * \| hash\n`````` * v v\n`````` * [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]\n`````` * ^\n`````` * maxBucket\n`````` \n`````` @param inputHash {String}\n`````` * @param inputHash {String}\n`````` * @param minBucket {int} The lower boundary, inclusive, of the range to check.\n`````` * @param maxBucket {int} The upper boundary, exclusive, of the range to check.\n`````` * @param bucketCount {int} The total number of buckets. Should be greater than\n`````` * or equal to maxBucket.\n`````` /\n`````` isHashInBucket(inputHash, minBucket, maxBucket, bucketCount) {\n`````` const minHash = Sampling.fractionToKey(minBucket / bucketCount);\n`````` const maxHash = Sampling.fractionToKey(maxBucket / bucketCount);\n`````` const maxHash = Sampling.fractionToKey(maxBucket / bucketCount);\n`````` return minHash <= inputHash && inputHash < maxHash;\n`````` },\n``````\n`````` /\n`````` @promise A hash of `data`, truncated to the 12 most significant characters.\n`````` * @promise A hash of `data`, truncated to the 12 most significant characters.\n`````` /\n`````` async truncatedHash(data) {\n`````` const hasher = crypto.subtle;\n`````` const input = new TextEncoder(\"utf-8\").encode(JSON.stringify(data));\n`````` const hash = await hasher.digest(\"SHA-256\", input);\n`````` const hash = await hasher.digest(\"SHA-256\", input);\n`````` // truncate hash to 12 characters (2^48), because the full hash is larger\n`````` // than JS can meaningfully represent as a number.\n`````` return Sampling.bufferToHex(hash).slice(0, 12);\n`````` },\n``````\n``````\n`````` /\n`````` Sample by splitting the input into two buckets, one with a size (rate) and\n`````` * another with a size (1.0 - rate), and then check if the input's hash falls\n`````` * into the first bucket.\n`````` \n`````` \n`````` * @param {object} input Input to hash to determine the sample.\n`````` * @param {Number} rate Number between 0.0 and 1.0 to sample at. A value of\n`````` * 0.25 returns true 25% of the time.\n`````` * @promises {boolean} True if the input is in the sample.\n`````` /\n`````` /\n`````` async stableSample(input, rate) {\n`````` const inputHash = await Sampling.truncatedHash(input);\n`````` const samplePoint = Sampling.fractionToKey(rate);\n``````\n`````` return inputHash < samplePoint;\n`````` },\n``````\n`````` /*\n`````` Sample by splitting the input space into a series of buckets, and checking\n`````` * Sample by splitting the input space into a series of buckets, and checking\n`````` * if the given input is in a range of buckets.\n`````` \n`````` The range to check is defined by a start point and length, and can wrap\n`````` * around the input space. For example, if there are 100 buckets, and we ask to\n`````` * check 50 buckets starting from bucket 70, then buckets 70-99 and 0-19 will\n`````` * check 50 buckets starting from bucket 70, then buckets 70-99 and 0-19 will\n`````` * be checked.\n`````` \n`````` @param {object} input Input to hash to determine the matching bucket.\n`````` * @param {integer} start Index of the bucket to start checking.\n`````` * @param {integer} count Number of buckets to check.\n`````` * @param {integer} count Number of buckets to check.\n`````` * @param {integer} total Total number of buckets to group inputs into.\n`````` * @promises {boolean} True if the given input is within the range of buckets\n`````` * we're checking. /\n`````` async bucketSample(input, start, count, total) {\n`````` const inputHash = await Sampling.truncatedHash(input);\n`````` const inputHash = await Sampling.truncatedHash(input);\n`````` const wrappedStart = start % total;\n`````` const end = wrappedStart + count;\n``````\n`````` // If the range we're testing wraps, we have to check two ranges: from start\n`````` // to max, and from min to end.\n`````` // to max, and from min to end.\n`````` if (end > total) {\n`````` return (\n`````` Sampling.isHashInBucket(inputHash, 0, end % total, total) \|\|\n`````` Sampling.isHashInBucket(inputHash, wrappedStart, total, total)\n`````` );\n`````` }\n``````\n``````\n`````` return Sampling.isHashInBucket(inputHash, wrappedStart, end, total);\n`````` },\n``````\n`````` /\n`````` Sample over a list of ratios such that, over the input space, each ratio\n`````` * Sample over a list of ratios such that, over the input space, each ratio\n`````` * has a number of matches in correct proportion to the other ratios.\n`````` \n`````` For example, given the ratios:\n`````` * For example, given the ratios:\n`````` \n`````` [1, 2, 3, 4]\n`````` \n`````` 10% of all inputs will return 0, 20% of all inputs will return 1, 30% will\n`````` * return 2, and 40% will return 3. You can determine the percent of inputs\n`````` * return 2, and 40% will return 3. You can determine the percent of inputs\n`````` * that will return an index by dividing the ratio by the sum of all ratios\n`````` * passed in. In the case above, 4 / (1 + 2 + 3 + 4) == 0.4, or 40% of the\n`````` * inputs.\n`````` \n`````` @param {object} input\n`````` * @param {object} input\n`````` * @param {Array<integer>} ratios\n`````` * @promises {integer}\n`````` * Index of the ratio that matched the input\n`````` * @rejects {Error}\n`````` * @rejects {Error}\n`````` * If the list of ratios doesn't have at least one element\n`````` */\n`````` async ratioSample(input, ratios) {\n`````` if (ratios.length < 1) {\n`````` throw new Error(\n`````` throw new Error(\n`````` `ratios must be at least 1 element long (got length: \\${ratios.length})`\n`````` );\n`````` }\n``````\n`````` const inputHash = await Sampling.truncatedHash(input);\n`````` const inputHash = await Sampling.truncatedHash(input);\n`````` const ratioTotal = ratios.reduce((acc, ratio) => acc + ratio);\n``````\n`````` let samplePoint = 0;\n`````` for (let k = 0; k < ratios.length - 1; k++) {\n`````` samplePoint += ratios[k];\n`````` samplePoint += ratios[k];\n`````` if (inputHash <= Sampling.fractionToKey(samplePoint / ratioTotal)) {\n`````` return k;\n`````` }\n`````` }\n``````\n``````\n`````` // No need to check the last bucket if the others didn't match.\n`````` return ratios.length - 1;\n`````` return ratios.length - 1;\n`````` },\n``````};\n``````" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5630728,"math_prob":0.9729201,"size":7711,"snap":"2020-45-2020-50","text_gpt3_token_len":2330,"char_repetition_ratio":0.1437654,"word_repetition_ratio":0.17439294,"special_character_ratio":0.36999092,"punctuation_ratio":0.15539183,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9947978,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-24T03:59:57Z\",\"WARC-Record-ID\":\"<urn:uuid:45ba9e96-f3ce-4998-9982-791ad98622e6>\",\"Content-Length\":\"131804\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:42a758bf-cca0-4553-b609-a40fd3eba9da>\",\"WARC-Concurrent-To\":\"<urn:uuid:013d5934-bfd1-4131-a2cd-a7ac2e636369>\",\"WARC-IP-Address\":\"63.245.208.205\",\"WARC-Target-URI\":\"https://dxr.mozilla.org/mozilla-central/source/toolkit/components/utils/Sampling.jsm\",\"WARC-Payload-Digest\":\"sha1:UTYA6BHWUZL476Z2L7254YSEAJIRAFF2\",\"WARC-Block-Digest\":\"sha1:XLF6ITEJKCN7TC3RVLBZ74FUCJJLWWS6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107881640.29_warc_CC-MAIN-20201024022853-20201024052853-00047.warc.gz\"}"}
https://resources.illuminateed.com/playlist/resource-sview/id/539f4320f07787e05ae148d2/rid/539f58ddf07787c45be148d6/bc0/user/bc1/playlist/bc0_id/51c48c2607121cee71a57150	[ "Resource VIDEO: Introduction to Mixed Numbers\nVIDEO: Introduction to Mixed Numbers\n• Created By Ashley Scott\n• ##### Resource Playlists\n• Description:\nThis video from Math Antics briefly (7:42 sec) gives students an introduction to the concept of a mixed number. By the end of this video, students should understand that a mixed number is the same as a whole number paired with a proper fraction. Students should also be able to represent mixed numbers using a number line and a ruler.\n• Purpose:\nTo understand and explain the concept of a mixed number.\n• Targets: Student, Parent, Teacher\n• Standards:\n• CACCCS.MA.3.3.NF.2.a" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85645366,"math_prob":0.7059037,"size":586,"snap":"2019-35-2019-39","text_gpt3_token_len":165,"char_repetition_ratio":0.13402061,"word_repetition_ratio":0.041666668,"special_character_ratio":0.2662116,"punctuation_ratio":0.21167883,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9885528,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-21T09:14:16Z\",\"WARC-Record-ID\":\"<urn:uuid:d1120e24-a65a-41e9-a131-e71eb6cc691e>\",\"Content-Length\":\"59367\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eef7e37c-624b-4913-a55f-c55f2466276d>\",\"WARC-Concurrent-To\":\"<urn:uuid:de752d67-c9af-424a-86b6-184239df7d46>\",\"WARC-IP-Address\":\"104.17.75.54\",\"WARC-Target-URI\":\"https://resources.illuminateed.com/playlist/resource-sview/id/539f4320f07787e05ae148d2/rid/539f58ddf07787c45be148d6/bc0/user/bc1/playlist/bc0_id/51c48c2607121cee71a57150\",\"WARC-Payload-Digest\":\"sha1:YASVUQNELU6AF4KOKCB4H6SSMFA57L2B\",\"WARC-Block-Digest\":\"sha1:AJF2BQM3HBYMV7ZS77B2ZNBP5W3TCSCV\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514574377.11_warc_CC-MAIN-20190921084226-20190921110226-00147.warc.gz\"}"}
https://www.icserankers.com/2020/11/icse-solutions-for-chapter1-periodic-table-periodic-properties-and-variations-of-properties-class10-chemistry.html	[ "### Intext Questions 1\n\n1. a) State modern periodic law. Name the scientist who stated the law.\nb) What is a periodic table? How many groups and periods does modern periodic table have?\n\nSolution\na) The modern periodic law states that \"The properties of elements are the periodic functions of their atomic number.\" Henry Moseley put forward the modern periodic law.\n\nb) A tabular arrangement of the elements in groups (vertical columns) and periods (horizontal rows) highlighting the regular trends in properties of elements is called a Periodic Table. Modern Periodic table has 7 periods and 18 groups.\n\n2. Why sodium element of group 1 and chlorine element of group 17 both have valency 1?\n\nSolution\nValency is the combining capacity of the atom of an element. It is equal to the number of electrons an atom can donate or accept or share. It is just a number and does not have a positive or negative sign.\nGroup 1 elements have 1 electron in their outermost orbital, while Group 7 elements have 7 electrons in their outermost orbital.\nValency depends on the number of electrons in the outermost shell (i.e. valence shell). If the number of electrons present in the outermost shell is 1, then it can donate one electron while combining with other elements to obtain a stable electronic configuration. If the number of electrons present in the outermost shell is 7, then its valency is again 1 (8 - 7 = 1) as it can accept 1 electron from the combining atom.\nIn a given period, the number of electrons in the valence (outermost) shell increases from left to right. But the valency increases only up to Group 14, where it becomes 4, and then it decreases, that is, it becomes 1 in Group 17.\n\n3. What are horizontal rows and vertical columns in a periodic table known as ?\n\nSolution\nThe horizontal rows are known as periods and vertical columns in the periodic table are known as groups\n\n4. Periodicity is observed due to the similar ..........\n(Number of valence electrons / atomic number / electronic configuration).\n\nSolution\nPeriodicity is observed due to the similar electronic configuration.\n(Number of valence electrons/atomic number/electronic configuration).\n\n5. How does the electronic configuration in atoms change\n(i) in a period from left to right?\n(ii) in a group top to bottom?\n\nSolution\n(i) Though the number of shells remain the same, number of valence electrons increases by one, as we move across any given period from left to right.\n(ii) While going from top to bottom in a group, the number of shells increases successively i.e. one by one but the number of valence electrons remains the same.\n\n6. Correct the statements.\n(i) Elements in the same periods have equal valency.\n(ii) Valency depends upon the number of shells in an atom.\n(iii) Copper and zinc are representative elements.\n(iv) Transition elements are placed at extreme right of the periodic table.\n\nSolution\n(i) Elements in the same group have equal valency.\n(ii) Valency depends upon the number of valence electrons in an atom.\n(iii) Copper and zinc are transition elements.\n(iv) Noble gases are placed at the extreme right of the periodic table.\n\n7. Name two elements in each case:\n (i) Alkali metals (ii) Alkaline earth metals (iii) Halogens (iv) Inert gas (v) Transition element (vi) Lanthanides (vii) Actinides\n\nSolution\n (i) Alkali metals Sodium and Potassium (ii) Alkaline earth metals Calcium and Magnesium (iii) Halogens Chlorine and Bromine (iv) Inert gas Neon and Argon (v) Transition element Iron and Cobalt (vi) Lanthanides Cerium and Europium (vii) Actinides Uranium and Neptunium\n\n8. What do you understand by?\n(i) Periodicity:\n(ii) Typical elements:\n(iii) Orbits:\n\nSolution\n(i) The properties that reappear at regular intervals, or in which there is a gradual variation at regular intervals, are called periodic properties and the phenomenon is known as the periodicity of elements.\n(ii) The third-period elements, Na, Mg, Al, Si, P and Cl summarize the properties of their respective groups and are called typical elements.\n(iii) The elements of the second period show resemblance in properties with the elements of the next group of the third period leading to a diagonal relationship. Such elements are called bridge elements.\n\n9. Why are noble gases placed in a separate group?\n\nSolution\nNoble gases are unreactive since they have their outermost orbit complete. Due to stable electronic configuration, they hardly react with other elements. So these elements are placed in a separate group i.e.18\n\n10. Name two elements you would expect to show chemical reactions similar to calcium. What is the basis of your choice?\n\nSolution\nBeryllium and magnesium will show similar chemical reactions as calcium. Since these elements belong to same group 2 and also have two electrons in their outermost shell like calcium.\n\n11. Name the metal(s) and non-metals in the first twenty elements.\nMetals:\nNon-\nmetals:\n\nSolution\nMetals: Lithium, Beryllium, Sodium, Magnesium, Aluminium, Potassium, Calcium.\nNon-metals: Hydrogen, Helium, Carbon, Nitrogen, Oxygen, Fluorine, Neon, Phosphorus, Sulphur, Chlorine, Argon.\n\n12. Name the type of elements, which have their:\n(i) Outermost shell complete –.........\n(ii) Outermost shell incomplete –.......\n(iii) two outermost shell incomplete – .......\n(iv)one electron short of octet – .......\n(v) two electrons in the outermost orbit - ..........\n\nSolution\n(i) Outermost shell complete – Noble gases\n(ii) Outermost shell incomplete – Representative elements\n(iii) two outermost shell incomplete – Transition elements\n(iv) one electron short of octet – Halogens\n(v) two electrons in the outermost orbit - Alkaline Earth metals\n\n13. An element has 2 electrons in its N shell.\n(i) What is its atomic number?\n(ii) State its position in periodic table\n(iii) Is is metal or non-metal?\n(iv) State the name assigned to this group?\n\nSolution\n(i) 30\n(ii) It belongs to group 12 and fourth period.\n(iii) It is a metal.\n(iv) The name assigned to this group is IIB\n\n14. State the valency of the elements of periods 3 and write the formula of their oxides.\n\nSolution\n Elements Valency Formula of oxides Na 1 Na2O Mg 2 MgO Al 3 Al2O3 Si 4 SiO2 P 5 P2O5 S 2 SO2 Cl 1 Cl2O\n\n15. An element A has atomic number 14. To which period does this element belong and how many elements are there in this period.\n\nSolution\nAn element A with atomic number 14 belongs to period three and there are eight elements in this period.\n\n16. Answer the following in respect of element 34/15 P\n(i) Give its electronic configuration\n(ii) To which group and period does it belong?\n(iii) What is its valency?\n(iv) Is it a metal or non - metal\n(v) Is it a reducing agent or oxidizing agent?\n(vi) Give its formula with chlorine.\n\nSolution\n(i) Electronic configuration of P: 2, 8, 5\n(ii) 15th Group and 3rd Period.\n(iii) Valency of P = 8 - 5 = 3\n(iv) Phosphorus is a non-metal.\n(v) It is an oxidizing agent.\n(vi) Formula with chlorine = PCl3\n\n### Intext Question 2\n\n1. Name any five periods properties.\n\nSolution\n(i) Electron affinity\n(ii) Atomic size\n(iii) Metallic character\n(iv) Non-metallic character\n(v) Ionization energy\n\n2. What do you understand by atomic size? State its unit.\n\nSolution\nAtomic size is the distance between the centre of the nucleus of an atom and its outermost shell.\nIt's measured in Angstrom and Picometre\n\n3. Give the trends in atomic size on moving:\n(i) down the group\n(ii) across the period right to left.\n\nSolution\n(i) The atomic size of an atom increases when we go down a group from top to bottom\n(ii) It increases as we move from right to left in a period\n\n4. Arrange the elements of second and third periods in increasing order of their atomic size.\n(i) Second Period\n(ii) Third Period\n\nSolution\n(i) Second Period: Fluorine <Neon< Oxygen< Nitrogen < Carbon < Boron< Beryllium\n< Lithium.\n(ii) Third Period: Chlorine < Argon < Sulphur < Phosphorus < Silicon < Aluminum < Magnesium < Sodium.\n\n5. Why is the size of (i) neon greater than fluorine? (ii) sodium is greater than magnesium?\n\nSolution\n(i) The size of Neon is bigger compared to fluorine because the outer shell of neon is complete(octet). As a result, the effect of nuclear pull over the valence shell electrons cannot\nbe seen. Hence the size of Neon is greater than fluorine.\n(ii) Since atomic number of magnesium is more than sodium but the numbers of shells are same, the nuclear pull is more in case of Mg atom. Hence its size is smaller than sodium.\n\n6. Which is greater in size?\n(i) an atom or a cation\n(ii) an atom or an anion\n(iii) Fe2+ or Fe3+\n\nSolution\n(i) An atom is always bigger than cation since cation is formed by the loss of electrons; hence protons are more than electrons in a cation. So the electrons are strongly attracted by the nucleus and are pulled inward.\n(ii) An anion is bigger than an atom since it is formed by gain of electrons and so the number of electrons are more than protons. The effective positive charge in the nucleus is less, so less inward pull is experienced. Hence the size expands.\n(iii) An anion is bigger than an atom since it is formed by gain of electrons and so the number of electrons are more than protons. The effective positive charge in the nucleus is less, so less inward pull is experienced. Hence the size expands.\n\n7. Metallic character and non-metallic character are periodic properties discuss.\n\nSolution\nThe periodic variation in electronic configuration as one move sequentially in increasing order of atomic number produces a periodic variation in properties.\nAs the elements are arranged in increasing order of atomic number, the metals with tendency to lose electrons are placed on the left and the metallic character decreases from left to right and increases down a group and non-metals with tendency to gain electrons are placed automatically on the right and the non-metallic character increase across a period and decreases down a group.\n\n8. Give the trend in metallic character:\n(i) across the period left to right,\n(ii) down the group top to bottom.\n\nSolution\n(i) The metallic character decreases as we go from left to right in a period.\n(ii) It increases as we go down a group\n\n9. State the trends in chemical reactivity:\n(i) across the periods left to right\n(ii) Down the group\n\nSolution\n(i) Across a period, the chemical reactivity of elements first decreases and then increases.\n(ii) Down the group, chemical reactivity increases as the tendency to lose electrons increases down the group.\n\n10. State the trends in physical properties on moving down the group. Give an example to illustrate.\n\nSolution\nThe melting and boiling points of metals decrease on going down the group.\nExample: Observe the trend in group 1 elements given in the following table:\n Metals m.p. b.p. Li 180.5°C 1347°C Na 94.5°C 883°C K 63.5°C 774°C\n\nFrom the above table, it is clear that m.p. and b.p. decrease from Li to K. The melting and boiling points of non-metals increase on going down the group.\nExample: Observe the trend in Group 17 elements given in the following table:\n Non-metals m.p. b.p. Physical State Fluorine -219.6°C -187°C Gas Chlorine -101°C -34.6°C Gas Bromide -7.2°C +58.8°C Liquid Iodine +113.6°C +183°C Solid\n\n11. An element X belong to 4th period and 17th group, state.\n(i) no of valence electrons in it\n(ii) name of the element.\n(iii) name the family to which it belong.\n(iv) Write the formula of the compound formed when it reacts with 27/13 y\n\nSolution\n(i) The element from the 17th group has 7 electrons in its outermost shell.\n(ii) The name of the element is bromine.\n(iii) Bromine belongs to the halogen family.\n(iv) The element 2713 Y has three electrons in its outermost shell which it can donate; hence, its valency is three. While the valency of bromine is\n1. Thus, 13' can donate three electrons, and bromine can accept 1 electron to get the stable electronic configuration.\nTherefore, the formula of the compound is AlBr3\n\n12. The given table shows elements with the same number of electrons in its valence shell.\nState:\n(i) Whether these elements belong to same group or period.\n(ii) Arrange them in order of increasing metallic character.\n\nSolution\n(i) Yes, these elements belong to the same group but are not from the same period.\n(ii) We know that m.p. decreases on going down the group. Hence, from the above table, the elements can be ordered according to their period as follows:", null, "The metallic character increases as one moves down the group.\nHence, the order of the given elements with increasing metallic character is as follows: B\n\n### Intext Question 3\n\n1. (a) Define the term ‘ionisation potential\n(b) Represent in the form of an equation. In which unit it is measured?\n\nSolution\n(a) The energy required to remove an electron from a neutral isolated gaseous atom and convert it into a positively charged gaseous ion is called Ionization energy or ionization potential.\n(b) M(g) + I.E - M+ (g) + e-\nM can be any element\nIt is measured in electron volts per atom. Its S.I unit kJmol-1\n\n2. What do you understand by successive ionization energies?\n\nSolution\nThe energy required to remove the residual electrons one by one is called successive ionization energy.\n\n3. State the trends in ionization energy:\n(a) across the period:\n(b) down the group.\n\nSolution\n(a) Ionization energy increases as we move from left to right across a period as the atomic size decreases.\n(b) Ionization energy decreases down a group as the atomic size increases.\n\n4. Name the elements with highest and lowest ionization energies.\n\nSolution\nHelium has the highest ionization energy of all the elements while cesium has the lowest ionization energy\n\n5. Arrange the elements of second and third period in increasing order of ionization energy.\n\nSolution\nSecond period: Neon > Fluorine > Oxygen > Nitrogen > Carbon > Boron > Beryllium > Lithium\nThird Period: Argon> Chlorine > Sulphur > Phosphorus > Silicon > Aluminum >Magnesium > Sodium\n\n6. (a) Define the term 'electron affinity'.\n(b) Arrange the elements of second period in increasing order of their electron affinity. Name the elements which do not follow the trend in this period.\n\nSolution\n(a) Electron affinity is the energy released when a neutral gaseous atom acquires an electron to form an anion.\n(b) Second period:\nLithium<Boron<Carbon<Oxygen<Fluorine\nNeon, Nitrogen and Beryllium do not follow the trend.\n\n7. State the factors on which electron affinity depends.\n\nSolution\nElectron affinity depends on:\n(a) Atomic size\n(b) Nuclear charge\n\n8. Electron affinity values generally ______across the periods left to right and_____down the group top to bottom.\n\nSolution\nElectron affinity values generally increases across the periods left to right and decreases down the group top to bottom.\n\n9. Give reason:\n(a) Electron affinity of halogens is comparatively high,\n(b) Electronegativity of chorine is higher than Sulphur.\n\nSolution\n(a) As we move from left to right the increase in atomic number and decrease in size results in a greater nuclear pull. As a result, the ability to attract the electrons increases, and so does the electron affinity.\nBut noble gases have complete stable octet configuration, hence their electron affinity is lower than halogens.\nHence halogens on extreme right have highest electron affinity in a period.\n(b) Chlorine is smaller than sulphur with a bigger atomic number. Since its nuclear pull is more, hence its electron affinity is higher than sulphur.\n\n10. Why fluorine has higher E.N. than chorine?\n\nSolution\nSince size of chlorine is bigger than fluorine hence the electrons being farther away from the nucleus experience a lesser force of attraction, hence electron negativity of chlorine is less than fluorine\n\n11. Define the term 'Electronegativity' state its unit.\n\nSolution\nElectronegativity measures an atom's tendency to attract shared pair of electrons towards itself. Its S.I unit is Pauling unit.\n\n12. (a) Name the elements with highest and lowest electronegativity,\n(b) State the character of the oxide of period 3.\n\nSolution\n(a) The element fluorine has the highest electronegativity and Cesium has the lowest electronegativity.\n(b) The nature of oxides changes from basic to acidic as we move from left to right in third period. Hence sodium forms most basic oxide while oxide of Aluminum is amphoteric and oxides of phosphorus, sulphur and chlorine are progressively acidic.\n\n13. Name the periodic property which relates to the:\n(a) Amount of energy required to remove an electron from an isolated gaseous atom,\n(b) character of element which loses one or more electrons when supplied with energy.\n(c) tendency of an atom to attract the shared pair of electron.\n\nSolution\n(a) Ionization energy\n(b) Metallic character\n(c) Electronegativity\n\n14.\nExplain the following:\n(a) Group 17 elements are strong non-metals, while group 1 elements are strong metals\n(b) Metallic character of elements decreases from left to right in a period while it increases in moving down a group.\n(c) Halogens have a high electron affinity.\n(d) The reducing power of element increases down in the group while decreases in a period.\n(e) Size of atom progressively becomes smaller when we move from sodium (Na) to chlorine(CI) in the third period of the periodic table.\n\nSolution\n(a) On moving across a period, nuclear pull increases because of the increase in atomic number, and thus, the atomic size decreases. Hence, elements cannot lose electrons easily. Hence, Group 17 elements are strong non-metals, while Group 1 elements are strong metals.\n\n(b) On moving across a period, nuclear pull increases because of the increase in atomic number, and thus, the atomic size decreases. Hence, elements cannot lose electrons easily. Hence, Group 17 elements are strong non-metals, while Group 1 elements are strong metals. Down a group, the atomic size increases and the nuclear charge also increases. The effect of an increased atomic size is greater as compared to the increased nuclear charge. Therefore, metallic nature increases as one moves down a group, i.e. they can lose electrons easily.\n\n(c) The atomic size of halogens is very small. The smaller the atomic size, the greater the electron affinity, because the effective attractive force between the nucleus and the valence electrons is greater in smaller atoms, and so the electrons are held firmly.\n\n(d) The reducing property depends on the ionisation potential and electron affinity of the\nelements. In a period, from left to right in a horizontal row of the periodic table, the atomic size decreases and the nuclear charge increases, so the electron affinity and ionisation energy both increase. Hence, the tendency to lose electrons decreases across the period from left to right and thus the reducing property also decreases across the period from left to right. The electron affinity and ionisation potential decreases along the group from top to bottom. Hence, the tendency to lose electrons increases, and thus, the reducing property also increases along the group from top to bottom.\n\n(e) In a period, the size of an atom decreases from left to right. This is because the nuclear charge, i.e. the atomic number increases from left to right in the same period, thereby bringing the outermost shell closer to the nucleus. Therefore, considering the third period given above, it has been found that sodium is the largest in size, while chlorine is the smallest.\n\n### Exercise -1\n\n1. (a) How does the electronic configuration of an atom relate to its position in the modern periodic table ?\n(b) Write the number of protons, neutrons and electronic configuration of 39/19 K, 31/15P.\nAlso state their position in periodic table.\n\nSolution\n(a) The total number of electron shells in an atom determines the period to which the element belongs, and the valence electrons determine the group to which it will belong. So with the help of electronic configuration we can figure out the period and group number of an element.\nElements with one and two valence electrons belong to group 1 and 2 respectively, while to determine the group number of elements with 3 to 8 valence electrons, we add 10 to their valence electrons.\nFor example an element X has atomic number 15\nIts configuration will be:\nK shell has 2 electrons, L will have 8, and the remaining 5 will be placed in M shell Since it has three shells it belongs to period 3 and with 5 valence electrons the element will be placed in five plus ten that is the 15th group\nSo with the help of electronic configuration we can figure out the period and group number of an element.\n\n(b) Atomic number = Number of protons\nHence, number of protons in K atom = 19\nNumber of neutrons = Mass number - Atomic number\nHence, number of neutrons in K atom = 39–19 = 20\nNumber of electrons = Number of protons\nHence, number of electrons = 19\nAnd electronic configuration of K atom = 2, 8, 8, 1\nSince K atom has 4 shells, hence it belongs to fourth period.\nWith one valence electron, it belongs to group 1\nNumber of protons in P atom = 15\nNumber of neutrons in P atom = 31-15 = 16\nNumber of electrons in P atom = 15\nAnd electronic configuration of P atom = 2, 8, 5\nSince it has three shells, it belongs to period 3 and with 5 valence electrons Phosphorus is found in five plus ten that is 15th group.\n\n2. Fluorine, chlorine and Bromine are put in one group on basis of their similar properties.\n(a) what are those similar properties?\n(b) What is the common name of this group or family?\n\nSolution\n(a) Fluorine, chlorine and bromine are non-metals with seven valence electrons. They are highly electronegative elements with valency of one. They exist as diatomic molecules. They form ionic compounds with alkali metals\n(b) They are known as halogens. The term means salt forming and therefore compounds\ncontaining these elements are called salts.\n\n3. What is the main characteristic of the last element in each period of the periodic table? What is the general name of such elements?\n\nSolution\nThe last element in each period of the periodic table is a gaseous element with its valence shell completely filled. Except for helium with complete duplet configuration, rest all the 5 gases have complete octet configuration.\nThese group 18 elements are commonly referred to as noble gases.\n\n4. According to atomic structure, what determines which element will be the first and which will be the last in a period?\n\nSolution\nThe electronic configuration of an element determines its position in Modern Periodic table. The element with one valence electron is the first while the element with 8 valence electrons is placed in the 18th group of a period.\n\n5. How does the number of:\n(i) valence electrons and\n(ii) valency vary on moving from left to right in the second period of the periodic table?\n\nSolution\n(i) The number of valence electrons increases by one as we move from left to right in a period. The group number 1 and 2 have 1 and 2 valence electrons respectively while group 13 to 18 have group number minus 10 = valence electrons. So, group 13 to 18 have 3, 4, 5, 6, 7 and 8 valence electrons respectively.\n(ii) Valency is determined by the number of valence electrons. For elements belonging\nto group 1, 2 and 13, the valency is equal to the number of valence electrons, so their valency is 1, 2 and 3 respectively.\nSince the elements in group 14 to 17 needs to gain electrons to complete their octet configuration. Their valency is 8 minus the number of valence electrons. So their valencies are 4, 3, 2 and 1 respectively.\n\n6. Fill in the blanks:\n(a) The horizontal rows in the periodic table are called ……….\n(b) On moving across a period from right to left in periodic table, the atomic size of the atom………….\n(c) on moving from right to left in the second period, the number of valence electrons.........\n\nSolution\n(a) The horizontal rows in the periodic table are called Periods.\n(b) On moving across a period from right to left in periodic table, the atomic size of the atom increases.\n(c) on moving from right to left in the second period, the number of valence electrons decreases\n\n7. An element barium has atomic number 56. Look up its position in the periodic table and answer the following questions\n(a) Is it a metal or a non- metal?\n(b) Is it more or less reactive than calcium?\n(c) What is its valency?\n(d) What will be the formula of its phosphate?\n(e) Is it larger or smaller than cesium (Cs)?\n\nSolution\n(a) Since it belongs to group II, it has 2 valence electrons and hence it is a metal.\n(b) Barium is placed below calcium in the group. Since, it has more number of shells; it is easier for it to lose its valence electrons to complete its octet configuration. Hence it is more reactive than calcium.\n(c) It needs to lose its 2 valence electrons to complete its octet configuration; therefore its valency is also 2.\n(d) The formula of its phosphate will be (Ba)3(PO4)2\n(e) As we move from left to right in a period, the size decreases, therefore, it will be smaller than Cesium\n\n8. How do the following change on moving from left to right in a period of the periodic table? Give examples in support of your answer.\n(a) atomic structure (electron arrangements)\n(b) Chemical reactivity of elements.\n(c) Nature of oxides of the elements.\n\nSolution\n(a) The number of valence electrons increases by one as we move across any given period.\nTherefore as we move from Lithium to Neon in period 2, the valence electrons will\nincrease from 1 to 7.\n\n(b) The metallic character decreases as we move from left to right while the non metallic\ncharacter increases.\nOngoing from left to right in a period, the chemical reactivity of elements first decreases and then increases.\nFor example in period 3, Sodium is the most reactive metal and Chlorine is the most reactive non-metal and Silicon is least reactive\n\n(c) The oxides of metals are basic and that of non-metals are acidic in general. Therefore since\nmetallic strength decreases and non-metallic strength increases on moving from left to right across a period, the strength of basic oxides decreases, while the strength of acidic oxides\nincreases.\nFor example, sodium forms a basic oxide, while sulphur and phosphorus form acidic oxides.\n\n9. This question refers to the elements of the periodic table with atomic number from 3 to 18. Some of the elements are shown by letters, but the letters are not the usual symbols of the elements.", null, "Which of these:\n(a) have most electronegative element.\n\n(b) is a halogen?\n(c) is an alkali metal?\n(d) is an element with valency 4?\n(e) have least ionization energy?\n(f) have least atomic size in period 3.\n\nSolution\n(a) Noble gases- H and P\n(b) Halogens- G and O\n(c) Alkali metals - A and I\n(d) D and L have valency of 4\n(e) I with atomic number 11.\n(f) Cl has the least atomic size in period 3 with atomic number 17.\n\n10. In group I of the periodic table, three elements X, Y and Z have ionic radii 1.33 A°, 0.95 A° and 0.60 A˚ respectively. Giving a reason, arrange them in the order of increasing atomic number in the group.\n\nSolution\nAs we move down a group, the numbers of shells increases and hence the atomic size increases.\nTherefore, Z will have the smallest atomic number followed by Y, while X will have the largest atomic number.\nSo the elements in order of increasing atomic number will be Z<Y<X.\n\n11. How does the chemical reactivity of:\n(a) alkali metals vary?\n(b) halogens vary?\n\nSolution\n(a) Since, the distance of the valence electrons from the nucleus keeps on increasing down the group, therefore, the ionization energy keeps on decreasing. Hence the reactivity of alkali metals increases from lithium to francium.\n(b) As we move down a group, the size keeps on increasing, so it becomes more difficult for\natoms to attract electrons. Thus reactivity of halogens decreases from Fluorine to Astatine.\n\n12. An element X belong to 3rd periods and group II of the periodic table state:\n(a) the number of valence electrons,\n(b) the valency,\n(c) name of the element,\n(d) whether it is a metal or a non-metal.\n\nSolution\n(a) Since it belongs to period 3 it has 3 shells, K, L and M. The outermost M shell will have 2 valence electrons as it is placed in group II\n(b) With 2 valence electrons, its valency will be 2.\n(c) Since it has electronic configuration of 2, 8, 2, its atomic number is 12 and hence X is Magnesium\n(d) It is a metal.\n\n13. The electronic configuration of an element T is 2, 8, 8, 1.\n(a) What is the group number of T?\n(b) What is the period number of T?\n(c) How many valence electrons are there in an atom of T?\n(d) What is the valency of T?\n(e) Is it a metal or a non-metal?\n\nSolution\n(a) Group 1 since the valence electrons is 1\n(b) With 4 shells T belong to period 4.\n(c) Number of electrons = 2+8+8+1 = 19\n(d) T needs to lose one electron to complete its octet hence its valency is 1\n(e) Since it has one valence electron, it is a metal.\n\n14. Arrange the elements of group 17 and group 1 according to the given conditions.\n(a) Increasing order of atomic size,\n(b) Increasing non - metallic character\n(c) Increasing ionization potential\n(d) Increasing electron affinity\n(e) Decreasing electronegativity.\n\nSolution\n(a) Group 1: Lithium< Sodium< Potassium< Rubidium < Caesium< Francium\nGroup 17: Fluorine < Chlorine < Bromine< Iodine < Astatine\n(b) Group 1: Francium\nGroup 17: Astatine< Iodine< Bromine< Chlorine< Fluorine\n(c) Group 1: Francium< Cesium< Rubidium< Potassium Sodium<\nLithium Group 17: Astatine< Iodine< Bromine< Chlorine< Fluorine\n(d) Group 1:\nFrancium Group\n17: Astatine\n(e) Group 1: Lithium>Sodium> Potassium> Rubidium> Cesium> Francium\nGroup 17: Fluorine > Chlorine> Bromine > Iodine > Astatine\n\n15. Complete the following sentences choosing the correct word or words from those given in brackets at the end of each sentence:\n(a) The properties of the elements are a periodic function of their ............... (atomic number, mass number, reative atomic mass).\n(b) Moving across a .............. of the periodic table the elements show increasing . . . . . . . . . . . . . . . . ....character (group, period, metallic, non-metallic). (c) The elements at the bottom of a group would be expected to show ........ metallic character than the element at the top. (less, more).\n(d) The similarities in the properties of a group of elements are because they have the same ...... (electronic configuration, number of outer electrons, atomic numbers).\n\nSolution\n(a) The properties of the elements are a periodic function of their atomic number (atomic number, mass number, reative atomic mass).\n(b) Moving across a period of the periodic table the elements show increasing non-metallic character (group, period, metallic, non-metallic).\n(c) The elements at the bottom of a group would be expected to show more metallic character\nthan the element at the top. (less, more).\n(d) The similarities in the properties of a group of elements are because they have the same number of outer electrons (electronic configuration, number of outer electrons, atomic numbers).\n\n16. Give reasons for the following:\n(a) The size of the anion is greater than the size of the parent atom.\n(b) argon atom is bigger than the chlorine atom.\n(c) Ionisation potential of the element increases across a period.\n\nSolution\n(a) Anion is formed by the gain of electrons. Thus the numbers of electrons are more than protons. The effective positive charge in the nucleus is less, so less inward pull is experienced. Hence the size expands. So the size of an atom is greater than the size of parent atom.\n\n(b) Since, Argon has stable octet configuration, so due to the inter - electronic repulsions the\neffect of nuclear pull over the valence shell electrons cannot be seen which results in the bigger size.\n\n(c) Since size of Bromine is bigger than chlorine, so it becomes more difficult for Br atoms to attract electrons. Thus, Cl is more reactive than Br.\n\n17. Which element has:\n(a) two shells, both of which are completely filled with electrons?\n(b) the electronic configuration 2, 8, 3?\n(c) a total of three shells with five electrons in its valence shell?\n(d) a total of four shells with two electrons in its valence shell?\n(e) twice as many electrons in its second shell as in its first shell?\n\nSolution\n(a) Neon\n(b) Aluminum\n(c) Phosphorus\n(d) Calcium\n(e) Carbon\n\n18. Name\n(a) An alkali metal in period 3 and halogen in period 2.\n(b) The noble gas with 3 shells.\n(c) The non-metals present in period 2 and metals in period 3.\n(d) The element of period 3 with valency 4\n(e) The element in period 3 which does not form oxide\n(f) The element of lower nuclear charge out of Be and Mg.\n(g) Which has higher E.A. fluorine or Neon.\n(h) Which has maximum metallic character Na, Li or K.\n\nSolution\n(a) Na and F\n(b) Argon\n(c) C, N, O and F are non-metals present in period 2 while Na, Mg and Al are metals in period 3.\n(d) Silicon\n(e) Argon\n(f) Mg\n(g) Fluorine\n(h) K\n\n19. Chorine in the periodic table is surrounded by the elements with atomic number 9, 16, 18 and 35.\n(a) Which of these have physical and chemical properties resembling chlorine.\n(b) Which is more electronegative than chlorine\n\nSolution\n(a) Element with atomic number 9 and 35\n(b) Element with atomic number 9.\n\n20. (a) State the number of elements in periods 1, Periods 2, and Period 3, of the periodic table.\n(b) name the elements in period 1.\n(c) What is the common feature of the electronic configuration of the elements at the end of period 2, and period 3?\n(d) if an element is in group 17, it is likely to be .............. (Metallic / non-metallic) in character while with one electron in its outermost energy level (shell), then it is likely to be ............... (Metallic/Non-metallic)\n\nSolution\n(a) Period 1 has 2 elements while period 2 and period 3 have 8 elements each.\n(b) Hydrogen and helium\n(c) The elements at the end of period 2 and Period 3 have 8 electrons in its outermost shell.\n(d) if an element is in group 17, it is likely to be Non metallic (Metallic/non-metallic) in character while with one electron in its outermost energy level (shell), then it is likely to be metallic (Metallic / Non-metallic)\n\n21. First ionization enthalpy of two elements X and Y are 500KJ mol-1 and 375KJ mol-1 respectively. Comment about their relative position in a group as well as in a period.\n\nSolution\nPosition in a group: X and Y\nPosition in a period: Y and X\n\n22. A metal M forms as oxide having the formula M2O3. It belongs to third period. Write the atomic number and valency of the metal.\n\nSolution\nPeriod no. = no. of shells, so n = 3\nFrom the formula M2O3 its valency is 3.\nSince it is a metal, its valence shell has 3 electrons.\nSo its electronic configuration is 2, 8, 3\nAtomic number = 13\nHence the metal is Aluminum with valency 3.\n\n23. Explain why are the following statements not correct:\n(a) All groups contain metals and non metals.\n(b) Atoms of elements in the same group have the same number of electron(s)\n(c) Non- metallic character decreases across a period with increase in atomic number\n(d) Reactivity increases with atomic number in a group as well as in a period.\n\nSolution\n(a) Since the elements in a group have same number of valence electrons, they can either contain metals or non-metals like alkali and alkaline metals have only metals whereas halogens are non-metals.\n\n(b) No two elements have the same number of electrons instead atoms of the same elements in the same group have the same number of valence electrons.\n\n(c) Non-metals have the tendency to gain electrons to attain stable configuration and therefore are said to be electronegative. As we move from left to right the increase in atomic number and decrease in size results in a greater nuclear pull. As a result the non-metallic character increases across a period.\n\n(d) On moving from left to right in a period, the reactivity first decreases and then increases since the tendency to lose electrons first decreases on going from left to right and then from P to Cl, tendency to gain electrons increases, so reactivity increases then. In case of a group, reactivity increases on going down since the tendency to lose electrons increases but for non-metals, reactivity decreases on going down the group as the tendency to gain electrons decreases down the group.\n\n24. Arrange the following in order of increasing radii:\n(a) Cl, Cl-\n(b) Mg2+, Mg, Mg+\n(c) N, O, P\n\nSolution\n(a) Cl< Cl-\n(b) Mg2+ < Mg+ < Mg\n(c) O < N < P\n\n25. Which element from the following has the highest ionization energy?\nExplain your choice. (a) P, Na, Cl\n(b) F, 0, Ne\n(c) Ne, He, Ar\n\nSolution\n(a) Cl\nMetals have low ionisation energy and non-metals have high ionisation energy. Also, across the period, ionisation energy tends to increase. The elements P, Na and Cl belong to the third period. Na - Group 1, P - Group 15 and Cl - Group 17.\n\n(b) Ne\nInert gases have zero electron affinity because of their stable electronic configuration.\n\n(c) He\nIonisation energy decreases with an increase in the atomic size, i.e. it decreases as one moves down a group. Ne, He and Ar are inert gases. He - Period 1, Ne - Period 2 and Ar - Period 3.\n\n(a) An element in period 3 whose electron affinity is zero\n(i) Sulphur\n(ii) Sodium\n(iii) Neon\n(iv) Argon\n\n(b) An alkaline earth metal\n(ii) potassium\n(iii) calcium\n(iv) Copper\n\n(c) An element with highest ionization potential\n(i) Calcium\n(ii) Fluorine\n(iii) Helium\n(iv) Neon\n\nSolution\n(a) (iv) Argon\n(b) (iii) Calcium\n(c) (iii) Helium\n\n27. The table given below represents the first three periods. Study the table and answer the question as given below\n(a) Write the formula of the sulphate of the element with atomic number 13 (b) what type of bonding will be present in the oxide of the element with atomic number 17?\n(c) Which feature of the atomic structure accounts for the similarities in the chemical properties of the elements in group 7A of the periodic table?\n(d) Name the element which has the highest ionization potential.\n(e) How many electrons are present in the valency shell of the element with atomic number 18?", null, "(f) What is the name given to the energy released when an atom in its isolated gaseous state accepts an electron to form an anion?\n(g) Fill in the blanks:\nThe atomic size .......... as we move from left to right across the periods, because the ......... increases but the ............... remains the same\n\nSolution\n(a) (Al)2(SO4)3\n(b) Covalent bonding\n(c) Same number of valence electrons\n(d) Helium\n(e) 8\n(f) Electron affinity\n(g) The atomic size Decreases as we move from left to right across the periods, because the\natomic number increases but the number of shells remains the same\n\n28. The electro negativities (according to pauling) of the elements in periodic table are as follows with the elements arranges in alphabetical order:\n\n(a) Arrange the elements in the order in which they occur in the periodic table from left to right.\n(The group 1 element first, followed by the group 2 element and so on, up to group 7).\n(b) Choose the word or phrase from the brackets which correctly completes each of the following statements:\n(i) The element below sodium in the same group would be expected to have a..........\n(lower/higher) electro-negativity than sodium and the element above chlorine would\nbe expected to have a ....... (lower/ higher) ionization potential than chlorine.\n(ii) On moving from left to right in a given period, the number of shells (remains the same/ increases/ decreases).\n(iii) On moving down a group, the number of valence electrons (remains the same/increases/ decreases).\n\nSolution\n(a) Na, Mg, Al, Si, P, S, Cl\n(b) (i) The element below sodium in the same group would be expected to have a Lower (lower/higher) electro-negativity than sodium and the element above chlorine would be expected to have a higher (lower/ higher) ionization potential than chlorine.\n(ii) remains the same\n(iii) remains the same\n\n29. Parts (a) to (e) refer to changes in the properties of elements on moving from left to right across a period of the periodic table. For each property, choose the correct answer.\n(a) The non-metallic character of the elements:\n(i) decreases\n(ii) increases,\n(iii) remains the same,\n(iv) depends on the period\n\n(b) The electronegativity:\n(i) depends on the number of valence electrons,\n(ii) remains the same,\n(iii) decreases,\n(iv) increases.\n\n(c) The ionization potential:\n(i) goes up and down\n(ii) decreases\n(iii) increases\n(iv) remains the same\n\n(d) The atomic size:\n(i) decreases,\n(ii) increases,\n(iii) remains the same,\n(iv) sometimes increases and sometimes decreases.\n\n(e) The electron affinity of the elements in groups 1 to 7:\n(i) goes up and then down.\n(ii) decreases and then increases,\n(iii) increases,\n(iv) decreases.\n\nSolution\n(a) Increases\n(b) Increases\n(c) Increases\n(d) Decreases\n(e) Increases\n\n30. The elements of one short period of the periodic table are given below in order from left to right:\nLi Be B C O F Ne\n(a) To which period do these elements belong?\n(b) One element of this period is missing. Which is the missing element and where should it be placed?\n(c) Which one of the elements in this period shows the property of catenation?\n(d) Place the three elements fluorine, beryllium and nitrogen in the order of increasing electronegativity.\n(e) Which one of the above elements belongs to the halogen series?\n\nSolution\n(a) Period 2\n(b) Nitrogen (N), between carbon and oxygen\n(c) Carbon\n(d) Be< N<F\n(e) Fluorine\n\n31. A group of elements in the periodic table are given below (boron is the first member of the group and thallium is the last).\nBoron, Aluminium, Gallium, Indium, Thallium.\nAnswer the following questions in relation to the above group of elements:\n(a) Which element has the most metallic character?\n(b) Which element would be expected to have the highest electronegativity?\n(c) If the electronic configuration of aluminium is 2, 8, 3, how many electrons are there in the outer shell of thallium\n(d) The atomic number of boron is 5. Write chemical formula of the compound formed when boron reacts with chlorine.\n(e) Will the elements in the group to the right of this boron group be more metallic or less metallic in character? Justify your answer.\n\nSolution\n(a) Thallium has the most metallic character since metallic character increases down the group\n(b) Boron has the highest electronegativity since it has the smallest size in the group.\n(c) 3. Since all the elements in a group have same number of valence electrons.\n(d) BCl3\n(e) The elements in the group to the right of boron group would be less metallic as with the decrease in size and increase in atomic number, it will be more difficult for them to lose electrons\n\n32. Select the correct answer from the choice A, B, C, D which are given. Write down only the letter corresponding to the correct answer. With reference to the variation of properties in the periodic table, which of the following is generally true?\nA. Atomic size increases from left to right across a period.\nB. Ionization potential increases from left to right across a period.\nC. Electron affinity increases going down a group.\nD. electro-negativity increases going down a group.\n\nSolution\nB. 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https://www.scirp.org/journal/PaperInformation.aspx?PaperID=71270&	[ "Torque Free Axi-Symmetric Gyros with Changing Moments of Inertia\n\nAbstract\n\nThe properties and characteristics of torque free gyros with rotational symmetry and changing moments of inertia are the subject of the subsequent discussion. It shall be understood that the symmetry can be expressed by the notation (A=B) which does not presuppose geometric symmetry, where A and B are the principle moments of inertia about x and y axes respectively. We study the case of a torque free gyro upon which no external torque is acting. The equations of motion are derived when the origin of the xyz-coordinate system coincides with the gyro’s mass center c. This study is useful for the satellites, which have rotational symmetry and changed inertia moments, the antennas and the solar power collector systems.\n\nShare and Cite:\n\nIsmail, A. and El-Haiby, F. (2016) Torque Free Axi-Symmetric Gyros with Changing Moments of Inertia. Applied Mathematics, 7, 1934-1942. doi: 10.4236/am.2016.716159.\n\n1. Introduction\n\nA gyro is a body of rotation which is set spinning at a large angular velocity around its axis of symmetry. The most important practical applications of gyros are met in devices for measuring the orientation or maintaining the stability of airplanes, spacecrafts and submarine vehicles in general. Various gyros are used as sensors in inertial guidance systems. Most textbooks in introductory mechanics explain the mysterious behavior of a spinning gyro by using Lagrange equations and severe mathematics . Other textbooks treat the problem of torque-induced precession of a top based on Euler equations, which are referred to the non-inertial reference frame rotating together with the body. The problem of the torque free inertial rotation of a symmetrical top is discussed, in particular, in and illustrated by a simulation computer program (free rotation of an axially symmetrical body) . For our problem, we consider a gyro body with xyz-coordinate system fixed of it, such that the z-axis of the body is the axis of symmetry and the inertia tensor is assumed to take the form", null, "(1.1)\n\nHere A and C are the principal moments of inertia in the x and z directions.\n\nThe rate of change of the inertia tensor with respect to time takes the form", null, "(1.2)\n\nWe note that the inertia products remain zeros.\n\nAssume that the angular velocity of the satellite or the gyros is", null, "(1.3)\n\nand the angular momentum is", null, "(1.4)\n\nwhere", null, "and", null, "are the unit vectors in the", null, "and", null, "directions.\n\nThus the gyro’s kinetic energy of rotation becomes", null, "(1.5)\n\n2. Equations of Motion\n\nApplying Euler’s equations of motion and putting the applied torque equal zero, we get", null, "(2.1)", null, "(2.2)", null, "(2.3)\n\nFor the considered problem (mentioned above) we apply the angular momentum principle to get", null, ", (2.4a)", null, ", (2.4b)", null, ". (2.4c)\n\n3. Components of Angular Momentum\n\nThe Equation (2.4c) can be integrated to give\n\nWe can conclude that the z-component of the angular moment is constant, that is", null, "(3.1)\n\nThe angular velocity is obtained by multiplying the Equations (2.4a) by and (2.4b) by and adding the resulted equations, we get\n\n, (3.2)\n\n(3.3)\n\nthat is or (3.4)\n\nwhere.\n\nThus the―component of angular momentum is constant\n\n(3.5)\n\nThe and can be shown, see Figure 1.\n\nFor torque free axi-symmetric gyro, angular velocity, angular momentum,\n\nFigure 1. The angular momentum components.\n\nand the gyro’s symmetry axis lie in one plane .\n\nSince the Equations (3.1) and (3.5) represent the angular momentum components, it can be deduced that the nutation angle remains constant when the inertia moments change as it when the inertia moments does not change.\n\nFor the components of angular velocity (3.4), we introduce the auxiliary frequency defined by\n\n, (3.6)\n\nthen\n\n. (3.7)\n\nThe two Equations (2.4a) and (2.4b) can be combined to yield\n\n. (3.8)\n\nThe solution of this differential equation can be obtained as\n\n. (3.9)\n\nIf the constant, and in order to find the value of constant we employ the Equation (3.5) to get and then\n\n, (3.10)\n\n, (3.11)\n\n, (3.12)\n\nwhere.\n\nWe can get the―component of the angular velocity at any instant by using\n\n(3.13)\n\n(3.14)\n\nwhere the subscript (0) refers to values of time\n\nThe components of angular velocity can be shown, see Figure 2.\n\nThe angular velocity component has a relative angular velocity in the xy- plane.\n\nIntroducing a floating coordinate system, we can express the angular velocity () by using the or the coordinate system.\n\nThe Equations (3.11) and (3.12) show that the component and consequently v-axis and u-axis are perpendicular, such that u-axis rotates with a relative angular velocity where\n\nIn the xy-plane, the z-component of the absolute angular velocity of v-axis is\n\nFigure 2. The components of angular velocity.\n\n(3.15)\n\nWe can find that, for a flattened gyro, the v-axis rotates faster than the x-axis, and for elongated gyro the v-axis rotates more slowly than the x-axis.\n\nThus, the coordinate system moves with an angular velocity\n\n(3.16)\n\nThe angular velocity of the gyro body and the angular velocity of the floating coordinate system are related by\n\n. (3.17)\n\nwhere is the spin of the gyro.\n\n4. Euler Frequencies\n\nThe frequency of the angular momentum remains constant since there is no external torque applied to the gyro , the direction of the angular momentum vector may be used to define as space-fixed coordinate axis L which assigned to Z-axis and is called the precession axis .\n\nIf the z-axis of the rotating coordinate system is the symmetry axis of the gyro, the v-axis lies in a plane formed by the Z- and z- axes, and we write\n\n. (4.1)\n\nThe nutation angle is the angle between the z-axis and Z-axis where Z-axis and L are coincide, see Figure 3.\n\nThe components of for elongated axis symmetric gyro are obtained from the shape, so\n\n(4.2)\n\n(4.3)\n\nThe nutation angle remains a constant and the gyro is carry out a steady precession about the angular momentum vector, since\n\n(4.4) we obtain\n\n(4.5)\n\n. (4.6)\n\nAlso from the figure\n\nFigure 3. Euler’s angles.\n\n(4.7)\n\nAlso for the angle between the z-axis and the angular velocity axis, we have\n\n. (4.8)\n\nThe motion of a torque free gyro with rotational symmetry and changing moment of inertia can be visualized by imaging a space cone and body cone as shown see Figure 4 and Figure 5.\n\nWe can obtain the relation between the precession and the spin as follows\n\n. (4.9)\n\nFor the elongated gyro we find that and and have the same sign then the precession is direct.\n\nFor the flattened gyro we find that and have the opposite signs, then the precession is retrograde.\n\n5. Conclusions\n\nThe system (2.4) is integrated to obtain the angular velocities and the angular momentum,\n\nFigure 4. Space cone and body cone for an elongated gyro.\n\nFigure 5. Space cone and body cone for a flattened gyro.\n\nthen Euler’s angles are deduced. The motions are classified into two cases:\n\n1) the elongated gyro.\n\n2) the flattened gyro.\n\nFor each case, we investigate the equations of motions, the precession, the nutation and the spin for these motions in detailed by using the analytical techniques and the illustrated shapes. The obtained results can be applied on the satellites with rotational symmetry and changed inertia moments, the antennas and the solar power collector systems .\n\nAcknowledgements\n\nThis project is supported by Institute of Scientific Research and Islamic Heritage Revival, Umm Al-Qura University, Saudi Arabia.\n\nConflicts of Interest\n\nThe authors declare no conflicts of interest.\n\n Goldstein, H. (1980) Classical Mechanics. 2nd Edition, Addison-Wesley, Reading. Goldstein, H., Poole, Ch.P. and Safko, J.L. (2002) Classical Mechanics. 3rd Edition, Addison-Wesley, Reading. Kittel, Ch., Knight, W.D. and Ruderman, M.A. (1965-1971) Mechanics. Berkeley Physics Course. McGraw-Hill, New York. Landau, L.D. and Lifschitz, E.M. (1976) Mechanics. Pergamon, New York. Butikov, E. (2006) Inertial Rotation of a Rigid Body. European Journal of Physics, 27, 913-922. Butikov, E. (2006) Free Rotation of an Axially Symmetrical Body. http://www.ifmo.ru/butikov/Applets/Precession.html Bruno, A.D. (2007) Analysis of the Euler-Poisson Equations by Methods of Power Geometry and Normal Form. Journal of Applied Mathematics and Mechanics, 71, 168-199. http://dx.doi.org/10.1016/j.jappmathmech.2007.06.002 Amer, T.S. (2004) Motion of a Rigid Body Analogous to the Case of Euler and Poinsot. Analysis, 24, 305-315. http://dx.doi.org/10.1524/anly.2004.24.14.305 Udwadia, F.E. and Kalaba, R.E. (2007) Analytical Dynamics: A New Approach. Eshagh, M. and Najafi Alamdari, M. (2007) Perturbations in Orbital Elements of a Low Earth Orbiting Satellite. Journal of the Earth & Space Physics, 33, 1-12. Kraus, J.D. and Marhefka, R.J. (2002) Antennas for all Applications. McGraw-Hill, New York. Balanis, C. (1997) Antenna Theory. John Wiley & Sons, New Jersey, Published simultaneously in Canada, 450-458. Price, H., Lupfert, E., Kearney, D., Zarza, E., Cohen, G., Gee, R. and Mahoney, R. (2002) Advances in Parabolic Trough Solar Power Technology. 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https://www.colorhexa.com/110433	[ "# #110433 Color Information\n\nIn a RGB color space, hex #110433 is composed of 6.7% red, 1.6% green and 20% blue. Whereas in a CMYK color space, it is composed of 66.7% cyan, 92.2% magenta, 0% yellow and 80% black. It has a hue angle of 256.6 degrees, a saturation of 85.5% and a lightness of 10.8%. #110433 color hex could be obtained by blending #220866 with #000000. Closest websafe color is: #000033.\n\n• R 7\n• G 2\n• B 20\nRGB color chart\n• C 67\n• M 92\n• Y 0\n• K 80\nCMYK color chart\n\n#110433 color description : Very dark violet.\n\n# #110433 Color Conversion\n\nThe hexadecimal color #110433 has RGB values of R:17, G:4, B:51 and CMYK values of C:0.67, M:0.92, Y:0, K:0.8. Its decimal value is 1115187.\n\nHex triplet RGB Decimal 110433 `#110433` 17, 4, 51 `rgb(17,4,51)` 6.7, 1.6, 20 `rgb(6.7%,1.6%,20%)` 67, 92, 0, 80 256.6°, 85.5, 10.8 `hsl(256.6,85.5%,10.8%)` 256.6°, 92.2, 20 000033 `#000033`\nCIE-LAB 4.02, 18.381, -27.021 0.872, 0.445, 3.172 0.194, 0.099, 0.445 4.02, 32.68, 304.225 4.02, 0.344, -12.207 6.671, 11.66, -23.521 00010001, 00000100, 00110011\n\n# Color Schemes with #110433\n\n• #110433\n``#110433` `rgb(17,4,51)``\n• #263304\n``#263304` `rgb(38,51,4)``\nComplementary Color\n• #040f33\n``#040f33` `rgb(4,15,51)``\n• #110433\n``#110433` `rgb(17,4,51)``\n• #290433\n``#290433` `rgb(41,4,51)``\nAnalogous Color\n• #0f3304\n``#0f3304` `rgb(15,51,4)``\n• #110433\n``#110433` `rgb(17,4,51)``\n• #332904\n``#332904` `rgb(51,41,4)``\nSplit Complementary Color\n• #043311\n``#043311` `rgb(4,51,17)``\n• #110433\n``#110433` `rgb(17,4,51)``\n• #331104\n``#331104` `rgb(51,17,4)``\n• #042633\n``#042633` `rgb(4,38,51)``\n• #110433\n``#110433` `rgb(17,4,51)``\n• #331104\n``#331104` `rgb(51,17,4)``\n• #263304\n``#263304` `rgb(38,51,4)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #010004\n``#010004` `rgb(1,0,4)``\n• #09021b\n``#09021b` `rgb(9,2,27)``\n• #110433\n``#110433` `rgb(17,4,51)``\n• #19064b\n``#19064b` `rgb(25,6,75)``\n• #210862\n``#210862` `rgb(33,8,98)``\n• #290a7a\n``#290a7a` `rgb(41,10,122)``\nMonochromatic Color\n\n# Alternatives to #110433\n\nBelow, you can see some colors close to #110433. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #050433\n``#050433` `rgb(5,4,51)``\n• #090433\n``#090433` `rgb(9,4,51)``\n• #0d0433\n``#0d0433` `rgb(13,4,51)``\n• #110433\n``#110433` `rgb(17,4,51)``\n• #150433\n``#150433` `rgb(21,4,51)``\n• #190433\n``#190433` `rgb(25,4,51)``\n• #1d0433\n``#1d0433` `rgb(29,4,51)``\nSimilar Colors\n\n# #110433 Preview\n\nText with hexadecimal color #110433\n\nThis text has a font color of #110433.\n\n``<span style=\"color:#110433;\">Text here</span>``\n#110433 background color\n\nThis paragraph has a background color of #110433.\n\n``<p style=\"background-color:#110433;\">Content here</p>``\n#110433 border color\n\nThis element has a border color of #110433.\n\n``<div style=\"border:1px solid #110433;\">Content here</div>``\nCSS codes\n``.text {color:#110433;}``\n``.background {background-color:#110433;}``\n``.border {border:1px solid #110433;}``\n\n# Shades and Tints of #110433\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #05010f is the darkest color, while #fcfbff is the lightest one.\n\n• #05010f\n``#05010f` `rgb(5,1,15)``\n• #0b0321\n``#0b0321` `rgb(11,3,33)``\n• #110433\n``#110433` `rgb(17,4,51)``\n• #170545\n``#170545` `rgb(23,5,69)``\n• #1d0757\n``#1d0757` `rgb(29,7,87)``\n• #23086a\n``#23086a` `rgb(35,8,106)``\n• #290a7c\n``#290a7c` `rgb(41,10,124)``\n• #2f0b8e\n``#2f0b8e` `rgb(47,11,142)``\n• #350da0\n``#350da0` `rgb(53,13,160)``\n• #3b0eb2\n``#3b0eb2` `rgb(59,14,178)``\n• #420fc5\n``#420fc5` `rgb(66,15,197)``\n• #4811d7\n``#4811d7` `rgb(72,17,215)``\n• #4e12e9\n``#4e12e9` `rgb(78,18,233)``\n• #5a21ee\n``#5a21ee` `rgb(90,33,238)``\n• #6733ef\n``#6733ef` `rgb(103,51,239)``\n• #7546f0\n``#7546f0` `rgb(117,70,240)``\n• #8258f2\n``#8258f2` `rgb(130,88,242)``\n• #906af3\n``#906af3` `rgb(144,106,243)``\n• #9d7cf5\n``#9d7cf5` `rgb(157,124,245)``\n• #ab8ef6\n``#ab8ef6` `rgb(171,142,246)``\n• #b9a0f8\n``#b9a0f8` `rgb(185,160,248)``\n• #c6b3f9\n``#c6b3f9` `rgb(198,179,249)``\n• #d4c5fa\n``#d4c5fa` `rgb(212,197,250)``\n• #e1d7fc\n``#e1d7fc` `rgb(225,215,252)``\n• #efe9fd\n``#efe9fd` `rgb(239,233,253)``\n• #fcfbff\n``#fcfbff` `rgb(252,251,255)``\nTint Color Variation\n\n# Tones of #110433\n\nA tone is produced by adding gray to any pure hue. In this case, #1b1b1c is the less saturated color, while #100235 is the most saturated one.\n\n• #1b1b1c\n``#1b1b1c` `rgb(27,27,28)``\n• #1a191e\n``#1a191e` `rgb(26,25,30)``\n• #1a1720\n``#1a1720` `rgb(26,23,32)``\n• #191522\n``#191522` `rgb(25,21,34)``\n• #181324\n``#181324` `rgb(24,19,36)``\n• #171126\n``#171126` `rgb(23,17,38)``\n• #160f28\n``#160f28` `rgb(22,15,40)``\n• #150c2b\n``#150c2b` `rgb(21,12,43)``\n• #140a2d\n``#140a2d` `rgb(20,10,45)``\n• #13082f\n``#13082f` `rgb(19,8,47)``\n• #120631\n``#120631` `rgb(18,6,49)``\n• #110433\n``#110433` `rgb(17,4,51)``\n• #100235\n``#100235` `rgb(16,2,53)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #110433 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5058546,"math_prob":0.80094665,"size":3639,"snap":"2019-35-2019-39","text_gpt3_token_len":1608,"char_repetition_ratio":0.12599725,"word_repetition_ratio":0.011090573,"special_character_ratio":0.5718604,"punctuation_ratio":0.23489933,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99328494,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-19T14:46:50Z\",\"WARC-Record-ID\":\"<urn:uuid:dff91f1f-17c3-4fd5-ab39-05daa19cde7a>\",\"Content-Length\":\"36151\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b239f2bc-e33d-499b-a958-25753c70ee86>\",\"WARC-Concurrent-To\":\"<urn:uuid:a729c074-262f-4802-913e-be8b65a5df51>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/110433\",\"WARC-Payload-Digest\":\"sha1:436DQBYPAE5QM6A4F5WDU5DHC2OJQNM7\",\"WARC-Block-Digest\":\"sha1:EDCOKLBH6QHDOY4KCGJKGTODV6UPWRLF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573533.49_warc_CC-MAIN-20190919142838-20190919164838-00254.warc.gz\"}"}
https://mail.python.org/pipermail/matrix-sig/1996-August/000733.html	[ "# [PYTHON MATRIX-SIG] Another problem?\n\[email protected] [email protected]\nFri, 16 Aug 1996 21:57:47 +0200 (MET DST)\n\n```>>>>> \"JH\" == Jim Hugunin <[email protected]> writes:\n\nJH> I doubt that my new version is more than a factor of 2 slower\nJH> than a C implementation of the same algorithm (though feel free\nJH> to show me wrong with a code sample).\n\nI decided to accept that challenge:\n\nnu~% cc -O4 -o x x.c -lm -non_shared\nnu~% time ./x\nCan divide by 2\nCan divide by 29\n./x 0.03s user 0.02s system 83% cpu 0.060 total\n\nRemember:\n\nnu~% python ./x.py\nCan divide by 2\nCan divide by 29\nfactor took 10.354 seconds\nCan divide by 2\nCan divide by 29\nafactor took 0.573 seconds\n\nand, yes the machine is \"empty\", so clock-times are comparable. And:\nchanging the array to type='b' makes the python version slower, so\nI'm not being unfair.... The only trick I've added is making the thing\nnon-recursive, but I doubt that change would make the python version\nmuch faster.\n\nnu~% python\nPython 1.4b2 (Aug 12 1996) [C]\nCopyright 1991-1996 Stichting Mathematisch Centrum, Amsterdam\n>>> 0.573/0.060 > 2.0\n1\n>>>\n\n-----------------------------------\n\n#include <stdio.h>\n#include <math.h>\n#include <malloc.h>\n\n#define bool char\nint sieve(int max)\n{\nbool sieve;\nint result;\nint i,j,limit;\nsieve=(bool )malloc((max+1)sizeof(bool));\nfor (i=0;i<=max;i++) sieve[i]=1;\nsieve=0;\nsieve=0;\nlimit=floor(sqrt((float)max));\nfor (i=2;i<=limit;i++) {\nif (sieve[i]) {\nfor (j=ii;j<=max;j+=i) sieve[j]=0;\n}\n}\nj=0;\nfor (i=2;i<=max;i++) j+=sieve[i];\n/printf(\"Sieve contains %d primes\\n\",j);/\nresult=(int )malloc((j+1)sizeof(int));\nj=0;\nfor (i=2;i<=max;i++) if (sieve[i]) result[j++]=i;\nresult[j]=0;\nfree(sieve);\nreturn(result);\n}\n\nvoid factor(int a,int max)\n{\nint *s;\nint i;\nif (max<2) return;\ns=sieve(max);\nfor (i=0;s[i];i++) {\nif (!(a%s[i])) {\nprintf(\"Can divide by %d\\n\",s[i]);\n}\n}\nfree(s);\n}\n\nint main() {\nfactor(129753308,99999);\nexit(0);\n}\n\n=================\nMATRIX-SIG - SIG on Matrix Math for Python\n\nsend messages to: [email protected]" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5566852,"math_prob":0.92148787,"size":2114,"snap":"2019-43-2019-47","text_gpt3_token_len":718,"char_repetition_ratio":0.12938389,"word_repetition_ratio":0.04109589,"special_character_ratio":0.39687794,"punctuation_ratio":0.18609408,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9983763,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-18T22:02:48Z\",\"WARC-Record-ID\":\"<urn:uuid:1eeb9fb2-9d09-4ceb-a2b9-be990b2247b7>\",\"Content-Length\":\"4662\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f9de73d6-f706-4a0a-91a3-d88fab80196a>\",\"WARC-Concurrent-To\":\"<urn:uuid:90ddeb8a-3b6b-44ac-9d3a-ea7bdacf7490>\",\"WARC-IP-Address\":\"188.166.95.178\",\"WARC-Target-URI\":\"https://mail.python.org/pipermail/matrix-sig/1996-August/000733.html\",\"WARC-Payload-Digest\":\"sha1:WBP7M25FAD27SUAY7N32NZXU7L7DDFNY\",\"WARC-Block-Digest\":\"sha1:S2Q4WL543SYHG5JGHFHAZ2GQ42EYAFZI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669847.1_warc_CC-MAIN-20191118205402-20191118233402-00350.warc.gz\"}"}
http://www.ltdstate.com/	[ "var x = 1200,y = 60 var xin = true, yin = true var step = 1 var delay = 40 var obj=document.getElementById(\"ad\") function floatAD() { var L=T=0 var R= document.body.clientWidth-obj.offsetWidth var B = document.body.clientHeight-obj.offsetHeight obj.style.left = x + document.body.scrollLeft obj.style.top = y + document.body.scrollTop x = x + step(xin?1:-1) if (x < L) { xin = true; x = L} if (x > R){ xin = false; x = R} y = y + step(yin?1:-1) if (y < T) { yin = true; y = T } if (y > B) { yin = false; y = B } } var itl= setInterval('floatAD()', delay) obj.onmouseover=function(){clearInterval(itl)} obj.onmouseout=function(){itl=setInterval('floatAD()', delay)}", null, "", null, "新闻文章", null, "", null, "", null, "" ]	[ null, "http://www.ltdstate.com/skin/default/1509lan/images/logo.png", null, "http://www.ltdstate.com/skin/default/1509lan/images/biaoshi.png", null, "http://www.ltdstate.com/skin/default/1509lan/images/xiaoyuanzixun.png", null, "http://www.ltdstate.com/skin/default/1509lan/images/remenhuati.png", null, "http://www.ltdstate.com/skin/default/1509lan/images/jiantou.png", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.76127285,"math_prob":0.9991598,"size":1097,"snap":"2021-04-2021-17","text_gpt3_token_len":1183,"char_repetition_ratio":0.116193965,"word_repetition_ratio":0.0,"special_character_ratio":0.31540564,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99301285,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-19T20:35:23Z\",\"WARC-Record-ID\":\"<urn:uuid:cd2e192d-2e14-4a0e-8852-b2e28c7e4c9f>\",\"Content-Length\":\"43714\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:94a36972-b408-4f67-8988-715f3275a13e>\",\"WARC-Concurrent-To\":\"<urn:uuid:0f498def-9694-4e9d-b522-367f4b7fa9fe>\",\"WARC-IP-Address\":\"45.194.147.57\",\"WARC-Target-URI\":\"http://www.ltdstate.com/\",\"WARC-Payload-Digest\":\"sha1:WPTDRWWW5X44OIHGAA6TKL4PREJCIZYB\",\"WARC-Block-Digest\":\"sha1:ZGSYCKOKKBO4QMXZAGLJOROQ4BHTUO3U\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703519784.35_warc_CC-MAIN-20210119201033-20210119231033-00095.warc.gz\"}"}
https://www.groundai.com/project/fluctuation-induced-noise-in-out-of-equilibrium-disordered-superconducting-films/	[ "Fluctuation-induced noise in out-of-equilibrium disordered superconducting films\n\n# Fluctuation-induced noise in out-of-equilibrium disordered superconducting films\n\nAleksandra Petković Laboratoire de Physique Théorique, IRSAMC, CNRS and Université de Toulouse, UPS, F-31062 Toulouse, France Laboratoire de Physique Théorique et Hautes Energies, Université Pierre et Marie Curie and CNRS UMR 7589, 4 place Jussieu, 75005 Paris, France Laboratoire de Physique Théorique-CNRS, Ecole Normale Supérieure, 24 rue Lhomond, 75005 Paris, France Valerii M. Vinokur Materials Science Division, Argonne National Laboratory, Argonne, Illinois 60439, USA\n19-07-2019 23:10\n###### Abstract\n\nWe study out-of-equilibrium transport in disordered superconductors close to the superconducting transition. We consider a thin film connected by resistive tunnel interfaces to thermal reservoirs having different chemical potentials and temperatures. The nonequilibrium longitudinal current-current correlation function is calculated within the nonlinear sigma model description and nonlinear dependence on temperatures and chemical potentials is obtained. Different contributions are calculated, originating from the fluctuation-induced suppression of the quasiparticle density of states, Maki-Thompson and Aslamazov-Larkin processes. As a special case of our results, close-to-equilibrium we obtain the longitudinal ac conductivity using the fluctuation-dissipation theorem.\n\n###### pacs:\n74.40.-n, 74.25.F-\n\n## I Introduction\n\nAn equilibrium in nature is rare and it is rather an exception than the rule. A vast majority of natural processes, ranging from large-scale flows in the atmosphere to the electric charge transfer found throughout the technological realm, are out-of-equilibrium processes. Yet, the physics of the equilibrium state is much more studied and better understood. A major difficulty impeding the similar progress in the research of the nonequilibrium processes is that while all thermodynamics-based science rests on the law that, in equilibrium, any system assumes the state with the minimal free energy, the equally powerful and fundamental principles that govern the far-from-equilibrium behaviors still wait to be revealed. There has been an important advance both theoretical and experimentalBlanter and Bütikker (2000); Naz (2003); Kogan (1996); de Jong and Beenakker (1996); Kopnin (2001) employing a variety of approaches to out-of-equilibrium problems in electronic systemsKopnin (2001); Khlus (1987); Lesovik (1989); Büttiker (1990); Beenakker and Büttiker (1992); Shulman and Kogan (1969); Nagaev (2000, 1991); Altshuler et al. (1994); Gutman and Gefen (2001); among them, those based on the Keldysh techniqueKeldysh (1965); Kamenev and Levchenko (2009) seem to appear the most promising as paving a way towards general method which would allow treating interacting nonequilibrium systems on a common fundamental ground.\n\nIn this work we undertake the study of a disordered superconducting system employing the theory that has been proven to effectively tackle the low-energy excitation physics, the Keldysh nonlinear sigma model Feigel’man et al. (2000). More specifically, we focus on the nonequilibrium phase transition between the normal and the superconducting state. In a vicinity of the transition, when the system is in the normal state, the behavior of the system is governed by fluctuations of the superconducting order parameter. Fluctuation-induced short-living Cooper pairs are formed and contribute critically both into the thermodynamic and transport characteristics of the systemsLarkin and Varlamov (2005). In disordered thin films the temperature range where fluctuations are essential is determined by the sheet resistance (being in any case significantly larger than the one of bulk superconductorsAslamazov and Larkin (1968a, b); Maki (1968); Larkin and Varlamov (2005)) and depends on the particular process involved, extending often to temperatures well above the superconducting transition temperatureLarkin and Varlamov (2005); Baturina et al. (2012).\n\nHere we study the influence of superconducting fluctuations on dynamic properties of a thin film in the fluctuational region of a normal state. The film is driven out of the equilibrium due to contacts with thermal reservoirs having different temperatures and chemical potentials. The Keldysh Ginzburg-Landau-like action under nonequilibrium conditions and several effects of nonequilibrium superconducting fluctuations were addressed in Refs. Chtchelkatchev and Vinokur, 2009; Petković et al., 2010, 2011 for this setting. Out-of-equilibrium fluctuation contributions to the dc electrical conductivity were calculated in Ref. Petković et al., 2011. Importantly, under nonequilibrium conditions the fluctuation-dissipation theorem (FDT) is generally violated and there is no fundamental relation between the current fluctuations (i.e. noise) and conductivity, in contrast to close-to-equilibrium conditions where this relation holds. Therefore, unlike in equilibrium systems, the nonequilibrium noise is not fully tied to the conductivity and can carry additional information, not contained in the conductivity. This poses a problem of the independent calculation of noise in out-of-equilibrium state.\n\nShot noise in noninteracting diffusive electron system was studied in Refs. Nagaev, 2000; Altshuler et al., 1994. The influence of the Coulomb interaction on shot noise in disordered systems was analyzed in Refs. Beenakker and Büttiker, 1992; Nagaev, 2000; Kozub and Rudin, 1995; Gutman and Gefen, 2001, while the influence of Bardeen-Cooper-Schrieffer (BCS) interaction on shot noise due to electric current flow above the critical temperature was considered only up to the second order in the electric field in Ref. Nagaev, 1991. In the present work, we focus on a film above the nonequilibrium superconducting transition and calculate nonlinear dependence of the Nyquist noise on the temperatures and chemical potentials of the thermal baths that are in contact with the film, see Fig. 1.\n\nThe paper is organized as follows. In Sec. II we introduce the nonlinear sigma model for superconductors within the Keldysh technique. In Sec. III we find the current correlation function for the case of noninteracting electrons in a nonequilibrium disordered thin film. We further consider the BCS interaction. In Sec. IV we introduce nonequilibrium fluctuation propagators. Then we proceed with calculations of different contributions to the current-current correlation function caused by superconducting fluctuations: the density of states contribution is calculated in Sec. V, the Maki-Thompson in Sec. VI and the Aslamazov-Larkin in Sec. VII. In Sec. VIII we summarize our results. Some calculation details are given in Appendices.\n\n## Ii Keldysh formalism: disordered superconductors\n\nIn this section we introduce the notation and provide the basic equations needed for the calculation of the current-current correlation function. We present the model, discuss its applicability and explain the procedure that allows us to analyze the fluctuations of the superconducting order parameter in the metallic state. In order to treat the nonequilibrium physics, we employ the Keldysh technique Keldysh (1965). We start with the nonlinear sigma model and then further discuss the calculation of the current-current correlations.\n\n### ii.1 Nonlinear sigma model\n\nThe nonlinear sigma model can be used to describe the low-energy physics for superconductors with BCS interaction in the presence of short-range quenched disorder (see Appendix A). The partition function takes the form , where the action consists of three partsFeigel’man et al. (2000); Kamenev and Andreev (1999)\n\n S[ˇQK,ˇΔK]=SΔ+Sϕ+SQ. (1)\n\nHere and in the following we set . The fields and are Hubbard-Stratonovich fields introduced to decouple the four-fermion terms originating from the Hamiltonian describing the BCS interaction and disorder, respectivelyFeigel’man et al. (2000); Kamenev and Andreev (1999). The contributions to the action (1) are given by\n\n SΔ= −ν2λTr[ˇΔ†KˇYˇΔK],Sϕ=e2ν2Tr[ˇϕKˇYˇϕK], (2) SQ= iπν4Tr[D(∂rˇQK)2−4ˇΞ∂tˇQK−4ieˇϕKˇQK+4iˇΔKˇQK]. (3)\n\nHere is the diffusion coefficient, and it carries information about the disorder. The bare single particle density of states at the Fermi level per one spin projection is denoted by . The superconductive coupling constant is positive. The matrix field satisfies the nonlinear relation . The check symbol denotes matrices that are defined in the tensor product of the Keldysh and Nambu spaces. The former and the latter are spanned by the Pauli matrices and , , respectively, and we define , . One uses different notation for the same matrices for convenience, and . We assume implicitly the multiplication in the time-space, and “Tr” includes the integration over the real space. The subscript denotes the gauge transformed fields , and . The fields and are defined in same way as , where and are the vector and the scalar potential, respectively. The field is given by , and . is defined in the same way. We have also defined . The quantum (q) and classical (cl) components of the fields are respectively defined as the half-sum and the half-difference of the field values at the lower and the upper branches of the Keldysh time-contour. The field becomes the superconducting order parameter at the mean-field (saddle-point) level, while the saddle point equation for produces the Usadel quasiclassical equations, where plays the role of the quasiclassical Greens function. The covariant spatial derivative is given by . We stress that the nonlinear sigma model action captures the low-energy physics at energy scales much smaller that the elastic scattering rate. It is valid in the limit when the lifetime of fluctuating Cooper pairs is much greater than the elastic scattering time.\n\nWe further explain the strategy to treat the superconducting fluctuations. First we find the saddle point equation for of the action (1), in the absence of the BCS interaction (i.e. ). It reads as Larkin and Ovchinnikov (1975, 1977); Kamenev and Andreev (1999); Feigel’man et al. (2000)\n\n ˇΛ =ˇUˇΛ0ˇU−1,ˇΛ0=^σz⊗^τz, (4) ˇUt,t′(r) =ˇU−1t,t′(r)=(δ(t−t′−0)^τ0^Ft,t′(r)0−δ(t−t′+0)^τ0), (5) ^Ft,t′(r) =(Fet,t′(r)00Fht,t′(r)). (6)\n\nAfter Wigner transforming we obtain that can be related to the quasiparticle electron/hole distribution functions via . One then considers massless fluctuations around the normal-metal saddle point solution, since massive modes can be integrated out in the Gaussian approximation and lead to unimportant renormalization of the parameters in the action. The massless fluctuations satisfy and are conveniently parameterized as Feigel’man et al. (2000)\n\n ˇQK(r) =e−ˇW(r)/2ˇΛ(r)eˇW(r)/2,ˇW=ˇUˇWˇU−1, (7) ˇW =(wτ+−w∗τ−w0τ0+wzτz¯w0τ0+¯wzτz¯wτ+−¯w∗τ−), (8)\n\nsuch that . Here we introduced four real fields with representing diffuson degrees of freedom and the two complex fields for Cooperon degrees of freedom. One now substitutes matrix given by Eq.(7) into the action (1) and requires that the terms linear in vanish. This leads to a kinetic equation that electron and hole distribution function have to satisfy. Next we switch on the BCS interaction assuming the system is in the normal state. In that case the average value of the superconducting order parameter is zero, and therefore one again obtains the normal-metal saddle point solution (4). In order to study the influence of the BCS interaction, one has to consider the fluctuations. Assuming that the system is not too close to the transition, we take into account quadratic fluctuations around the metallic saddle point solution. Integrating them out, one obtains the effective action depending only on the superconducting order parameter and electromagnetic fields. This action allows us to treat the superconducting fluctuations in the normal metallic state. Levchenko and Kamenev (2007); Petković et al. (2011)\n\n### ii.2 Current-current correlation function\n\nOur aim in the following is to calculate the symmetrized two-operator current correlation function\n\n S(r,t;r′,t′)=12⟨jx(r,t)jx(r′,t′)+jx(r′,t′)jx(r,t)⟩ (9)\n\nunder nonequilibrium conditions. It can be obtained differentiating the nonlinear sigma model partition function\n\n S(r,t;r′,t′)=−14∂2Z∂Aqx(r,t)∂Aqx(r′,t′)∣∣Aq=0,Acl=0. (10)", null, "Figure 1: Thin superconducting film is connected by tunneling interfaces to the gate and the substrate. The interfaces are characterized by the resistances R1 and R2. The substrate temperature is T1, the gate temperature is T2 and the gate voltage is VG. The parameters T1, T2, VG, and the resistances of the tunneling interfaces determine the quasiparticle distribution in the film and allow us to change it in a controlled way.\n\nIn the rest of the paper we focus on the particular system shown in Fig. 1. We consider a thin superconducting film connected by resistive interfaces to thermal reservoirs having different chemical potentials and temperatures. We assume that the system is in zero magnetic field and in the normal state but close to the transition into the superconducting state. We study a stationary situation (). Therefore, the noise depends only on the time difference . In the following we use the notation for the energy dependence of the distribution function.\n\nThere are four different contributions to the current correlation function\n\n S(r−r′,t−t′)= S0(r−r′,t−t′)+SDOS(r−r′,t−t′)+SMT(r−r′,t−t′)+SAL(r−r′,t−t′), (11)\n\nwhere denotes the noise in the noninteracting case (), and the other terms are contributions induced by superconducting fluctuations. The main processes are: i)coherent Andreev reflections of quasiparticles on the local fluctuations of the superconducting order parameter resulting into the so-called Maki-Thompson (MT) contributionsMaki (1968); Thompson (1970); ii) formation of Cooper pairs and their involvement into charge transfer is described by the Aslamazov-Larkin (AL) correctionsAslamazov and Larkin (1968a), and iii) the suppression of the single-particle density of states (DOS) due to quasiparticle participation in Cooper-pairing Larkin and Varlamov (2005). Randomness in the production and dissociation of fluctuating Cooper pairs, and in other closely related processes discussed above, gives the corresponding contributions to the current noise. In the following sections we evaluate and analyze the individual contributions. We focus on the regime where the superconducting fluctuations can be treated perturbatively, i.e., the fluctuation contribution are small compared to the noise in the noninteracting case. This implies that the film is not too close to the superconducting transition and that the physics is dominated by Gaussian fluctuations. Then one finds . Here is the Ginzburg number, is the film thickness and denotes the sum of the contributions induced by fluctuations of the superconducting order parameter.\n\n## Iii Noninteracting electrons\n\nWe consider noninteracting electrons first and evaluate Eq. (10) taking into account nonequilibrium conditions. After differentiating the partition function with respect to the vector potential field in Eq. (10), we substitute the field with its saddle-point solution given by Eq. (4). We do not take here into account the fluctuations around , since they are responsible for weak-localization effects. Since the main aim of this paper is to find the noise close to the superconducting transition, and since the weak-localization contribution is a non-singular function in the vicinity of the transition, we do not consider it here. Then, we obtain the longitudinal current-current correlation function to beNagaev (2000)\n\n S0(r−r′,ϵ)= 2σDT0(ϵ)δ(r−r′), (12) T0(ϵ)= 14∫dΩ{1−12[Fe(Ω)Fe(Ω+ϵ)+Fh(Ω)Fh(Ω+ϵ)]}. (13)\n\nHere the Drude conductivity is and the integration is always from to if not specified differently. In equilibrium, the distribution function is and the expression (12) simplifies to\n\n Seq0(r−r′,ϵ)=δ(r−r′)σDϵcoth(ϵ2T). (14)\n\nThis universal relation between the conductivity and the noise is known as the FDT. The microscopic details about the disorder strength are hidden in the diffusion constant, i.e., in . We stress that the range of applicability of the nonlinear sigma model is , where is the elastic scattering time. This explains why in Eq. (14) appears the frequency independent Drude conductivity .\n\nNext we analyze Eq. (12) in out-of-equilibrium conditions. Then, the FDT is generally violated. We assume that the film is thin such that the Thouless energy corresponding to diffusion across the film , well exceeds all the relevant energy scales. Here denotes the film thickness. The current across the interface separating the substrate and the film is , where is the tunneling resistance per unit area characterizing the interface and the subscript denotes the substrate. A similar equation holds for the interface between the film and the gate. From the continuity equation for the current follows , where . Here and denote the distributions in the substrate and in the gate, respectively. The gauge invariant distribution in the film is defined as and takes the form\n\n ~Fe/h(ϵ)=xtanh(ϵ±(1−x)eVG2T1)+(1−x)tanh(ϵ∓xeVG2T2). (15)\n\nThe upper (lower) signs correspond to electrons (holes). We assumed very resistive interfaces, such that the resistance of the film can be neglected with respect to the resistance of the interfaces. Also, we assumed .\n\nNow we can proceed with the evaluation of the expression (12). In the case one can calculate it exactly and obtains\n\n S0(r−r′,ϵ)= Seq0(r−r′,ϵ)[x2+(1−x)2]+Seq0(r−r′,ϵ+eVG)x(1−x)+Seq0(r−r′,ϵ−eVG)x(1−x), (16)\n\nwhere is the equilibrium noise, see Eq. (14). When one of the tunneling resistances is infinitely large, then effectively the system is in contact only with one reservoir and therefore in an equilibrium. In that case or , and as expected Eq. (16) reduces to the equilibrium noise . Moreover, notice that by increasing the temperature of the system, thermal fluctuations increase and the current noise increases. Similarly, in Eq. (16), the gate voltage also leads to an increase of the noise with respect to the equilibrium one (). In the case of zero frequency, the noise was found in Ref. Petković et al., 2010. Eq. (16) also describes a diffusive bridge placed between two reservoirs having different chemical potentialsAltshuler et al. (1994). There, plays the role of the coordinate along the bridge, and after performing the integration over in Eq. (16) one finds the shot noise of that system.\n\nNext we discuss the influence of the electron-electron interaction on the distribution function (15). In general, the interaction leads to smearing of a noninteracting distribution function. However, when the inelastic length is large in comparison to the system dimensions, inelastic processes can be neglected. The inelastic length is expected to increase as the temperature and/or applied voltage decrease, allowing one to tune the ratio of the characteristic system length and the inelastic relaxation length. Coulomb-interaction induced corrections to the noise in the equilibrium have been studied by Altshuler and AronovAltshuler and Aronov (1985). Out-of-equilibrium, the influence of the the Coulomb interaction on the shot noise in diffusive contacts has been studied in the limit of large inelastic length in Refs. Beenakker and Büttiker, 1992; Nagaev, 2000; Gutman and Gefen, 2001, while contributions due to inelastic collisions have been studied in Ref. Kozub and Rudin, 1995. In the present paper, we assume that time-of-flight of the quasiparticles through the film is much smaller than the typical energy-relaxation time. We consider a thin film characterized by the large inelastic scattering length and focus on the influence of the BCS interaction on the longitudinal transport in the normal state.\n\n## Iv Superconducting fluctuations\n\nHaving discussed the noninteracting case in the previous section, we start the analysis of the influence of the BCS interaction on the current noise. The system is assumed to be in the normal state, but in the vicinity of the nonequilibrium superconducting transition. To be more precise, we do not discuss the Berezinskii-Kosterlitz-Thouless transition, but the crossover that in the equilibrium happens at the BCS critical temperature . In the following we analyze this crossover in nonequilibrium conditions, studying influence of the gate voltage and temperatures of the reservoirs on the Cooper pair lifetime and the processes discussed in Sec. II.2. In this and in the following sections we consider the case where the Ginzburg-Landau rate is much smaller than the effective temperature , defined below by Eq. (22).\n\nThe saddle point equation of the action (1) for the superconducting order parameter has the trivial solution with the average value of the superconducting order parameter being zero. Then, one recovers the normal-metal saddle point (4) for the field . Therefore, now we include the massless fluctuations around it. We substitute Eq. (7) into the action (1) and expand it to the second order in , Eq. (8). Since we are neither interested in the weak localization correction nor in the Altshuler-Aronov type corrections, but in fluctuations caused by the fluctuating superconducting order parameter , in the following we consider only the Cooperon degrees of freedom. After integrating out the Cooperon degrees of freedom, we obtain a Ginzburg-Landau-like action which depends only on the order parameter . This calculation was performed in Ref. Petković et al., 2011 and the correlation functions of Cooperon fields have been obtained. They read as 111Note that in Ref. Petković et al., 2011 it was used the notation and and similarly for the Fourier transform in the time-domain. This notation is confusing since then the convention is that complex conjugation includes the change of momentum and energy, and . In the following, we will use and . ,Petković et al. (2011)\n\n ⟨⟨wϵ1,ϵ2(q)w∗−ϵ3,−ϵ4(q)⟩⟩=2iνδϵ1−ϵ2,ϵ4−ϵ3−L−1KLA,1−2LR,1−2+Fh(ϵ3)LR,1−2+Fe(ϵ1)LA,1−2[Dq2−i(ϵ1+ϵ2)][Dq2−i(ϵ3+ϵ4)], (17) ⟨⟨¯wϵ1,ϵ2(q)¯w∗−ϵ3,−ϵ4(q)⟩⟩=2iνδϵ1−ϵ2,ϵ4−ϵ3−L−1KLA,1−2LR,1−2−Fh(ϵ2)LA,1−2−Fe(ϵ4)LR,1−2[Dq2+i(ϵ1+ϵ2)][Dq2+i(ϵ3+ϵ4)], (18) ⟨⟨¯wϵ1,ϵ2(q)w∗−ϵ3,−ϵ4(q)⟩⟩=2iνδϵ1−ϵ2,ϵ4−ϵ3L−1KLA,1−2LR,1−2+Fh(ϵ2)LA,1−2−Fh(ϵ3)LR,1−2[Dq2+i(ϵ1+ϵ2)][Dq2−i(ϵ3+ϵ4)]. (19)\n\nThe average is with respect to the action given by Eq. (1) and it includes averaging over the fluctuations of , and . Here denotes retarded/advanced fluctuation propagators and is the Keldysh propagator. Their low frequency and low momentum behavior is given by the following expressionsPetković et al. (2011):\n\n L−1K =iπ2[1−~Fh(0)~Fe(0)], (20) L−1R/A(q,ϵ) =π8Te{−(τGLzcp)−1+[∓4iTe~FR(0)−Dq2±iϵ∓2ieVG(1−x)](1±iTeΩ)}. (21)\n\nHere and in Eqs. (17-19) we set . The parameters appearing in the retarded and advanced propagators are functionals of . They are given by the following expressions , , and . The symbol denotes the principal value of the integral. The nonequilibrium Ginzburg-Landau (GL) rate is defined as , where is the Debye energy. The GL time denotes the lifetime of the fluctuation induced Cooper pairs. It carries the information how far the system is from the transition and at the transition 222More precisely, in two dimensions this condition does not correspond to a real superconducting transition, but to a crossover. it becomes infinitely large, signaling that the Cooper pairs become long living. We point out that the fluctuation propagators given above are valid when the system is in normal state but in the vicinity of the transition such that . Substituting the distribution function (15) in the previous equations, one evaluates all these parameters and finds that they are functions of temperatures of the reservoirs and , the ratio of tunneling resistances and the gate voltage Petković et al. (2011):\n\n Te= ⎡⎢ ⎢⎣xT1cosh2(1−x)eVG2T1+(1−x)T2cosh2xeVG2T2⎤⎥ ⎥⎦−1, (22) Ω−1= 2xT1π2Im[Ψ′(12−ieVG(1−x)2πT1)]+2(1−x)T2π2Im[Ψ′(12+ieVGx2πT2)], (23) τ−1GL= 8πzcpTe{xRe[Ψ(12+i(1−x)eVG2πT1)]+(1−x)Re[Ψ(12+ixeVG2πT2)]+2ln2 +xlnT1T2−lnTcT2+γ}+4zcpT2eΩ~FR(0). (24)\n\nHere is the digamma function, defined as , where is the gamma function. We expressed the critical temperature as , where is the Euler constant. Now one easily finds the parameter .\n\nWe are now equipped to start the calculation of different fluctuation contributions to the current-current correlation function. However, before doing it, we make a brief digression in order to stress the importance of the above given fluctuation propagators. In close-to-equilibrium conditions, the Aslamazov-Larkin contribution to the conductivity can be obtained by employing together the linear response theory and the phenomenological time-dependent GL (TDGL) equation, while the Maki-Thompson and the DOS, should be derived starting from the microscopic theory. The phenomenological TDGL equation reads as\n\n [−8π(T−Tc)−D(q−2eˇAK)2+i(ω−2eϕK)]ΔclK(q,ω)+ζ(q,ω)=0, (25)\n\nwhere the thermal noise satisfies\n\n ⟨ζ(r,t)ζ∗(r′,t′)⟩=16πνT2δ(r−r′)δ(t−t′). (26)\n\nHowever, if one is interested in a nonlinear dependence on the drive, then the previous equation is not a good starting point. Notice that even for the simple setup here considered (see Fig. 1), the TDGL equation takes different formPetković et al. (2011) than the usual phenomenological TDGL equation (25). It is given by\n\n 8TeπL−1R(q,ω)ΔclK(q,ω)+ζ(q,ω)=0, (27)\n\nwhere the nonequilibrium noise satisfies the following condition\n\n ⟨ζ(r,t)ζ∗(r′,t′)⟩= 16πνT2e[1−~Fh(0)~Fe(0)]δ(r−r′)δ(t−t′). (28)\n\nIn Eq. (27), the retarded fluctuation propagator is given by Eq. (21) with many drive-dependent parameters that we discussed above. Only in the equilibrium, and , Eqs. (27) and (28) coincide with Eqs. (25) and (26), while otherwise they are different. Therefore, we point out that in order to study a nonlinear dependence of some observable on the drive, one should derive a corresponding TDGL equation starting from the microscopic theory and not to use the phenomenological one (25), as is usually the case in literature.\n\n## V Density of states contribution\n\nIn this section we start the calculation of the current correlation function (10) in the presence of the BCS interaction. We consider linear response to the in-plane electric field, while the system is driven out-of-equilibrium by thermal baths having different temperatures, between which it is sandwiched, or by an electric field perpendicular to the film, see Fig. 1.\n\nFirst we consider the density of state contribution to the noise. Taking into account the massless fluctuations around the normal-metal saddle point Eq. (4) to the second order in the Cooperon fields, we collect all the terms of the type and in Eq. (10). They constitute the DOS contribution and together give\n\n SDOS(r−r′,ϵ)= −νDe216πδ(r−r′)∫dϵ1dϵ2{⟨⟨¯wϵ1,ϵ2(r)¯w∗−ϵ2,−ϵ1(r)+wϵ1,ϵ2(r)w∗−ϵ2,−ϵ1(r)⟩⟩ ×[1−Fh(ϵ2−ϵ)Fh(ϵ2)−Fe(ϵ1−ϵ)Fe(ϵ1)]−⟨⟨wϵ1,ϵ2(r)w∗−ϵ2−ϵ,−ϵ1−ϵ(r)+¯w∗−ϵ2,−ϵ1(r)¯wϵ1+ϵ,ϵ2+ϵ(r)⟩⟩ ×Fh(ϵ2)Fe(ϵ1+ϵ)}+ϵ→−ϵ. (29)\n\nThe leading contribution in Eq. (V) close to the transition into superconducting state is given by\n\n SDOS(r−r′,ϵ)= +δ(r−r′)De216π3df∫d2qdEdωIm(L−1KLA,ωLR,ω{2Fh(E−ϵ)Fh(E)[Dq2−i(2E+ω)]2 +Fh(E−ω)Fe(E+ϵ)[Dq2−i(2E+2ϵ−ω)][Dq2−i(2E−ω)]})+ϵ→−ϵ. (30)\n\nWe use the notation . Notice that Eq. (V) is the only singular part of Eq. (V) for and therefore the most dominant close to the transition. The holes and electrons give the same contribution. This is expected, since the system is in the normal state and there is no long-living Cooper-pair condensate that could allow for a charge imbalance between the holes and electrons. We further evaluate Eq. (V) and find up to logarithmic accuracy:\n\n SDOS(r−r′,ϵ)≈ −e22π2dfδ(r−r′)[1−~Fh(0)~Fe(0)]T2ezcpln(Teτ−1GL)Re{2∫dE~Fh(E−ϵ)~Fh(E)(E+i0)2 +∫dE~Fh(E)~Fe(E+ϵ)(E+ϵ+i0)(E+i0)}+ϵ→−ϵ. (31)\n\nHere and in the following denotes . In order to discuss the assumptions made in Eq. (V), we introduce\n\n ϵ0=−4Te~FR(0)+Teτ−1GL/Ω. (32)\n\nWe assume that the system is close to the transition such that and are much smaller than relevant energy scales of the distribution function, i.e., . For the special case and , is exactly zero. Eq. (V) is also valid for and then the condition becomes . When obtaining Eq. (V), we used the fact that the most important contribution in Eq. (V) originates from small momenta and that therefore we can safely cut the momentum integration at the upper limit .\n\nIn the case of equal temperatures of the reservoirs , substituting the distribution function (15) in Eq. (V), we obtain the nonlinear dependence of the noise on the gate voltage , temperature and frequency . Integrating out spatial coordinates, one obtains the final result\n\n SDOS(ϵ)≈ −e2π3df[1−~Fh(0)~Fe(0)]T2ezcpln(Teτ−1GL)[x2D(eVG(1−x),eVG(1−x)) +x(1−x)D(eVG(1−x),−eVGx)+x(1−x)D(−eVGx,eVG(1−x)) +(1−x)2D(−eVGx,−eVGx)+ϵ→−ϵ] (33)\n\nThe function is defined as\n\n D(x,y)= 1Tcoth(y−x+ϵ2T)Im[Ψ′(12−iϵ−x2πT)−Ψ′(12+iy2πT)]−πϵcoth(x+y+ϵ2T) ×Re[Ψ(12−ix2πT)−Ψ(12−ix+ϵ2πT)+Ψ(12+iy2πT)−Ψ(12+iy+ϵ2πT)], (34)\n\nwhere the frequency satisfies . However, for but arbitrary with respect to , one obtains the result by setting to zero in Eq. (V). The expressions for , , and are provided in Sec. IV. In Appendix B we discuss the case of different temperatures of the reservoirs, .\n\nNext we analyze the expression (V). For (or ) the system is effectively in an equilibrium and Eq. (V) simplifies to\n\n SeqDOS(ϵ)= −2e2Tπ3dfln(Tτ−1GL)coth(ϵ2T){Im[Ψ′(12−iϵ2πT)]−2πTϵRe[Ψ(12)−Ψ(12−iϵ2πT)]}. (35)\n\nFor we also get the same expression from Eq. (V), since it also corresponds to the equilibrium situation. In the case of noninteracting equilibrium electrons, studied in Sec. III, the source of the current fluctuations is thermal fluctuations. Here we obtained the additional contribution (V) to the current noise caused by superconducting fluctuations. The Cooper pair density fluctuates due to randomness in the formation and dissociation of Cooper pairs, and these fluctuations affect the single-particle density of states leading to the noise .\n\nIn equilibrium the FDT holds and relates the current fluctuations (noise) to the real part of the ac conductivity that provides information about absorbed energy in the sample:\n\n Seq(ϵ)=Re[σ(ϵ)]ϵcoth(ϵ2T), (36)\n\nwhere . Using it we obtain the density of states contribution to the in-plane ac conductivity in equilibrium:\n\n Re[σeqDOS(ϵ)]= −2e2π3dfTϵln(Tτ−1GL){Im[Ψ′(12−iϵ2πT)]−2πTϵRe[Ψ(12)−Ψ(12−iϵ2πT)]} (37) ≈ ⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩−21e2ζ(3)π4dfln(Tτ−1GL),ϵ≪T−4e2π2dfln(Tτ−1GL)(Tϵ)2ln(ϵT),ϵ≫T. (38)\n\nThis result is in the agreement with findings of Ref. Petković and Vinokur, 2013, where it is obtained in a different manner, i.e. directly considering the ac conductivity in equilibrium within the nonlinear sigma model approach. In Eqs. (35,37,38) the GL rate takes its equilibrium value . One obtains that suppression of the single-particle density of states due to superconducting fluctuations gives the negative contribution to the Drude conductivity, see Fig. 2 where as the function of frequency is shown by the red dashed line. We remind the reader that our calculations apply when the frequency, temperatures and the gate voltage are much smaller than the elastic scattering rate, since we use the nonlinear sigma model that captures low-energy physics.", null, "Figure 2: Current noise SDOS(ϵ) is shown as a function of frequency ϵ for the case of equal temperatures of the reservoirs T1=" ]	[ null, "https://storage.googleapis.com/groundai-web-prod/media%2Fusers%2Fuser_275292%2Fproject_322164%2Fimages%2Fx1.png", null, "https://storage.googleapis.com/groundai-web-prod/media%2Fusers%2Fuser_275292%2Fproject_322164%2Fimages%2Fx2.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8884307,"math_prob":0.9850629,"size":27567,"snap":"2020-45-2020-50","text_gpt3_token_len":5852,"char_repetition_ratio":0.174727,"word_repetition_ratio":0.033341214,"special_character_ratio":0.20462872,"punctuation_ratio":0.10675563,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9888832,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-22T17:22:50Z\",\"WARC-Record-ID\":\"<urn:uuid:5523ac45-2982-4fd7-94c4-4ce39ff745ae>\",\"Content-Length\":\"1049303\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4b19ee1e-7dfd-4b2f-baf9-835bdcb406bb>\",\"WARC-Concurrent-To\":\"<urn:uuid:bdc0ecc0-fd12-4416-852d-a2071edbaa78>\",\"WARC-IP-Address\":\"35.186.203.76\",\"WARC-Target-URI\":\"https://www.groundai.com/project/fluctuation-induced-noise-in-out-of-equilibrium-disordered-superconducting-films/\",\"WARC-Payload-Digest\":\"sha1:PVFZMEP464G7TH5EY6MUEMZR7QYFKT5K\",\"WARC-Block-Digest\":\"sha1:DZSOG5AW5WWQHCEJLFHF3KUQI6XYCRX3\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107880014.26_warc_CC-MAIN-20201022170349-20201022200349-00701.warc.gz\"}"}
https://worksheets.site/math-mazes-addition-negatives.html	[ "# Math Mazes with addition and subtraction and Negative Numbers\n\nAddition and Subtraction Math Mazes with Integers. Free Printable Math Mazes. Fun math activities. Free printable worksheets to practice operations with negative numbers. Fun math worksheets to learn math operations with the whole range of Integer numbers. Pastimes and activities for the middle school math class. 7th Grade Math Activity Sheets. Number Snake No2. Number Snake PDF.", null, "## 1. Description\n\nHow to master addition and subtraction across zero in a fun way? Try solving these puzzle mazes. Explore the mazes calculating the value of each cell as you move forward. Get all the coins (colored circles). Every time you reach a coin you can compare your result so far with the correct answer and receive immediate feedback.\n\n### 1.1. Importance of immediate feedback\n\nImmediate feedback, and the possibility of correcting errors as soon as we make them, are very effective when we do not have a teacher or tutor present.\n\nFeedback promotes autonomy and self-learning. It is the student who discovers errors and makes correction the moment they arise. Students can work on their own, and at their own pace, making the materials more adaptable to each individual.\n\nFeedback can be a very powerful motivator and promote personal goals to overcome the next task.\n\nIn summary, immediate feedback has the potential to help academic performance, promote motivation, self-regulation and self-efficiency, allowing students to narrow the gap between their current performance and their desired performance.\n\nThe file has 40 Number Snakes for Addition and Subtraction, one per sheet. Sometimes the student will need to calculate a value by adding, sometimes by subtracting. The numbers range from -20 to 20. Each worksheet requires about 60 calculations.\n\nThe booklet is in PDF format", null, "This is a pencil and paper activity that does not require preparation. Early middle school students only needs the printed sheets and a pencil to use these personal puzzles. The students write their calculations in the boxes as they explore every chamber of the maze." ]	[ null, "https://worksheets.site/math-mazes/mazes-addition-negatives.png", null, "https://worksheets.site/innards/pdf.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.95013434,"math_prob":0.88269305,"size":1595,"snap":"2020-10-2020-16","text_gpt3_token_len":304,"char_repetition_ratio":0.1062225,"word_repetition_ratio":0.0,"special_character_ratio":0.18808778,"punctuation_ratio":0.096551724,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9552312,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-02T15:38:36Z\",\"WARC-Record-ID\":\"<urn:uuid:78406a7c-055c-4f64-ac7c-43d3ef7fcbd0>\",\"Content-Length\":\"13868\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5a29074e-3156-4ec6-b225-150d354e0a87>\",\"WARC-Concurrent-To\":\"<urn:uuid:5007ae9c-7642-437c-b587-d42f894dd980>\",\"WARC-IP-Address\":\"107.191.126.132\",\"WARC-Target-URI\":\"https://worksheets.site/math-mazes-addition-negatives.html\",\"WARC-Payload-Digest\":\"sha1:QLKDW6P2LJOPKLURHLQKI5V43QHUCYCW\",\"WARC-Block-Digest\":\"sha1:3HIBBRUHHPUS3QYHMK2VG6QQRMTKMNX2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370506988.10_warc_CC-MAIN-20200402143006-20200402173006-00525.warc.gz\"}"}
https://forum.allaboutcircuits.com/threads/rc-integrator.116646/	[ "# RC Integrator\n\n#### jag1972\n\nJoined Feb 25, 2010\n71\nI am looking at the passive RC Integrator circuit, the Vout equation for this circuit is: $$Vout= \\frac {1}{RC} \\int Vin \\: dt$$. This only works for certain conditions e.g. when $$RC > T$$. This is however misleading as most text books show a square wave which equals $$5tau$$. This is totally confusing IMO as the vout equation does not match the output voltage at all. Is it just me that thinks this or is it misleading or have I made a mistake in my assumption.\n\nJag.\n\n#### StayatHomeElectronics\n\nJoined Sep 25, 2008\n1,070\nThe equation for Vout is valid when time, t, is less than the time constant, RC. That is t < RC = tau. This represents only the first portion of the response to the squarewave transitions. The circuits use as an integrator is limited in time. After time t it is no longer integrating.\n\nAnother equation for the capacitor charging is an exponential voltage rise (or decay depending on initial conditions) to the applied voltage. The amount of time it takes to get over 99% of the way there is 5tau. The first one fifth of this equation looks very much like a straight line.\n\n#### jag1972\n\nJoined Feb 25, 2010\n71\n\n#### jag1972\n\nJoined Feb 25, 2010\n71\nI have a series RC Circuit where $$R= 1 k\\Omega\\: and C = 10 nF$$, the input to the circuit is a square wave with a period 1 ms.\nThe input square wave is: $$\\: x(t)=\\: 5,\\: 0\\geq\\:t\\:> 0.5 ms\\: 0,\\: 0.5 ms\\geq\\:t\\:> 1 ms$$.\n\nWhen I use the integrator formulae: $$\\frac{1}{RC} \\int_{0}^{0.5 ms} \\: x(t}\\:dt$$\n\n$$tau = 0.1 ms$$, therefore after 5 tau, vC is 99.9% of 5 V. This can be determined from the following equation/model for vC:\n$$vC=5(1-\\exp^{-5})$$\nHowever when I use the integrator formulae I do not get that value: $$Vout = \\frac{1}{RC} \\int_{0}^{0.5 ms} 5\\: dt$$\nThe value using the integrator formulae is 25 V which is clearly wrong can some please tell me where I am going wrong. I am aware that the integrator only works as a true integrator when RC is large compared to T. But I would have thought that Vout should always be right.\n\n#### StayatHomeElectronics\n\nJoined Sep 25, 2008\n1,070\n\ntau = R * C = 1000 ohms * (10 * 10^-9 F) = 1 * 10^-5 s = 0.01 ms = 10 us.\n\nIn order that your circuit is integrating the time over which you are integrating must be less than tau. The time over which you are integrating is 50 * tau. Even at 5 * tau in your initial calculation, the integration is no longer valid. You can only do your integral from time = 0 -> 10 us. With times greater than 10 us, only the exponential equation is valid.\n\nLook at the exponential response of the integrator to the input square wave. At the transitions the output starts off rising linearly, that portion of the response is what is equivalent to the integral. Once the graph starts to change slope, the integration will begin to have errors. So, time < tau is stated as begin the only valid time for integration because the output is very close to linear, the expected result of the integration of a square wave.\n\nThanks for your reply, I thought that was the answer. BTW I made a mistake with the reistor value it was supposed to be $$10 k\\Omega$$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9074552,"math_prob":0.9983977,"size":481,"snap":"2020-45-2020-50","text_gpt3_token_len":121,"char_repetition_ratio":0.102725364,"word_repetition_ratio":0.0,"special_character_ratio":0.26611227,"punctuation_ratio":0.12,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984768,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-25T11:47:42Z\",\"WARC-Record-ID\":\"<urn:uuid:ddcfc134-9322-4da2-a8ab-a52ed9f6bcde>\",\"Content-Length\":\"121938\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1ffc21ea-0ac0-4a56-85bf-451772bd926a>\",\"WARC-Concurrent-To\":\"<urn:uuid:f14f4ba8-f918-479b-9968-ed2e7d7b290a>\",\"WARC-IP-Address\":\"172.67.34.128\",\"WARC-Target-URI\":\"https://forum.allaboutcircuits.com/threads/rc-integrator.116646/\",\"WARC-Payload-Digest\":\"sha1:UEGV4XFN4KIOZA57DDACGLBZWRRHUIA4\",\"WARC-Block-Digest\":\"sha1:KPJ22RD3SQYYFEEFAVW74BPPKMS6W5GW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107888931.67_warc_CC-MAIN-20201025100059-20201025130059-00371.warc.gz\"}"}
https://www.vedantu.com/question-answer/tangents-can-a-circle-have-class-10-maths-cbse-5ed6ab1184c7204b9f97921d	[ "", null, "", null, "", null, "", null, "", null, "Question", null, "Answers\n\n# How many tangents can a circle have?", null, "", null, "Hint:- Go with definitions of tangents and circle.\nAs we know that for any conic,\nLocus is the set of all points whose coordinates satisfy a given equation or condition.\nAnd as we know that,\nCircle is the locus of points equidistant from a given point, which is the centre of the circle.\nAnd, tangent is the line which intersects a circle at one point only.\nThe circle has infinite points. So, there will be infinite tangents that can be drawn\non these points which touches at only one point.\nHence, a circle can have infinite tangents.\nNote:- Whenever we came up with this type of questions, then remember that tangents intersect\nthe circle at one point and chords cut the circle at two points. So, there can be infinite number of\nchords and tangents that can be drawn on a circle. And remember that diameter is also a chord to\nthe circle with length as 2r.\n\nView Notes\nTangent to a Circle", null, "", null, "Construction of Tangent to a Circle", null, "", null, "Tangent of a Circle", null, "", null, "How to Draw a Perfect Circle", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "" ]	[ null, "https://www.facebook.com/tr", null, "https://static.vedantu.com/cdn/images/seo-templates/seo-qna.svg", null, "https://static.vedantu.com/cdn/images/seo-templates/arrow-left.png", null, "https://static.vedantu.com/cdn/images/seo-templates/arrow-right.png", null, "https://static.vedantu.com/cdn/images/seo-templates/qna-tab-arrow.svg", null, "https://static.vedantu.com/cdn/images/seo-templates/qna-tab-arrow.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark-filled.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark-filled.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark-filled.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark-filled.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark-filled.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark-filled.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark-filled.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark-filled.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark-filled.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark-filled.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark.svg", null, "https://www.vedantu.com/cdn/images/seo-templates/bookmark-filled.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94503313,"math_prob":0.97666276,"size":916,"snap":"2021-21-2021-25","text_gpt3_token_len":202,"char_repetition_ratio":0.16557017,"word_repetition_ratio":0.024390243,"special_character_ratio":0.2117904,"punctuation_ratio":0.12105263,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99921256,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-09T12:32:01Z\",\"WARC-Record-ID\":\"<urn:uuid:9402843a-6148-4691-87d1-21018720aafa>\",\"Content-Length\":\"565282\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:24473e59-e382-4f5f-ab87-9ba358de5f7f>\",\"WARC-Concurrent-To\":\"<urn:uuid:a9ac53fe-c310-4901-84a4-f7fd6bc0f2bb>\",\"WARC-IP-Address\":\"13.32.200.57\",\"WARC-Target-URI\":\"https://www.vedantu.com/question-answer/tangents-can-a-circle-have-class-10-maths-cbse-5ed6ab1184c7204b9f97921d\",\"WARC-Payload-Digest\":\"sha1:VFIKALZD3RMWOAESSC5EMLHOOKKB36OV\",\"WARC-Block-Digest\":\"sha1:TROA4XTAQGGCPPJ2H2ORKMBKDJVX4A3X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988986.98_warc_CC-MAIN-20210509122756-20210509152756-00029.warc.gz\"}"}
https://economics.stackexchange.com/questions/43353/elasticity-of-substitution-of-goods-in-canonical-new-keynesian-model	[ "# Elasticity of substitution of goods in canonical New Keynesian Model\n\nIn the context of a New Keynesian Model (NKM) with imperfect competition, the aggregate demand for good can be represented using Dixit-Stiglitz aggregator $$C_t=\\Bigg(\\sum^1_0C_t(i)^{\\varepsilon-1\\over \\varepsilon}\\;di\\Bigg)^{\\varepsilon\\over\\varepsilon-1}$$ where $$\\varepsilon$$ is the substitutability of intermediate goods and which determines the \"market power\" of monopolistically competitive firms. Theoretically speaking, if $$\\varepsilon \\to \\infty$$, the price markup in the product market, i.e. $$\\mathcal{M}={\\varepsilon\\over \\varepsilon-1}$$, approaches unity and the market is perfectly competitive while goods are perfect substitutes. From Gali (2008), and other articles on NKM that uses calibration techniques, $$\\varepsilon$$ is usually within range 6 to 10 which translates to an average markup of 1.2 to 1.1 or a 20-10% markup over marginal costs.\n\nHowever, it is a bit unclear to me as to how the value of $$\\varepsilon$$ is chosen, which leads to my question: what is a \"sensible\" value of $$\\varepsilon$$ for a highly competitive market? And what about imperfectly competitive market with few major firms?\n\n• Hi. Welcome to Econ Stack Exchange. Try not to use abbreviations (like NKM) unless you first define them. This helps with discoverability with search engines, not to mention the average reader. Apr 6 at 15:44\n• Thank you, I'll keep that in mind and modify it\n– Rei\nApr 6 at 18:35" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86417454,"math_prob":0.9642165,"size":1115,"snap":"2021-31-2021-39","text_gpt3_token_len":292,"char_repetition_ratio":0.16471647,"word_repetition_ratio":0.0,"special_character_ratio":0.23587444,"punctuation_ratio":0.102564104,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9879799,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-17T09:46:31Z\",\"WARC-Record-ID\":\"<urn:uuid:d17dbc0a-7789-4b51-b7ab-6794e5c4a5bd>\",\"Content-Length\":\"155339\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bcb181ae-115d-488d-8aaa-9e77630016d9>\",\"WARC-Concurrent-To\":\"<urn:uuid:a026e2b5-d727-497e-a5d6-6fff208f0ab5>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://economics.stackexchange.com/questions/43353/elasticity-of-substitution-of-goods-in-canonical-new-keynesian-model\",\"WARC-Payload-Digest\":\"sha1:MD3BPNCP46FXDGE4H4RFS3G4UX7RSUWO\",\"WARC-Block-Digest\":\"sha1:HS2NHHIEYPH6QOE6KN3IXOZRNOFA4ZEX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780055632.65_warc_CC-MAIN-20210917090202-20210917120202-00166.warc.gz\"}"}
http://opelm.com/qspevdu_t1002002016	[ "", null, "• 湖南\n• 长沙市\n• 常德市\n• 郴州市\n• 衡阳市\n• 怀化市\n• 娄底市\n• 邵阳市\n• 湘潭市\n• 湘西土家族苗族自治州\n• 益阳市\n• 永州市\n• 岳阳市\n• 张家界市\n• 株洲市\n• 山西\n• 长治市\n• 大同市\n• 晋城市\n• 晋中市\n• 临汾市\n• 吕梁市\n• 朔州市\n• 太原市\n• 忻州市\n• 阳泉市\n• 运城市\n• 安徽\n• 安庆市\n• 蚌埠市\n• 亳州市\n• 巢湖市\n• 池州市\n• 滁州市\n• 阜阳市\n• 合肥市\n• 淮北市\n• 淮南市\n• 黄山市\n• 六安市\n• 马鞍山市\n• 宿州市\n• 铜陵市\n• 芜湖市\n• 宣城市\n• 广西\n• 百色市\n• 北海市\n• 崇左市\n• 防城港市\n• 贵港市\n• 桂林市\n• 河池市\n• 贺州市\n• 来宾市\n• 柳州市\n• 南宁市\n• 钦州市\n• 梧州市\n• 玉林市\n• 河南\n• 安阳市\n• 鹤壁市\n• 焦作市\n• 开封市\n• 洛阳市\n• 漯河市\n• 南阳市\n• 平顶山市\n• 濮阳市\n• 三门峡市\n• 商丘市\n• 新乡市\n• 信阳市\n• 许昌市\n• 郑州市\n• 周口市\n• 驻马店市\n• 吉林\n• 白城市\n• 白山市\n• 长春市\n• 吉林市\n• 辽源市\n• 四平市\n• 松原市\n• 通化市\n• 延边朝鲜族自治州\n• 广东\n• 潮州市\n• 东莞市\n• 佛山市\n• 广州市\n• 河源市\n• 惠州市\n• 江门市\n• 揭阳市\n• 茂名市\n• 梅州市\n• 清远市\n• 汕头市\n• 汕尾市\n• 韶关市\n• 深圳市\n• 阳江市\n• 云浮市\n• 湛江市\n• 肇庆市\n• 中山市\n• 珠海市\n• 辽宁\n• 鞍山市\n• 本溪市\n• 朝阳市\n• 大连市\n• 丹东市\n• 抚顺市\n• 阜新市\n• 葫芦岛市\n• 锦州市\n• 辽阳市\n• 盘锦市\n• 沈阳市\n• 铁岭市\n• 营口市\n• 湖北\n• 鄂州市\n• 恩施土家族苗族自治州\n• 黄冈市\n• 黄石市\n• 荆门市\n• 荆州市\n• 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https://ask.libreoffice.org/t/error-using-indirect-in-a-lookup-function/8973	[ "", null, "# Error using INDIRECT in a LOOKUP function\n\nI am having a problem in using INDIRECT in a LOOKUP function in Calc 4.1.3.2 in Linux (Kubuntu 13.10).\n\nIn my worksheet, I have column of values (0-255) in column A, in the cell range A3:A258. Column B has a corresponding set of values (which are image grey scale pixel counts in this case).\n\nFrom this data, I want to do some anaylsis based on maximum and minimum values in column B. However the nature of the data is that can be more than one maximum/minimum in the data in column B so I want to set a range of cells to find local maximum/minimum valies, rather than simply looking at the entire range of cells B3:B258. For example, if I know that there is a local maximum around cell B200, then I want to set the search range to B180:B220 or similar.\n\nOnce the local maximum/minimum has been found (that part of the process works just fine), and is placed into cell B268, I want to use that value in my LOOKUP to tell me what the corresponding grey scale value in column A is.\n\nThe formula I am trying to use looks like this:\n\n=LOOKUP(B268,INDIRECT(“B”&B266):INDIRECT(“B”&B267),INDIRECT(\"\\$A\\$\"&B266):INDIRECT(\"\\$A\\$\"&B267))\n\nwhere:\ncell B268 holds the local maximum/minimum value,\ncell B266 holds the lower bound of the row in the range of cells in column B where the max/min value is,\ncell B267 holds the upper bound of the row in the range of cells in column B where the max/min value is\n\nSo I want the above function to effectively translate for example into:\n\n=LOOKUP(B268;B180:B220;\\$A\\$180:\\$A\\$220)\n\nThe absolute cell references are necessary for column A as there are multiple data columns which will change as I copy the formula across into the data columns but they will all be matched against column A.\n\nI hope that all makes sense.\n\nWhen I try to use the above formula, I get a #N/A error and don’t know why. What am I doing wrong? Any advice would be helpful.\n\nThanks in advance and sorry if this question is a bit long.\n\nLOOKUP() will expect a data range for “looking up” sorted in ascending order. The help text doesn’t mention it. It should be of no meaning whether you use INDIRECT() or a direct reference." ]	[ null, "https://ask.libreoffice.org/uploads/asklibo/original/1X/1eec1ce28d4605f25e751aea59dbef2bc0782151.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86225003,"math_prob":0.7765825,"size":1917,"snap":"2022-05-2022-21","text_gpt3_token_len":484,"char_repetition_ratio":0.12859383,"word_repetition_ratio":0.07058824,"special_character_ratio":0.2649974,"punctuation_ratio":0.09952607,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9777893,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-22T19:32:51Z\",\"WARC-Record-ID\":\"<urn:uuid:f9a81220-d91b-4d15-8bd5-7b5c44d91986>\",\"Content-Length\":\"25580\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4268fc0b-3561-4cfa-b596-d8f4f90a71e1>\",\"WARC-Concurrent-To\":\"<urn:uuid:885e4878-c068-45e3-812d-cb1a8069d77f>\",\"WARC-IP-Address\":\"89.238.68.222\",\"WARC-Target-URI\":\"https://ask.libreoffice.org/t/error-using-indirect-in-a-lookup-function/8973\",\"WARC-Payload-Digest\":\"sha1:6ROXDIOH5XZVPT4MR5E4MI2ULWFEOLYA\",\"WARC-Block-Digest\":\"sha1:Z4EQUYKPYRBGS4ADG3QRHMBM3GNJF4V6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662546071.13_warc_CC-MAIN-20220522190453-20220522220453-00329.warc.gz\"}"}
https://stackoverflow.com/questions/10361974/de-interleave-and-interleave-buffer-with-vdsp-ctoz-and-vdsp-ztoz	[ "# De-interleave and interleave buffer with vDSP_ctoz() and vDSP_ztoz()?\n\nHow do I de-interleave the `float newAudio` into `float channel1` and `float* channel2` and interleave it back into `newAudio`?\n\n``````Novocaine audioManager = [Novocaine audioManager];\n\n__block float channel1;\n__block float channel2;\n[audioManager setInputBlock:^(float newAudio, UInt32 numSamples, UInt32 numChannels) {\n// Audio comes in interleaved, so,\n// if numChannels = 2, newAudio is channel 1, newAudio is channel 2, newAudio is channel 1, etc.\n\n// Deinterleave with vDSP_ctoz()/vDSP_ztoz(); and fill channel1 and channel2\n// ... processing on channel1 & channel2\n// Interleave channel1 and channel2 with vDSP_ctoz()/vDSP_ztoz(); to newAudio\n}];\n``````\n\nWhat would these two lines of code look like? I don't understand the syntax of ctoz/ztoz.\n\nWhat I do in Novocaine's accessory classes, like the Ringbuffer, for de-interleaving:\n\n``````float zero = 0.0;\nvDSP_vsadd(data, numChannels, &zero, leftSampleData, 1, numFrames);\nvDSP_vsadd(data+1, numChannels, &zero, rightSampleData, 1, numFrames);\n``````\n\nfor interleaving:\n\n``````float zero = 0.0;\nvDSP_vsadd(leftSampleData, 1, &zero, data, numChannels, numFrames);\nvDSP_vsadd(rightSampleData, 1, &zero, data+1, numChannels, numFrames);\n``````\n\nThe more general way to do things is to have an array of arrays, like\n\n``````int maxNumChannels = 2;\nint maxNumFrames = 1024;\nfloat arrays = (float )calloc(maxNumChannels, sizeof(float ));\nfor (int i=0; i < maxNumChannels; ++i) {\narrays[i] = (float )calloc(maxNumFrames, sizeof(float));\n}\n\n[[Novocaine audioManager] setInputBlock:^(float data, UInt32 numFrames, UInt32 numChannels) {\nfloat zero = 0.0;\nfor (int iChannel = 0; iChannel < numChannels; ++iChannel) {\nvDSP_vsadd(data, numChannels, &zero, arrays[iChannel], 1, numFrames);\n}\n}];\n``````\n\nwhich is what I use internally a lot in the RingBuffer accessory classes for Novocaine. I timed the speed of vDSP_vsadd versus memcpy, and (very, very surprisingly), there's no speed difference.\n\nOf course, you can always just use a ring buffer, and save yourself the hassle\n\n``````#import \"RingBuffer.h\"\n\nint maxNumFrames = 4096\nint maxNumChannels = 2\nRingBuffer ringBuffer = new RingBuffer(maxNumFrames, maxNumChannels)\n\n[[Novocaine audioManager] setInputBlock:^(float data, UInt32 numFrames, UInt32 numChannels) {\n}];\n\n[[Novocaine audioManager] setOuputBlock:^(float data, UInt32 numFrames, UInt32 numChannels) {\nringBuffer->FetchInterleavedData(data, numFrames, numChannels);\n}];\n``````\n\nHope that helps.\n\n• Thanks, that looks like a clean way to do it! Apr 29, 2012 at 19:29\n• Alex, please take a look at this question, i'm trying to add to your novacaine example by allowing it to read VBR data (in SInt16 rather than float) Nov 5, 2012 at 8:49\n\nHere is an example:\n\n``````#include <Accelerate/Accelerate.h>\n\nint main(int argc, const char * argv[])\n{\n// Bogus interleaved stereo data\nfloat stereoInput ;\nfor(int i = 0; i < 1024; ++i)\nstereoInput[i] = (float)i;\n\n// Buffers to hold the deinterleaved data\nfloat leftSampleData [1024 / 2];\nfloat rightSampleData [1024 / 2];\n\nDSPSplitComplex output = {\n.realp = leftSampleData,\n.imagp = rightSampleData\n};\n\n// Split the data. The left (even) samples will end up in leftSampleData, and the right (odd) will end up in rightSampleData\nvDSP_ctoz((const DSPComplex )stereoInput, 2, &output, 1, 1024 / 2);\n\n// Print the result for verification\nfor(int i = 0; i < 512; ++i)\nprintf(\"%d: %f + %f\\n\", i, leftSampleData[i], rightSampleData[i]);\n\nreturn 0;\n}\n``````\n\nsbooth answers how to de-interleave using vDSP_ctoz. Here's the complementary operation, namely interleaving using vDSP_ztoc.\n\n``````#include <stdio.h>\n#include <Accelerate/Accelerate.h>\n\nint main(int argc, const char argv[])\n{\nconst int NUM_FRAMES = 16;\nconst int NUM_CHANNELS = 2;\n\n// Buffers for left/right channels\nfloat xL[NUM_FRAMES];\nfloat xR[NUM_FRAMES];\n\n// Initialize with some identifiable data\nfor (int i = 0; i < NUM_FRAMES; i++)\n{\nxL[i] = 2i; // Even\nxR[i] = 2i+1; // Odd\n}\n\n// Buffer for interleaved data\nfloat stereo[NUM_CHANNELSNUM_FRAMES];\nvDSP_vclr(stereo, 1, NUM_CHANNELSNUM_FRAMES);\n\n// Interleave - take separate left & right buffers, and combine into\n// single buffer alternating left/right/left/right, etc.\nDSPSplitComplex x = {xL, xR};\nvDSP_ztoc(&x, 1, (DSPComplex)stereo, 2, NUM_FRAMES);\n\n// Print the result for verification. Should give output like\n// i: L, R\n// 0: 0.00, 1.00\n// 1: 2.00, 3.00\n// etc...\nprintf(\" i: L, R\\n\");\nfor (int i = 0; i < NUM_FRAMES; i++)\n{\nprintf(\"%2d: %5.2f, %5.2f\\n\", i, stereo[2i], stereo[2*i+1]);\n}\nreturn 0;\n}\n``````" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5222459,"math_prob":0.95991635,"size":1718,"snap":"2022-40-2023-06","text_gpt3_token_len":488,"char_repetition_ratio":0.1966161,"word_repetition_ratio":0.057692308,"special_character_ratio":0.26018625,"punctuation_ratio":0.26870748,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95115167,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-30T17:25:33Z\",\"WARC-Record-ID\":\"<urn:uuid:8ccc39a3-c88d-4d1a-84f7-2d62e2239ef3>\",\"Content-Length\":\"165421\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:49c2afb0-a33b-4dae-ae4b-b7807b068aba>\",\"WARC-Concurrent-To\":\"<urn:uuid:6fef707a-4132-444a-93c6-03ad53c3cbb4>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/10361974/de-interleave-and-interleave-buffer-with-vdsp-ctoz-and-vdsp-ztoz\",\"WARC-Payload-Digest\":\"sha1:AJ7FZAASFIIKUTHV46CIVT526DKJUIP3\",\"WARC-Block-Digest\":\"sha1:WK23AOIBR5DNYV3VS4A56TWT6K5GNI3G\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335491.4_warc_CC-MAIN-20220930145518-20220930175518-00022.warc.gz\"}"}
https://www.geeksforgeeks.org/python-program-to-arrange-given-numbers-to-form-the-biggest-number/	[ "# Python Program to Arrange given numbers to form the biggest number\n\n• Last Updated : 22 Dec, 2021\n\nGiven an array of numbers, arrange them in a way that yields the largest value. For example, if the given numbers are {54, 546, 548, 60}, the arrangement 6054854654 gives the largest value. And if the given numbers are {1, 34, 3, 98, 9, 76, 45, 4}, then the arrangement 998764543431 gives the largest value.\n\nA simple solution that comes to our mind is to sort all numbers in descending order, but simply sorting doesn’t work. For example, 548 is greater than 60, but in output 60 comes before 548. As a second example, 98 is greater than 9, but 9 comes before 98 in output.\n\nSo how do we go about it? The idea is to use any comparison based sorting algorithm.\nIn the used sorting algorithm, instead of using the default comparison, write a comparison function myCompare() and use it to sort numbers.\n\nGiven two numbers X and Y, how should myCompare() decide which number to put first – we compare two numbers XY (Y appended at the end of X) and YX (X appended at the end of Y). If XY is larger, then X should come before Y in output, else Y should come before. For example, let X and Y be 542 and 60. To compare X and Y, we compare 54260 and 60542. Since 60542 is greater than 54260, we put Y first.\n\nFollowing is the implementation of the above approach.\nTo keep the code simple, numbers are considered as strings, the vector is used instead of a normal array.\n\nBelow is the implementation of the above approach:\n\n## Python3\n\n `# Python3 Program to get the maximum``# possible integer from given array``# of integers...`` ` ` ` `# Custom comparator to sort according``# to the ab, ba as mentioned in description``def` `comparator(a, b):`` ``ab ``=` `str``(a) ``+` `str``(b)`` ``ba ``=` `str``(b) ``+` `str``(a)`` ``return` `((``int``(ba) & gt`` ``int``(ab)) ``-`` ``(``int``(ba) & lt`` ``int``(ab)))`` ` `def` `myCompare(mycmp):`` ` ` ``# Convert a cmp= function into a `` ``# key= function`` ``class` `K(``object``):`` ``def` `__init__(``self``, obj, ``*``args):`` ``self``.obj ``=` `obj`` ` ` ``def` `__lt__(``self``, other):`` ``return` `mycmp(``self``.obj, other.obj) & `` ``lt ``0`` ` ` ``def` `__gt__(``self``, other):`` ``return` `mycmp(``self``.obj, other.obj) & `` ``gt ``0`` ` ` ``def` `__eq__(``self``, other):`` ``return` `mycmp(``self``.obj, other.obj) ``=``=` `0`` ` ` ``def` `__le__(``self``, other):`` ``return` `mycmp(``self``.obj, other.obj) & `` ``lt ``=` `0`` ` ` ``def` `__ge__(``self``, other):`` ``return` `mycmp(``self``.obj, other.obj) & `` ``gt ``=` `0`` ` ` ``def` `__ne__(``self``, other):`` ``return` `mycmp(``self``.obj, other.obj) !``=` `0`` ``return` `K`` ` ` ` `# Driver code``if` `__name__ ``=``=` `\"__main__\"``:`` ``a ``=` `[``54``, ``546``, ``548``, ``60``]`` ``sorted_array ``=` `sorted``(a, key ``=` ` ``myCompare(comparator))`` ``number ``=` `\"\".join([``str``(i) ``for` `i ``in` `sorted_array])`` ``print``(number)`` ` `# This code is Contributed by SaurabhTewary`\n\nOutput:\n\n`6054854654`\n\nTime Complexity: O(nlogn) ,sorting is considered to have running time complexity of O(nlogn) and the for loop runs in O(n) time.\nAuxiliary Space: O(1).\n\nAnother approach:(using itertools\n\nUsing the inbuilt library of the Python, itertools library can be used to perform this task.\n\n## Python3\n\n `# Python3 implementation this is to ``# use itertools. permutations as ``# coded below:`` ` `from` `itertools ``import` `permutations``def` `largest(l):`` ``lst ``=` `[]`` ``for` `i ``in` `permutations(l, ``len``(l)):`` ` ` ``# Provides all permutations of the `` ``# list values, store them in list to `` ``# find max`` ``lst.append(\"\".join(``map``(``str``,i))) `` ``return` `max``(lst)`` ` `print``(largest([``54``, ``546``, ``548``, ``60``])) ``# This code is contributed by Raman Monga`\n\nOutput:\n\n`6054854654`\n\nTime Complexity: O(nlogn)\nAuxiliary Space: O(1).\n\nPlease refer complete article on Arrange given numbers to form the biggest number \| Set 1 for more details!\n\nMy Personal Notes arrow_drop_up" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.73421985,"math_prob":0.93275166,"size":3476,"snap":"2022-05-2022-21","text_gpt3_token_len":1014,"char_repetition_ratio":0.11981567,"word_repetition_ratio":0.088188976,"special_character_ratio":0.31731877,"punctuation_ratio":0.16923077,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9975572,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-28T02:33:46Z\",\"WARC-Record-ID\":\"<urn:uuid:d76a4a15-eba1-4001-8294-eb402aaa5b82>\",\"Content-Length\":\"141457\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5993355f-0e6d-4485-8c90-97e494100528>\",\"WARC-Concurrent-To\":\"<urn:uuid:c40d271f-dc46-434f-9863-7599e963035a>\",\"WARC-IP-Address\":\"104.104.244.41\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/python-program-to-arrange-given-numbers-to-form-the-biggest-number/\",\"WARC-Payload-Digest\":\"sha1:HZJMTKVOOELZYUIFQI3ISMPDILGA2VBQ\",\"WARC-Block-Digest\":\"sha1:4X3RYNHPJE25M6422EUEDZIPD62A4PHV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663011588.83_warc_CC-MAIN-20220528000300-20220528030300-00366.warc.gz\"}"}
https://www.inchcalculator.com/convert/gigahertz-to-cycle-per-second/	[ "# Gigahertz to Cycles per Second Converter\n\nEnter the frequency in gigahertz below to get the value converted to cycles per second.\n\nResults in Cycles per Second:", null, "1 GHz = 1,000,000,000 cps\n\nDo you want to convert cycles per second to gigahertz?\n\n## How to Convert Gigahertz to Cycles per Second\n\nTo convert a measurement in gigahertz to a measurement in cycles per second, multiply the frequency by the following conversion ratio: 1,000,000,000 cycles per second/gigahertz.\n\nSince one gigahertz is equal to 1,000,000,000 cycles per second, you can use this simple formula to convert:\n\ncycles per second = gigahertz × 1,000,000,000\n\nThe frequency in cycles per second is equal to the frequency in gigahertz multiplied by 1,000,000,000.\n\nFor example, here's how to convert 5 gigahertz to cycles per second using the formula above.\ncycles per second = (5 GHz × 1,000,000,000) = 5,000,000,000 cps\n\n### How Many Cycles per Second Are in a Gigahertz?\n\nThere are 1,000,000,000 cycles per second in a gigahertz, which is why we use this value in the formula above.\n\n1 GHz = 1,000,000,000 cps\n\n## What Is a Gigahertz?\n\nGigahertz is a measure of frequency equal to one billion cycles per second.\n\nThe gigahertz is a multiple of the hertz, which is the SI derived unit for frequency. In the metric system, \"giga\" is the prefix for billions, or 109. Gigahertz can be abbreviated as GHz; for example, 1 gigahertz can be written as 1 GHz.\n\n## What Are Cycles per Second?\n\nCycles per second are a measure of the number of oscillations, or cycles, that occur per second.\n\nCycles per second can be abbreviated as cps, and are also sometimes abbreviated as c/s or cycles/second. For example, 1 cycle per second can be written as 1 cps, 1 c/s, or 1 cycles/second.\n\nCycles per second can be expressed using the formula: cps = Cycles / Times\n\n## Gigahertz to Cycle per Second Conversion Table\n\nTable showing various gigahertz measurements converted to cycles per second.\nGigahertz Cycles Per Second\n0.000000001 GHz 1 cps\n0.000000002 GHz 2 cps\n0.000000003 GHz 3 cps\n0.000000004 GHz 4 cps\n0.000000005 GHz 5 cps\n0.000000006 GHz 6 cps\n0.000000007 GHz 7 cps\n0.000000008 GHz 8 cps\n0.000000009 GHz 9 cps\n0.0000000001 GHz 0.1 cps\n0.000000001 GHz 1 cps\n0.00000001 GHz 10 cps\n0.0000001 GHz 100 cps\n0.000001 GHz 1,000 cps\n0.00001 GHz 10,000 cps\n0.0001 GHz 100,000 cps\n0.001 GHz 1,000,000 cps\n0.01 GHz 10,000,000 cps\n0.1 GHz 100,000,000 cps\n1 GHz 1,000,000,000 cps" ]	[ null, "data:image/gif;base64,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", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8196645,"math_prob":0.9989051,"size":2843,"snap":"2023-40-2023-50","text_gpt3_token_len":863,"char_repetition_ratio":0.30081016,"word_repetition_ratio":0.033472802,"special_character_ratio":0.32641575,"punctuation_ratio":0.14262024,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97974837,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-01T15:44:30Z\",\"WARC-Record-ID\":\"<urn:uuid:23f165fc-cf88-4dd3-913f-a54439c62ced>\",\"Content-Length\":\"39146\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:390c02e7-9862-4ffb-a0f9-30fc874af6d6>\",\"WARC-Concurrent-To\":\"<urn:uuid:80e45641-28a5-4fae-b151-93a7e6ed5005>\",\"WARC-IP-Address\":\"172.67.15.221\",\"WARC-Target-URI\":\"https://www.inchcalculator.com/convert/gigahertz-to-cycle-per-second/\",\"WARC-Payload-Digest\":\"sha1:RMVMJL3QLCLKZDWVMOOFD76VWP5Z3Y6O\",\"WARC-Block-Digest\":\"sha1:MCYAWURT33TPQZRSFGGOLYKC5LMYNDQT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100290.24_warc_CC-MAIN-20231201151933-20231201181933-00637.warc.gz\"}"}
https://www.includehelp.com/java/wrapper-classes-in-java-with-example.aspx	[ "# Wrapper Classes in Java with Example\n\nLearn: Wrapper Classes in Java - in this article we will be learning about the introduction of Wrapper Classes, Why they are used? And Why they were added in Java in the first place?\nSubmitted by Mayank Singh, on June 20, 2017\n\nAs we know,\nJava is an Object-Oriented language, i.e. it closely follows the principles of Classes and Objects, but it is also true that Java is not 100% OOP Language, reason is that still Java uses Primitive data types such as int, char, float, long, double, etc. A need was felt to convert these Primitive Data Types into Classes and Objects, thus Java introduced a concept known as Wrapper Classes.\n\nThe Objects of Wrapper Classes wraps the Primitive data types, this comes in handy when we need to use more methods on a primitive data type like for example suppose we want to convert a Non-String Object to String type we use toString() method , the toString() method will return the String representations of the Objects. Similarly, we can have many other examples.\n\nComing back to Java's Wrapper Classes, let’s see what are the available Wrapper Classes in Java.\n\nData Type Wrapper Class\nint Integer\nfloat Float\nlong Long\nbyte Byte\nshort Short\nchar Character\ndouble Double\nboolean Boolean\n\nLet's us discuss two concepts related to Wrapper Classes, these are pretty straight forward:\n\n### 1) Boxing\n\nConversion of a Primitive Data type to Corresponding Object is known as Boxing, it is handled by the Compiler by the help of Constructors.\n\nExample:\n\n```System.out.println(\"Enter an Integer:\");\nint n=KB.nextInt();\nInteger I=new Integer(n); //Boxing : Creating an Integer Object\nSystem.out.println(I);\n```\n```Input: 8\nOutput: 8\n```\n\n### 2) Unboxing\n\nIt can be considered as opposite to Boxing, when the Object needs to be converted back into corresponding primitive data type, it is then known as Unboxing.\n\nExample:\n\n```Integer I=new Integer(n);\ninti=I; //Unboxing : Converting Object to Primitive type\nSystem.out.println(i);\n```\n```Input: 8\nOutput: 8\n```\n\nConsider the program:\n\n```import java.util.*;\n\nclass Wrapper\n{\npublic static void main(String args[])\n{\nScanner KB=new Scanner(System.in);\n\n//int-\t\tInteger\n\nSystem.out.println(\"Enter an Integer:\");\nint n=KB.nextInt();\nInteger I=new Integer(n);\nSystem.out.println(I);\n\n//long-\t\tLong\n\nSystem.out.println(\"Enter a Long Integer:\");\nlong l=KB.nextLong();\nLong L=new Long(l);\nSystem.out.println(L);\n\n//float- \tFloat\n\nSystem.out.println(\"Enter a Float Value:\");\nfloat f=KB.nextFloat();\nFloat F=new Float(f);\nSystem.out.println(F);\n\n//char-\t\tCharacter\n\nSystem.out.println(\"Enter a Character:\");\nchar c=KB.next().charAt(0);\nCharacter C=new Character(c);\nSystem.out.println(C);\n\n//double-\tDouble\n\nSystem.out.println(\"Enter a Double:\");\nDouble d=KB.nextDouble();\nDouble D=new Double(d);\nSystem.out.println(D);\n\n}\n}\n```\n\nOutput\n\n```Enter an Integer:\n8\n8\nEnter a Long Integer:\n1234567898745\n1234567898745\nEnter a Float Value:\n8.8\n8.8\nEnter a Character:\nc\nc\nEnter a Double:\n12.55\n12.55\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7530317,"math_prob":0.5132871,"size":3025,"snap":"2021-31-2021-39","text_gpt3_token_len":711,"char_repetition_ratio":0.16219795,"word_repetition_ratio":0.01847575,"special_character_ratio":0.2509091,"punctuation_ratio":0.1943128,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9604381,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-16T10:39:42Z\",\"WARC-Record-ID\":\"<urn:uuid:89073812-923b-46ab-af77-6b2b8bfe82b2>\",\"Content-Length\":\"141115\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c90b13bc-6c3e-45f3-91ac-32dc42d26a8e>\",\"WARC-Concurrent-To\":\"<urn:uuid:6c661127-d3a1-4feb-b59b-f0d0d634f2d3>\",\"WARC-IP-Address\":\"172.67.140.245\",\"WARC-Target-URI\":\"https://www.includehelp.com/java/wrapper-classes-in-java-with-example.aspx\",\"WARC-Payload-Digest\":\"sha1:NJNOFXO6VNCFH77UUEPGXY47MIEE6UVR\",\"WARC-Block-Digest\":\"sha1:4TMXGJKYP6USJ6HKL6RIZLOBMKHQUQCL\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780053493.41_warc_CC-MAIN-20210916094919-20210916124919-00158.warc.gz\"}"}
https://tracks-movie.com/what-is-the-formula-to-find-steric-number/	[ "# What is the formula to find steric number?\n\n## What is the formula to find steric number?\n\nThe steric number is calculated using the following formula: Steric Number = (number of lone electron pairs on the central atom) + (number of atoms bonded to the central atom)\n\n## How do you calculate steric number in hybridization?\n\nOne way to determine the hybridization of an atom is to calculate its steric number, which is equal to the number of sigma bonds surrounding the atom plus the number of lone pairs on the atoms.\n\nWhat is the hybridization of steric number 3?\n\nSTEP-5: Assign hybridization and shape of molecule\n\nSteric number hybridization Structure\n2 sp linear\n3 sp2 trigonal planar\n4 sp3 tetrahedral\n5 sp3d trigonal bipyramidal\n\nWhat is the AXE formula?\n\nAXE method The electron pairs around a central atom are represented by a formula AXnEm, where A represents the central atom and always has an implied subscript one. Each X represents a ligand (an atom bonded to A). Each E represents a lone pair of electrons on the central atom.\n\n### What is the steric number of xef4?\n\nThe steric number, or the number of atoms and lone pairs, of Xe is 6 . This means that it has a hybridisation of d2sp3 .\n\n### What does the steric number do?\n\nIllustrated Glossary of Organic Chemistry – Steric number. Steric number: The number of atoms, groups, or lone pairs (i.e., electron clouds) around a central atom. The steric number determines molecular geometry.\n\nHow do I calculate bond order?\n\nIf there are more than two atoms in the molecule, follow these steps to determine the bond order:\n\n1. Draw the Lewis structure.\n2. Count the total number of bonds.\n3. Count the number of bond groups between individual atoms.\n4. Divide the number of bonds between atoms by the total number of bond groups in the molecule.\n\nWhat shape is co32?\n\ntrigonal planar\n3 that the molecular geometry of CO 3 2− is trigonal planar with bond angles of 120°.\n\n#### What is the formula to calculate hybridization?\n\nHybridization=1/2(valency electron in central atom+no. Of atom attached to central atom by single bond+negative charge-positive charge).\n\n#### What is steric molecule number?\n\nSteric number: The number of atoms, groups, or lone pairs (i.e., electron clouds) around a central atom. The steric number determines molecular geometry. Carbon dioxide.\n\nWhat is steric number?\n\nKey Takeaways for Steric Number In chemistry, a molecule’s steric number is the number of atoms bonded to the central atom plus the number of lone electron pairs surrounding the central atom. The steric number is used in VSEPR theory to predict molecular geometry.\n\nHow do you find the steric number of a compound?\n\nThe steric number is a property of an atom, not a compound. You need to know what an atom connected to a given atom to know its steric number. For simple compounds, you can usually determine these connections because the formula suggests a central atom and surrounding groups.\n\n## How do you calculate steric number and hybridisation?\n\nShow activity on this post. The steric number is equal to the number of σ -bonds + the number of lone pairs of electrons on the central atom. It gives us the number of hybridised orbitals.\n\n## What is steric number in VSEPR theory?\n\nIn chemistry, a molecule’s steric number is the number of atoms bonded to the central atom plus the number of lone electron pairs surrounding the central atom. The steric number is used in VSEPR theory to predict molecular geometry. Molecular Geometry Introduction. Valence Shell Electron Pair Repulsion Theory." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8280253,"math_prob":0.99599016,"size":3464,"snap":"2023-40-2023-50","text_gpt3_token_len":798,"char_repetition_ratio":0.2248555,"word_repetition_ratio":0.22030981,"special_character_ratio":0.20842956,"punctuation_ratio":0.10534351,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9950191,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-08T16:36:49Z\",\"WARC-Record-ID\":\"<urn:uuid:90de844f-4b86-4c1f-b0a6-8da6ab6769fb>\",\"Content-Length\":\"52171\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:90275926-3a87-4b02-92a5-a149fa3353e4>\",\"WARC-Concurrent-To\":\"<urn:uuid:de06fe7f-f1e9-4988-8a8a-d008414cc907>\",\"WARC-IP-Address\":\"172.67.151.13\",\"WARC-Target-URI\":\"https://tracks-movie.com/what-is-the-formula-to-find-steric-number/\",\"WARC-Payload-Digest\":\"sha1:SJD2QDAX5FXT24J524ZVVPWG4H6KYFAM\",\"WARC-Block-Digest\":\"sha1:PYWTRZ4GTHEZRP747BFW7GL2BSVWUD7Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100762.64_warc_CC-MAIN-20231208144732-20231208174732-00773.warc.gz\"}"}
https://www.ademcetinkaya.com/2023/04/aef-abrdn-emerging-markets-equity.html	[ "Outlook: abrdn Emerging Markets Equity Income Fund Inc. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating.\nDominant Strategy : Hold\nTime series to forecast n: 23 Apr 2023 for (n+1 year)\nMethodology : Multi-Instance Learning (ML)\n\n## Abstract\n\nabrdn Emerging Markets Equity Income Fund Inc. Common Stock prediction model is evaluated with Multi-Instance Learning (ML) and Pearson Correlation1,2,3,4 and it is concluded that the AEF stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Hold\n\n## Key Points\n\n1. How accurate is machine learning in stock market?\n2. Should I buy stocks now or wait amid such uncertainty?\n3. Stock Forecast Based On a Predictive Algorithm\n\n## AEF Target Price Prediction Modeling Methodology\n\nWe consider abrdn Emerging Markets Equity Income Fund Inc. Common Stock Decision Process with Multi-Instance Learning (ML) where A is the set of discrete actions of AEF stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4\n\nF(Pearson Correlation)5,6,7= $\\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \\dots & {p}_{1n}\\\\ & ⋮\\\\ {p}_{j1}& {p}_{j2}& \\dots & {p}_{jn}\\\\ & ⋮\\\\ {p}_{k1}& {p}_{k2}& \\dots & {p}_{kn}\\\\ & ⋮\\\\ {p}_{n1}& {p}_{n2}& \\dots & {p}_{nn}\\end{array}$ X R(Multi-Instance Learning (ML)) X S(n):→ (n+1 year) $\\begin{array}{l}\\int {r}^{s}\\mathrm{rs}\\end{array}$\n\nn:Time series to forecast\n\np:Price signals of AEF stock\n\nj:Nash equilibria (Neural Network)\n\nk:Dominated move\n\na:Best response for target price\n\nFor further technical information as per how our model work we invite you to visit the article below:\n\nHow do AC Investment Research machine learning (predictive) algorithms actually work?\n\n## AEF Stock Forecast (Buy or Sell) for (n+1 year)\n\nSample Set: Neural Network\nStock/Index: AEF abrdn Emerging Markets Equity Income Fund Inc. Common Stock\nTime series to forecast n: 23 Apr 2023 for (n+1 year)\n\nAccording to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Hold\n\nX axis: Likelihood% (The higher the percentage value, the more likely the event will occur.)\n\nY axis: Potential Impact% (The higher the percentage value, the more likely the price will deviate.)\n\nZ axis (Grey to Black): Technical Analysis%\n\n## IFRS Reconciliation Adjustments for abrdn Emerging Markets Equity Income Fund Inc. Common Stock\n\n1. Unless paragraph 6.8.8 applies, for a hedge of a non-contractually specified benchmark component of interest rate risk, an entity shall apply the requirement in paragraphs 6.3.7(a) and B6.3.8—that the risk component shall be separately identifiable—only at the inception of the hedging relationship.\n2. If a financial instrument is designated in accordance with paragraph 6.7.1 as measured at fair value through profit or loss after its initial recognition, or was previously not recognised, the difference at the time of designation between the carrying amount, if any, and the fair value shall immediately be recognised in profit or loss. For financial assets measured at fair value through other comprehensive income in accordance with paragraph 4.1.2A, the cumulative gain or loss previously recognised in other comprehensive income shall immediately be reclassified from equity to profit or loss as a reclassification adjustment.\n3. An entity applies IAS 21 to financial assets and financial liabilities that are monetary items in accordance with IAS 21 and denominated in a foreign currency. IAS 21 requires any foreign exchange gains and losses on monetary assets and monetary liabilities to be recognised in profit or loss. An exception is a monetary item that is designated as a hedging instrument in a cash flow hedge (see paragraph 6.5.11), a hedge of a net investment (see paragraph 6.5.13) or a fair value hedge of an equity instrument for which an entity has elected to present changes in fair value in other comprehensive income in accordance with paragraph 5.7.5 (see paragraph 6.5.8).\n4. There are two types of components of nominal amounts that can be designated as the hedged item in a hedging relationship: a component that is a proportion of an entire item or a layer component. The type of component changes the accounting outcome. An entity shall designate the component for accounting purposes consistently with its risk management objective.\n\nInternational Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.\n\n## Conclusions\n\nabrdn Emerging Markets Equity Income Fund Inc. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating. abrdn Emerging Markets Equity Income Fund Inc. Common Stock prediction model is evaluated with Multi-Instance Learning (ML) and Pearson Correlation1,2,3,4 and it is concluded that the AEF stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Hold\n\n### AEF abrdn Emerging Markets Equity Income Fund Inc. Common Stock Financial Analysis\n\nRating Short-Term Long-Term Senior\nOutlookBa1Ba1\nIncome StatementBaa2C\nBalance SheetBaa2Baa2\nLeverage RatiosCCaa2\nCash FlowCC\nRates of Return and ProfitabilityCB2\n\n*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.\nHow does neural network examine financial reports and understand financial state of the company?\n\n### Prediction Confidence Score\n\nTrust metric by Neural Network: 72 out of 100 with 629 signals.", null, "## References\n\n1. Bottou L. 2012. Stochastic gradient descent tricks. In Neural Networks: Tricks of the Trade, ed. G Montavon, G Orr, K-R Müller, pp. 421–36. Berlin: Springer\n2. Hoerl AE, Kennard RW. 1970. Ridge regression: biased estimation for nonorthogonal problems. Technometrics 12:55–67\n3. D. Bertsekas. Min common/max crossing duality: A geometric view of conjugacy in convex optimization. Lab. for Information and Decision Systems, MIT, Tech. Rep. Report LIDS-P-2796, 2009\n4. M. Sobel. The variance of discounted Markov decision processes. Applied Probability, pages 794–802, 1982\n5. D. Bertsekas. Nonlinear programming. Athena Scientific, 1999.\n6. Wan M, Wang D, Goldman M, Taddy M, Rao J, et al. 2017. Modeling consumer preferences and price sensitiv- ities from large-scale grocery shopping transaction logs. In Proceedings of the 26th International Conference on the World Wide Web, pp. 1103–12. New York: ACM\n7. Mazumder R, Hastie T, Tibshirani R. 2010. Spectral regularization algorithms for learning large incomplete matrices. J. Mach. Learn. Res. 11:2287–322\nFrequently Asked QuestionsQ: What is the prediction methodology for AEF stock?\nA: AEF stock prediction methodology: We evaluate the prediction models Multi-Instance Learning (ML) and Pearson Correlation\nQ: Is AEF stock a buy or sell?\nA: The dominant strategy among neural network is to Hold AEF Stock.\nQ: Is abrdn Emerging Markets Equity Income Fund Inc. Common Stock stock a good investment?\nA: The consensus rating for abrdn Emerging Markets Equity Income Fund Inc. Common Stock is Hold and is assigned short-term Ba1 & long-term Ba1 estimated rating.\nQ: What is the consensus rating of AEF stock?\nA: The consensus rating for AEF is Hold.\nQ: What is the prediction period for AEF stock?\nA: The prediction period for AEF is (n+1 year)" ]	[ null, "https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEie6zmojlVkFtzhnSDPsL0ofN0Qf8imsRuJLmBsuPhvw7a_V8sO4akz1ZmrC1z138iTEnVBz4WOe7nRaUku8PCVLHIr3blhvleBHEbt1VcY4D4zSFZNC02CJ-SsnJxURqjoZxkZaeFVd_lKWZF2n-hK-At4y-ts0RYd4tsXgAkh6yfcsMyVn7NLarhmsw/s1600/footerlogo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87747145,"math_prob":0.8472487,"size":7432,"snap":"2023-14-2023-23","text_gpt3_token_len":1733,"char_repetition_ratio":0.102180935,"word_repetition_ratio":0.17322835,"special_character_ratio":0.21986006,"punctuation_ratio":0.13357401,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96211433,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-07T12:52:55Z\",\"WARC-Record-ID\":\"<urn:uuid:fec2b1b0-6d0d-438d-b6e5-fd01e1e734b7>\",\"Content-Length\":\"321284\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bd6c5c36-c457-49c5-b7b0-df83f74e7fac>\",\"WARC-Concurrent-To\":\"<urn:uuid:754ffc42-a95e-492f-9035-96713ff97540>\",\"WARC-IP-Address\":\"142.251.167.121\",\"WARC-Target-URI\":\"https://www.ademcetinkaya.com/2023/04/aef-abrdn-emerging-markets-equity.html\",\"WARC-Payload-Digest\":\"sha1:P2EISUOWSHYPGTUY2RHO2QNHV4P3GFV5\",\"WARC-Block-Digest\":\"sha1:7WZEGMHGACAOTV4M2GB7HLAL7V27JGMA\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224653764.55_warc_CC-MAIN-20230607111017-20230607141017-00210.warc.gz\"}"}