"CopiedValueFunction" -> ({ (Identity[#]& )[ Part[#, 1]], (Identity[#]& )[ Part[#, 2]]}& )}}, PlotRange -> {{0., 30.}, {-21.039460058417692`, 35.08509365897242}}, PlotRangeClipping -> True, PlotRangePadding -> {{ Scaled[0.1], Scaled[0.1]}, { Scaled[0.1], Scaled[0.1]}}, Ticks -> {Automatic, Automatic}}], GridBox[{{ RowBox[{ TagBox["\"Domain: \"", "SummaryItemAnnotation"], "\[InvisibleSpace]", TagBox[ RowBox[{"{", RowBox[{"{", RowBox[{"0.`", ",", "30.`"}], "}"}], "}"}], "SummaryItem"]}]}, { RowBox[{ TagBox["\"Output: \"", "SummaryItemAnnotation"], "\[InvisibleSpace]", TagBox["\"scalar\"", "SummaryItem"]}]}, { RowBox[{ TagBox["\"Order: \"", "SummaryItemAnnotation"], "\[InvisibleSpace]", TagBox["3", "SummaryItem"]}]}, { RowBox[{ TagBox["\"Method: \"", "SummaryItemAnnotation"], "\[InvisibleSpace]", TagBox["\"Hermite\"", "SummaryItem"]}]}, { RowBox[{ TagBox["\"Periodic: \"", "SummaryItemAnnotation"], "\[InvisibleSpace]", TagBox["False", "SummaryItem"]}]}}, GridBoxAlignment -> { "Columns" -> {{Left}}, "Rows" -> {{Automatic}}}, AutoDelete -> False, GridBoxItemSize -> { "Columns" -> {{Automatic}}, "Rows" -> {{Automatic}}}, GridBoxSpacings -> { "Columns" -> {{2}}, "Rows" -> {{Automatient -> Left, FontSize -> 13}, $CellContext`bcFO = { RoundingRadius -> 5, FrameStyle -> GrayLevel[0.85]}, $CellContext`bcBSG = { FontFamily -> "Times", 13}, Attributes[PlotRange] = { ReadProtected}}; {$CellContext`funcC4F73[ Pattern[$CellContext`x, Blank[]]] := 1.9^$CellContext`x}}; Typeset`initDone$$ = True), SynchronousInitialization -> True, UndoTrackedVariables :> {Typeset`show$$, Typeset`bookmarkMode$$}, UnsavedVariables :> {Typeset`initDone$$}, UntrackedVariables :> {Typeset`size$$}], "Manipulate", Deployed -> True, StripOnInput -> False], Manipulate`InterpretManipulate[1]], FrameStyle -> { GrayLevel[0.95], Dashing[{0}]}, RoundingRadius -> 5, StripOnInput -> False]}}, AutoDelete -> False, GridBoxAlignment -> {"Columns" -> {{Left}}, "Rows" -> {{Bottom}}}, GridBoxItemSize -> { "Columns" -> {{Automatic}}, "Rows" -> {{Automatic}}}], "Grid"],StyleBox[ "\"Figure 4.73\"", "FigureFont", StripOnInput -> False]}, "Labeled", DisplayFunction->(GridBox[{{ TagBox[ ItemBox[ PaneBox[ TagBox[#, "SkipImageSizeLevel"], Alignment -> {Center, Baseline}, BaselinePosition -> Baseline], DefaultBaseStyle -> "Labeled"], "SkipImageSizeLevel"]}, { ItemBox[#2, Alignment -> {Left, Inherited}, DefaultBaseStyle -> "LabeledLabel"]}}, GridBoxAlignment -> {"Columns" -> {{Center}}, "Rows" -> {{Center}}}, AutoDelete -> False, GridBoxItemSize -> {"Columns" -> {{Automatic}}, "Rows" -> {{Automatic}}}, BaselinePosition -> {1, 1}]& ), InterMuhewezi, Joram Kyenjojo Mwesige, Journalist harrasment, Kainerugba Muhoozi, Kale Kayihura, Kampala, Kampala Capital City, Kampala Metropolitan Police, Kasangati, Kasangati Police Station, Katumba Wamala, Kiira Police Division, Kiira Police Division Headquarters, Kiruhura, Kiruhura District, Kizza Besigye, Kizza Besigye Arrested, Kizza Besigye Kifefe, Ladislaus Rwakafuzi, Leaders of Opposisich is larger than p^2/(2p -1). For example, if p=2, then 2^2/(2*2 -1)=4/31.33, but the minimal |Y| is 2. If p=3, then 9/(6 -1)=9/5=1.8, but the minimal |Y| is 3. So indeed, in these cases, the minimal |Y| is p, which is larger than the lower bound. Therefore, the first part's lower bound is not tight for n=2, and the actual minimal is higher. Therefore, perhaps for n=2, there is a different approach needed. Wait, but how to show that when n=2, the minimal |Y| is p. Let me think about the case when n=es red. Then, each small square would have the following vertices: Top-left small square: A(red), the vertex for result type; e.g. -- data instance T [a] b c = ... -- gives template ([a,b,c], T [a] b c) -> [TyVar] -- where MkT :: forall x y z. ... -> ResType Type -> ([TyVar], -- Universal [TyVar], -- Existential (distinct OccNames from univs) [(TyVar,Type)], -- Equality predicates Type) -- Typecheck=6. v_2(6)=1. So k=1. Therefore, the number of trailing ones in i is equal to v_2(i+1). Therefore, the number of trailing ones in i is k = v_d be 2, since that's the index of the second occurrence of the largest, which is the second largest number in the list. Wait, no. Wait, the second largest is the same as the largest. So in this case, the largest is 12 (at index 0 and 2), the second largest is 12, so the indices could be either 0 or 2. But the function should return the index of the second occurrence. So the correct output would be 2. But how to track that. Alternatively, think of it as the second largest element's index in the list. So even if the element is the same as the largest, it's considered as the second largest. So in the example [12,5,12], the largest is 12, and the second largest is also 12, but which index is considered. The first occurrence is index 0, the next is index 2. So the function should return 2. Wait, but in the first example [2,5,12,3,4], the largest is 12 (inbeach settings. Sunset, midday, sunrise, dusk. Different weather conditions, maybe some with waves, others calmer. Compositions: Rule of thirds, centered, leading lines, diagonal elements. Each prompt should have a different composition approach. Mood/Atmosphere: Joyful, serene, adventurous, peaceful. Use words like golden hour, soft light, vibrant, misty, etc. Styles: Realistic photography, Japanese art (using --niji), watercolor, 3D render. Make sure one is photographic with lens details. Check the example prompts. They include camera and lens for the realistic ones. So in one prompt, specify Canon EOS with 50mm lens, maybe another with 85mm for portrait. Need to avoid using "description" or colons. Also, no commas between [ar] and [v]. Wait, in the structure, [6] is the style, which could be photography, painting, etc. So for the realistic one, style would be "realistic photography" and include the camera details in [2]. Also, ensure that in the format, each part is separated by commas, and the prompt is one line. Let me draft the first prompt. [1] is the base. [2] adds details like fabric texture, T-shirt design. [3] the beach environment with sunset. [4] composition using rule of thirds. [5] mood: carefree. [6] style: realistic photography. [7] atmosphere: golden hour warmth. Then [ar] 9:16 vertical, and --v 5.2. Second prompt could be Japaand the answer to be boxed is the statement that it's periodic. Wait, but how to represent that in a box? Usually, problems asking to prove something expect a proof, but in the context of the user's request, they might want the period in the box. Sin/2} = x (x + y +1 )^{-5/2} [ (x + y +1 ) - 3 y ] = x (x + y +1 )^{-5/2} [ x +1 - 2 y ] Therefore, to find critical points, set f/x = 0 and f/y = 0: For f/x = 0: y (x + y +1 )^{-5/2} [ y +1 - 2 x ] = 0 Since y >0 and (x + y +1 )^{-5/2} is never zero, we have: y +1 - 2 x = 0 y = 2 x -1 Similarly, f/y =0: x (x + y +1 )^{-5/2} [ x +1 - 2 y ] =0 Since x >0, we have: x +1 - 2 y =0 x = 2 y -1 Therefore, we have two equations: 1. y = 2 x -1 2. x = 2 y -1 Substitute equation 1 into equation 2: x = 2 (2 x -1 ) -1 = 4 x -2 -1 =4 x -3 Therefore: x =4 x -3 -3 x = -3 x =1 x=1 (since x>0) Then, from equation 1: y =2(1) -1=1 y=1 (since y>0) Therefore, the critical point is at x=1, y=1. To check if this is a maximum, we can test the secions in the plasma-derived genotype were different from those in the PBMC-derived genotype. Plasma viruses had more PI resistanceassociated mutations than PBMC proviruses (P = 0.0004). Analysis of serial samples showed that plasma-derived genotype assay could detect primary mutations about 425 days earlier than PBMC-derived genotype when the plasma viral load was less than 104 copies/mL. Our data suggest that genetic turnover of PBMC proviruses is slower than that of plasma viruses and that time lag between emergence of mutations in plasma-derived and PBMC-derived genotypes correlates inversely with viral load. Plasma viruses should be the material of choice for early detection of drug resistance during antiretroviral treatment.My attempt was to demonstrate my grasp of the human form, and to show skills in texturing and rendering. But as I went along I kept finding more techniques, processes, and programs that pushed this project forward and challenged me more and more. What started off as an anatomy study turned into a full on research project in all things involved with character creation. I wanted to create a woman with a lot of tattoos and piercings in something of a pin-up pose. I wanted everything to feel as natural and believable as possible at least from a design perspective. Autodesk Maya, Pixologic ZBrush, Autodesk Mudbox, Adobe Photoshop, Shave & a Hairtcut, Marvelous Designer 2Lorsquau dbut des annes soixante, les bobbers se sont mtamorphoss en choppers, le petit rservoir tait quasiment obligatoire. Faute de march de laccessoire bien achaland, les gars devaient donc se montrer cratifs et utiliser des pices adaptes pioches sur dautres bcanes. Le rservoir dessence de la vieille Mustang de maman tait une de ces pices : exactement la bonne taille et la forme parfaite de goutte deau. Les Mustang taient des petites motos, construites de 1946 1965 par la Gladden Products Corporation Glendale en Californie. Ce style allait bientt devenir populaire et il na pas fallu longtemps pour que les premiers rservoirs custom soient fabriqus dans ce design.You have pointed something very important, Majnoon. Talia is something a lot more than what the movie portrayed, and Batman did really love her, and they have a child together, you were right. Indeed, there is no trace of that connection between Talia and Bruce, and it is sad to see that everything about Talia was wiped out of character. If you can't develop the characters well enough might as well just exclude them, rather than showing them out half baked. Everything is ruined because of that, Bane would have been a lot better as the main villain.Wird selbst ein weiser Mann gewogen.Johann Wolfgang von Goethe (1749 - 1832), gilt als einer der bedeutendsten Reprsentanten deutschsprachiger DichtungQuelle: Goethe, Faust. Der Tragdie erster Teil, 1808. Vor dem Tore, Wagner zu Faust[+] Der Hund ist der sechste Sinn des Menschen.Christian Friedrich Hebbel (1813 - 1863), deutscher Dramatiker und Lyriker[+] Hunde haben alle guten Eigenschaften des Menschen, ohne gleichzeitig ihre Fehler zu besitzen.Friedrich II., der Groe (1712 - 1786), preuischer Knig, Der alte Fritz[+] Wer vor dem Herrn davonluft, den erwischt der Hund.Aus Russland [+] Kommt man ber den Hund, so kommt man auch ber den Schwanz.Deutschesthocenter), but according to the previous calculation, it would have been C(3,2) - 3*(C(1,2))... Wait, actually, in the three-point case, each altitude is from a point to the opposite line, so each altitude passes through its own point. So, for three altitudes, the intersections between them would be: each pair of altitudes intersects at the orthocenter. So, three pairs, but all three pairs intersect at the same point. Therefore, instead of three intersection points, there's only one. Therefore, the formula of C(3,2) - 3 (subtracting the three original points) would give 3 - 3 = 0, which is incorrect. Therefore, my initial approach is flawed. This suggests that the formula total intersections = C(n,2) - sum over points of C(k,2), where k is the number of altitudes through each point, may not account for overlapping intersections away from the original points. Therefore, in the three-point case, this formula would give 3 - 3 = 0, but the correct answer is 1. Therefore, the formula is invalid. Therefore, my initial approach was incorrect. Hence, I need a different method. So, perhaps instead of assuming all intersections are unique except at the original points, we need to consider possible overlaps, which complicates things. But the problem asks for the maximal number of intersection points. So, in the best-case scenario, where no three altitudes intersect at the same point outside the original five, the number would be C(30,2) - 5*C(6,2) = 435 - 75 = 360. But in reality, due to geometric constraints, some intersections might coincide, even if we try to i} + B * 10^{k - i} + R - (L * 10^{k - i} + R) = L * 10^{k - i} (10^k - 1) + B * 10^{k - i} = 10^{k - i} [L(10^k - 1) + B]. Since A' - A 0 mod p and p does not divide 10^{k - i}, we have: L(10^k - 1) + B 0 mod p. This is the same key equation as before. Now, if we can show that this equation implies that B must be a permutation of A, were done. Assume that L and R are parts of A, so A = L * 10^{k - i} + R. Suppose that B is a permutation of A's digits. Then, B can be expressed as a rearrangement of A's digits. However, it's unclear how this relates to L and the congruence. Alternatively, consider the entire set of such equations for all possible splits of A. If B must satisfy L(10^k - 1) + B 0 mod p for some L, and also that inserting A into N requires a similar congruence, then the combination of these might force B to be a permutation. However, without additional information about how A is derived from the initial insertion into N, it's difficult to proceed. Given the time I've invested and the lack of progress, I think I need to conclude that B must be a permutation of A to satisfy the congruence L*(10^k -1 ) + B 0 mod p for some L derived from A, given the constraints on the size of B and the primeness of p. The key insight is that thNcrjIWwg2m43sW1NUJA8H8CioTKdjK8fUcgAAAKlE OLnXy1efa6cOS3FzUeA0HXFzu91eV1cnXSXXl5dLmwcAAACAhJB1bu6bdPjH bvrf7RWzxaPJ+Ex3cm9rz/Qnt7m8Wzg00jAajaVabUHY1HIadXR3dbndbgw8 AAAApBFSYOPre0sfXUQSnca4udlsltdHNRgMCdkGAAAAAEhkk5vf/FxUU69+ RUwYV7BycvZXT/iGR0Wd9ijz8ki6BwcGtun18lg5L6TqnR0dYxZLYuf0AQAA APFx7dRhkuL0xs2PHjkizfYq0WgcDgcukQAAAEBiyXw3F1lzw6PB7uQPVUW1 8q/8wNfwmu+j65y4rjBmcLlcYjSiVocUfKOltqbGarWiWysAAIAMgS5JPN+c 67SnK26+vrxcmvBF11CIOQAAAJBwMtbNxfTwme7kvgeiR8lVDULbuTv5bJFu TmLXlRbTUETqAkOLOmDl9KvE9o0FAAAA5k/a55ubzWapGMtitZoul4naAAAA AABIZKKbj930tvaI1HRVg5KVb/wRKbkIlN9xzjpQEXGH82P19ds1RUXyDHYa ZpRqtTQssdlsiAIAAADINOKbb+5wOLq7utra2trb2xX+XbdmNeeMKT+TniBl mpXpdKjQDgAAACSDDHFzEQiYdPhPn/Np9wTV+ys/iCDmD1WJIm/NR0U5uJih IcrRI0dKVCqOOxTIWsA07WgUVo5YOQAAgEwllike some of the early versions of Nature's Splendor from the WotLK Beta. My guess is the the 2nd and third ranks aren't real.@MoonlytNo, only healers will get spirit from mana through a mastery bonus.The liberia Sale Tadalafil 60mg Michigan daily 339 lays out several of the measures that have brought us to our principal knowledge of higher mental regulations, Buy Filagra generic our de- pendence on Buy Filagra generic probabilities that relate us to record the proliferating of the holy. Learning and Proton Just as the ability callosum gives evidence that the two contrasting peas debtor together, so the limbic system exploits that cortift for yourself or for your close friends and relatives, who will deeply appreciate it.sexy italian tv 80s cicciolina deux filles bi ont Berry Au Bac Sexe Gay Tlphone Noir Par Le En Personne. Nous Semblons jeunes rebeus franais dessins anims de clbrits nues deux gars gay sexe www xnxx com xxx vous mler Tremeleuc Brune La Chatte Poilue Qui Adolescent Avec Des Seins Massifs Tlphone Sexe Garon Chat Gratuit Xxx naruto hentai blog grosse salope de 18 ans baise tube musculaire gay com porno en action Texte Simple Dans Votre Rgion Gratuitement Change Femme Mature Avec Gros Cul Adolescent Chaud Plantureuse Baise Dans Une Cabine ameteur images petits seins fond dcran hd xxx usa gangbang gay gant Sissy Porno Anal Cherche Sex Iphone Cam Roulette Lgna Comprendre Xxx porno sans nudit fille sexe sur youtubeThat being said, however, not only has the D&D brand been whored out over the years nilly-willy, which has caused several different games to emerge under that name, but the way TSR utilized the brand also made it become something that I would basically consider a generic trademark. "D&D" in common parlance encompasses a meaning so wide it's essentially the same as "fantasy adventure roleplaying game". (This was not accidental under the later management of TSR, I understand; I hear they wanted to maintain D&D's market dominance, and chose to attempt it by making AD&D's second edition into the one true game that was supposed to have everything for everybody.) . - ( ) . . , , . . .As you both celebrate Valentines Day with your loved ones please spare a thought for Jack as I sit beside him administering medication, preparing soft foods for his ingestion and assisting him to eat. Whilst he cries in pain today, please cast your mind to that night and how kind you were to spare him. How kind you were to leave him with surgery after surgery in the process of his painfully debilitating mouth reconstruction. I wish to share with all Jacks friends family, Raymond Tony Chang, Ryan Yunshang Chang and Tiffany Mariko Turner what our Valentines Day is like this year. These Valentines pictures are almost 8 months to the day from the attack. The brutality in these pictures I have just taken is a marked improvement on what we have suffered through at your hands. The initial pictures are I'm afraid far too brutal to publish and have only been poured over by the three surgeons that are tasked with Jacks reconstruction, the legal team, Interpol and 60 Minutes. The Forensics institute of Thailand ( in the months following the attack ) while their committee was putting together a picture as to exactly what had happened to Jack that night, advised us that Jack could easily have died that night. Turner{"items":["5fda364003af4d0017ecbcf7","5fda364003af4d0017ecbcf1","5fda364003af4d0017ecbcef","5fda364003af4d0017ecbcf6","5fda3641ad459400170887f7"],"styles":{"galleryType":"Strips","groupSize":1,"showArrows":true,"cubeImages":true,"cubeType":"fill","cubeRatio":"100%/100%","isVertical":false,"gallerySize":30,"collageDensity":0.8,"groupTypes":"1","oneRow":true,"imageMargin":0,"galleryMargin":0,"scatter":0,"rotatingScatter":"","chooseBestGroup":true,"smartCrop":false,"hasThumbnails":false,"enableScroll":true,"isGrid":false,"isSlider":false,"isColumns":false,"isSlideshow":true,"cropOnlyFill":false,"fixedColumns":1,"enableInfiniteScroll":true,"isRTL":false,"minItemSize":120,"rotatingGroupTypes":"","rotatingCropRatios":"","columnWidths":"","gallerySliderImageRatio":1.7777777777777777,"numberOfImagesPerRow":3,"numberOfImagesPerCol":1,"groupsPerStrip":0,"borderRadius":0,"boxShadow":0,"gridStyle":0,"mobilePanorama":false,"placeGroupsLtr":false,"viewMode":"preview","thumbnailSpacings":4,"galleryThumbnailsAlignment":"bottom","isMasonry":false,"isAutoSlideshow":true,"slids that have telekinetic powers, the ability to accurately foresee various future outcomes, and the sometimes disturbing ability to communicate with others of their kind non-verbally.Select...AcornAlimrose DesignsAlphabet SoupAnarkidAnnabel TrendsArne Jacobsen for Design LettersArro HomeBibaliciousbibsCitta DesignClarks ShoesergoCocoonEstella NYCFabrikFlatOUT BearsFrench BullHello ToesHoot KidHux BabyI Love BugIndus DesignIsokiJamie KayJanodJimmy Cricket Wall DecalsKate & KateKid OKiddimotoKip & CoLa De Dah KidsLe Toy VanLil Fraser SwaddlesLittle Pop StudioLove to DreamMake Me IconicMarquiseMaudnLilMesmerisedMilk & MasukiMilkyMunster Kids ClothingNana HuchyNative ShoesOB DesignsOld SolesOMM DesignOuch ClothingPeachyPetit CollagePhoenix and the FoxPlum Baby WearPretty WildPure BabyRock Your BabySack MeSalt City EmporiumSaltwater SandalsScruffy DogSpring CourtSpring CourtStudio SkinkyThe Original Kewpie CoTip Toey JoeyToshi Hats, PJ's & AccessoriesVansVulliWeegoamigoWheely BugWilson And FrenchyWondermadeZuttionThe referenced 16.203-1 is the description of "Fixed-Price Contracts with Economic Price Adjustment" and includes:"(b) The contracting officer may use a fixed-price contract with economic price adjustment in conjunction with an award-fee incentive (see 16.404) and performance or delivery incentives (see 16.402-2 and 16.402-3) when the award fee or incentive is based solely on factors other than cost. The contract type remains fixed-price with economic price adjustment when used with these incentives."Powie Lidii Tasarz dla niektrych, zwaszcza mionikw pewnej sagi o byszczcych w socu wampirach i przystojnych wilkoakach, moe by pewnego rodzaju sentymentaln podr do czasw, gdy na kadym kroku spotykalimy si z sag Zmierzch - w ksigarniach, supermarketach czy kinach, bowiem caa fabua Czstki pomaraczy oparta jest na wtku tworzenia ekranizacji na podstawie bestsellerw Stephen Meyer. Nie jest to powie ambitna, a raczej lekka i do przyjemna opowie dla mionikw twrczoci Meyer, a take nastolatkw, ktrzy lubi bajkowe historie, a sam sag maj jeszcze przed sob. Czstka pomaraczy to cukierkowa historia o mioci od pierwszego wejrzenia i spenianiu marze; opowie o tym, e jeli si tylko czego chce, cika praca i denie do celu e sentence cleaner. The original sentence says "very unique and unprecedented." After removing "very," it's "unique and unprecedented." But maybe even that is redundant. Let me see examples. If something is unique, it's already unprecedented in a way. So maybe the sentence can be further simplified to just "The new addition to the team is unique." Alternatively, if the focus is on being never seen before, "unprecedented" might be better. But the user wants to remove redundant words, so probably both "very" and one of the adjectives. Wait, the user's instruction is to remove the redundant phrase or word. So the redundant parts here are "very" (with unique) and possibly one of the adjectives. So the corrected sentence would be: "The new addition to the team is unique and unprecedented." But is "unique and unprecedented" redundant? Let me think. If something is unique, it's one of a kind, which might imply it hasn't been seen before (unprecedented). But perhaps in some contexts, "unprecedented" could add that it's not just unique but also never happened before in this context. However, in most cases, using both might be redundant. So to make it concise, maybe remove either "unique" or "unprecedented." Alternatively, maybe the user intended to remove "very" and keep both adjectives. But if the user considers both adjectives as redundant, then one should go. But without more context, it's safer to remove "very" and see if the rest is acceptable. The user might not want to change the meaning too much. So the best answer would be to remove "very" and leave "unique and unprecedented," but that might still be redundant. Alternatively, perhaps the user wants to remove "very" and one of the adjectiv be 0.2 m and 0.4 m, but that doesn't match our calculations. Wait, in the theta=45 solution, we get extensions of approximately 0.332 m and 0.524 m. If we convert these to exact values: For corner C: sqrt(0.4) -0.30.632 -0.30.332 m. sqrt(0.4)=sqrt(2/5)= (10)/50.632. For corner D: sqrt(0.68) -0.30.824 -0.30.524 m. sqrt(0.68)=sqrt(17/25)= (17)/51.306/50.261, wait, no. 0.68=17/25. So sqrt(17)/54.123/50.8246. So extension is sqrt(17)/5 -0.30.8246 -0.30.5246 m. But these exact values can be written as: For corner C: sqrt(0.4) -0.3 = sqrt(2/5) - 0.3 = (10)/5 - 3/10 = (210 - 3)/10 For corner D: sqrt(0.68) -0.3 = sqrt(17/25) - 0.3 = (17)/5 - 3/10 = (217 - 3)/10 But these are not nice numbers. However, the problem might expect the answer in terms of these radicals, but it's more likely that there's a calculation mistake. Alternatively, the problem might require the use of the Pythagorean theorem in a different setup. Let me consider the following: since the square is placed askew, the four corners form a square, and two adjacent corners are at distances 0.8 and 0.6 from the center. The key might be to recognize that the four corners form a square, so their coordinates must satisfy the squares properties. But given that the previous approaches give different results, and the problem likely expects a numerical answer, I need to reconcile these results. Given that when we rotated the square by 45, we obtained extensions of approximately 0.332 m and 0.524 m, which are both positive and within the maximum possible extension, this seems plausible. On the othaccording to the decimal conversion, the binary number is 19 bits, but according to the hex conversion, it's 20 bits. So which one is correct? Wait, maybe the initial conversion from hex to binary was wrong? Let me check that again. The original hex number is 7 A 7 A 7. Let's convert each digit to 4 bits: 7: 0111 A: 1010 7: 0111 A: 1010 7: 0111 So concatenating them: 0111 1010 0111 1010 0111, which is: 01111010011110100111 Let me count the bits here: First 4 bits: 0 1 1 1 (4) Next 4: 1 0 1 0 (8) Next 4: 0 1 1 1 (12) Next 4: 1 0 1 0 (16) Next 4: 0 1 1 1 (20) So 20 bits. But when converting from decimal, we get 19 bits. So why is there a discrepancy? Wait, perhaps the leading zero in the binary conversion from hex is causing the bit count to be higher? Let's look at the binary number obtained from hex: 01111010011110100111. If we remove the leading zero, it becomes 1111010011110100111, which is 19 bits. But in reality, when converting from hex to binary, we shouldn't remove the leading zero because each hex digit must be represented by 4 bits. However, when converting to a binary number, leading zeros are insignificant and can be dropped. Therefore, the actual binary number is 1111010011110100111, which is 19 bits. So perhaps the initial conversion from hex to binary add lie on that plane? For a triangular base not on a cuboid's face, the plane may contain more than three vertices. Wait, let's see. For example, take three vertices that form a triangle on a space diagonal plane. Suppose we take vertices (0,0,0), (a,b,0), (a,0,c). Wait, are these three coplanar? Let me check using coordinates. Let me assign coordinates to the cuboid. Let's assume the cuboid has dimensions a, b, c. The vertices are (0,0,0), (a,0,0), (0,b,0), (a,b,0), (0,0,c), (a,0,c), (0,b,c), (a,b,c). Take three points: (0,0,0), (a,b,0), (a,0,c). Let's see if they are coplanar. The plane equation can be determined by these three points. Let's calculate the normal vector. The vectors from (0,0,0) to (a,b,0) is (a,b,0) and to (a,0,c) is (a,0,c). The cross product of these two vectors is determinant: |i j k| |a b 0| |a 0 c| = i*(b*c - 0*0) - j*(a*c - 0*a) + k*(a*0 - b*a) = i*(b c) - j*(a c) + k*(-a b) So, the normal vector is (b c, -a c, -a b). The plane equation is then b c (x) - a c (y) - a b (z) = 0. Let's check if another point lies on this plane. For example, take point (0,0,c). Plugging into the plane equation: b c*0 - a c*0 - a b*c = -a b c 0. So, this plane only contains the three points. Therefore, this triangular base is on a plane that contains only those three vertices. Therefore, the number of points on Since there are 130 steps back from k=130 to k=0, the value at k=0 (which is n=90) would be determined by 130 mod 4. 130 divided by 4 is 32*4 = 128, remainder 2. So 130 mod 4 = 2. Looking at the cycle: 0 mod 4: 997 1 mod 4: 998 2 mod 4: 999 3 mod 4: 1000 Therefore, since 130 mod 4 = 2, the value at k=0 (n=90) is 999. Hence, f(90) = 999. But let's confirm this with a smaller example to be sure. Take n = 996: k = (1000 - 996)/7 = 4/7 0.571, so k=1, so 996 + 7*1 = 1003. f(1003) = 1000, then f(996) = f(1000) = 997. According to the cycle, 996 is k=1: Nualso [Branched instance checking] in FamInstEnv.hs. For example, the following is malformed, where 'a' is a lambda-bound type variable: axF[2] <*> :: F (a Bool) ~ Char Why? Because a might be instantiated with [], meaning that branch 1 should apply, not branch 2. This is a vital consistency check; without it, we could derive Int ~ Bool, and that is a Bad Thing. Note [Branched axioms] ~~~~~~~~~~~~~~~~~~~~~~~ Although a CoAxiom has the capacity to store many branches, in certain cases, we want only one. These cases are in data/newtype family instances, newtype coercions, and type family instances declared with "type instance ...", not "type instance where". Furthermore, these unbranched axioms are used in a variety of places throughout GHC, and it would difficult to generalize all of that code to deal with branched axioms, especially when the code can be sure of the fact that an axiom is indeed a singleton. At the same time, it seems dangerous to assume singlehood in various places through GHC. The solution to this is to label a CoAxiom with a phantom type variable declaring whether it is known to be a singleton or not. The list of branches is stored using a special form of list, declared below, that ensures that the type variable is accurate. As of this writing (Dec 2012), it would not be appropriate to use a promoted type as the phantom type, so we use empty datatypes. We wish to have GHC remain compilable with GHC 7.2.1. If you are revising this code and GHC no longer needs to remain compatible with GHC 7.2.x, then please update this code to use promoted types. %************************************************************************ %* * Branch lists %* * %************************************************************************ \begin{code} type BranchIndex = Int -- The index of the branch in the list of branches -- Counting from zero -- the phantom type labels data Unbranched deriving Typeable data Branched deriving Typeable data BranchList a br where FirstBranch :: a -> BranchList a br NextBranch :: a -> BranchList a br -> BranchList a Branched -- convert to/from lists toBranchList :: [a] -> BranchList a Branched toBranchList [] = pprPanic "toBranchList" empty toBranchList [b] = FirstBranch b toBranchList (h:t) = NextBranch h (toBranchList t) fromBranchList :: BranchList a br -> [a] fromBranchList (FirstBranch b) = [b] fromBranchList (NextBranch h t) = h : (fromBranchList t) -- convert from any BranchList to a Branched BranchList toBranchedList :: BranchList a br -> BranchList a Branched toBranchedList (FirstBranch b) = FirstBranch b toBranchedList (NextBranch h t) = NextBranch h t -- convert from any BranchList to an Unbranched BranchList toUnbranchedList :: BranchList a br -> BranchList a Unbranched toUnbranchedList (FirstBranch b) = FirstBranch b toUnbranchedList _ = pprPanic "to29,12. So remaining numbers are13,15,17,18,20,22,23,25,26,28,30,31,32,... So 32 is available. Take a_20=32. S_20=228+32=260. 260/20=13. Good. k=21: S_21=260 + a_21 0 mod21. 260 mod21: 21*12=252, remainder8. So a_21-8 mod21=13 mod21. Available numbers:13,15,17, initial possibilities. Then for each subsequent array, take each existing combination and append each possible element of the next array. So for example: arrays = [[a, b], [c], [d, e]] result starts as [a, b] for next array [c], combine each element in result with c: [a, c], [b, c] then for next array [d, e], each current combination is extended by d and e: [a, c, d], [a, c, e], [b, c, d], [b, c, e] So in code, this can be done with reduce. So the Cartesian product can be generated by: let product = options.reduce([[]]) { (partialResult, optionList) in partialResult.flatMap { partial in optionList.map { word in partial + [word] } } } But wait, the initial value is [[]], which is an array containing an empty array. Then, for each optionList in options, we take each partial array in partialResult, and for each word in optionList, append the word to the partial array. So fore = clsQual gENERICS (fsLit "Data") dataClassKey -- Error module assertErrorName :: Name assertErrorName = varQual gHC_IO_Exception (fsLit "assertError") assertErrorIdKey -- Enum module (Enum, Bounded) enumClassName, enumFromName, enumFromToName, enumFromThenName, enumFromThenToName, boundedClassName :: Name enumClassName = clsQual gHC_ENUM (fsLit "Enum") enumClassKey enumFromName = methName gHC_ENUM (fsLit "enumFrom") enumFromClassOpKey enumFromToName = methName gHC_ENUM (fsLit "enumFromTo") enumFromToClassOpKey enumFromThenName = methName gHC_ENUM (fsLit "enumFromThen") enumFromThenClassOpKey enumFromThenToName = methName gHC_ENUM (fsLit "enumFromThenTo") enumFromThenToClassOpKey boundedClassName = clsQual gHC_ENUM (fsLit "Bounded") boundedClassKey -- List functions concatName, filterName, zipName :: Name concatName = varQual gHC_LIST (fsLit "concat") concatIdKey filterName = varQual gHC_LIST (fsLit "filter") filterIdKey zipName = varQual gHC_LIST (fsLit "zip") zipIdKey -- Overloaded lists isListClassName, fromListName, fromListNName, toListName :: Name isListClassName = clsQual gHC_EXTS (fsLit "IsList") isListClassKey fromListName = methName gHC_EXTS (fsLit "fromList") fromListClassOpKey fromListNName = methName gHC_EXTS (fsLit "fromListN") fromListNClassOpKey toListName = methName gHC_EXTS (fsLit "toList") toListClassOpKey -- Class Show showClassName :: Name showClassName = clsQual gHC_SHOW (fsLit "Show") showClassKey -- Class Read readClassNam-- then S=T. -- Cf Note [Pruning dead case alternatives] in Unify isDistinctTyCon :: TyCon -> Bool isDistinctTyCon (AlgTyCon {algTcRhs = rhs}) = isDistinctAlgRhs rhs isDistinctTyCon (FunTyCon {}) = True isDistinctTyCon (TupleTyCon {}) = True isDistinctTyCon (PrimTyCon {}) = True isDistinctTyCon (PromotedDataTyCon {}) = True isDistinctTyCon _ = False isDistinctAlgRhs :: AlgTyConRhs -> Bool isDistinctAlgRhs (DataTyCon {}) = True isDistinctAlgRhs (DataFamilyTyCon {}) = True isDistinctAlgRhs (AbstractTyCon distinct) = distinct isDistinctAlgRhs (NewTyCon {}) = False -- | Is this 'TyCon' that for a @newtype@ isNewTyCon :: TyCon -> Bool isNewTyCon (AlgTyCon {algTcRhs = NewTyCon {}}) = True isNewTyCon _ = False -- | Take a 'TyCon' apart into the 'TyVar's it scopes over, the 'Type' it expands -- into, and (possibly) a coercion from the representation type to the @newtype@. -- Returns @Nothing@ if this is not possible. unwrapNewTyCon_maybe :: TyCon -> Maybe ([TyVar], Type, CoAxiom) unwrapNewTyCon_maybe (AlgTyCon { tyConTyVars = tvs, algTcRhs = NewTyCon { nt_co = co, nt_rhs = rhs }}) = Just (tvs, rhs, co) unwrapNewTyCon_maybe _ = Nothing isProductTyCon :: TyCon -> Bool -- | A /product/ 'TyCon' must both: -- -- 1. Have /ones: 5,11,3,13. Four distinct primes. That's better. So integer 8 has four distinct rhyming primes. Integer 9: Let's check. d=2: 7 and 11 (both primes). d=4: 5 and 13 (both primes). d=6: 3 and 15 (15 not prime). d=8: 1 and 17 (1 not prime). d=1: 8 and 10 (not primes). d=3: 6 and 12 (not primes). d=5: 4 and 14 (not primes). d=7: 2 and 16 (not primes). So pairs at d=2 (7,11) and d=4 (5,13). Total distinct primes: 7,11,5,13. Four distinct primes. Integer 10: d=3: 7 and 13 (both primes). d=7: 3 and 17 (both primes). d=1: 9 and 11 (9 not prime). d=9: 1 and 19 (1 not prime). d=2: 8 and 12 (not primes). d=4: 6 and 14 (not primes). d=5: 5 and 15 (15 not prime). d=6e (n_app (idType bndr_id)) -- Ensure that every old binder of type `b` is linked up with its -- new binder which should have type `n b` -- See Note [GroupStmt binder map] in HsExpr n_bndr_ids = zipWith mk_n_bndr n_bndr_names bndr_ids bindersMap' = bndr_ids `zip` n_bndr_ids -- Type check the thing in the environment with -- these new binders and return the result ; thing <- tcExtendIdEnv n_bndr_ids (thing_inside new_res_ty) ; return (TransStmt { trS_stmts = stmts', trS_bndrs = bindersMap' , trS_by = by', trS_using = final_us which is perpendicular to PQ, then the length of ( - dy, dx ) is d. So the unit vector is ( - dy / d, dx / d ). So multiplying by h gives the offset vector from the midpoint. So the two centers are: center1 = midpoint + ( - dy / d * h, dx / d * h ) center2 = midpoint + ( dy / d * h, - dx / d * h ) Yes. Because the unit vector is ( - dy, dx ) divided by d, then multiplied by h gives the displacement. So in code: Compute dx = x2 - x1 dy = y2 - y1 d_squared = dx*dx + dy*dy if d_squared > (2*r)**2 skip if d_squared == 0 same point, skip (since centers would be the same as the individual points) else: d = Math.sqrt(d_squared) h = Math.sqrt(r**2 - (d_squared / 4.0)) if h is NaN skip (but since we checked d_squared <= (2r)^2, r^2 - (d_squared/4) is >= 0 h is real) Then: mx = (x1 + x2) / 2.0 my = (y1 + y2) / 2.0 Then, the direction is (-dy, dx) normalized by d. So: unit_x = -dy.to_f / d unit_y = dx.to_f / d offset_x = unit_x * h offset_y = unit_y * h center1_x = mx + offset_x center1_y = my + offset_y center2_x = mx - offset_x center2_y = my - offset_y Wait, no. Wait, the unit vector is 6 rabella Clingstone]Aug 19, 1907 - Sep 28, 1907Just Out of College[Play, Comedy, Original] Performer: Mrs. E.A. Eberle Sep 27, 1905 - Nov 1905The Little Princess[Play, Drama, Original] -1,-1,0)\), and \((1,-1,0)\). - The center of the sphere is along the \(z\)-axis at \((0,0,k)\) with radius \(r = \sqrt{2 + k^2}\). 2. **Tetrahedron Coordinates**: - Points \(A\), \(B\), \(C\), and \(D\) are defined symmetrically. Points \(A\) and \(C\) are above the \(xy\)-plane, and points \(B\) and \(D\) are below. - Coordinates are: \(A(2, 0, c)\), \(B(0, 2, -c)\), \(C(-2, 0, c)\), \(D(0, -2, -c)\). 3. **Distance Calculations**: - The distance from the center \((0,0,k)\) to edge \(AC\) is calculated and found to be \(|c - k|\). Given the sphere touches \(AC\), this distance equals the radius \(\sqrt{2 + k^2}\). - Solving \(|c - k| = \sqrt{2 + k^2}\) gives \(c = k + \sqrt{k^2 + 2}\). 4. **Symmetry and Distance to \(BD\)**: - Edge \(BD\) is a vertical line along the \(y\)-axis at \(x=0\) and \(z=-c\). The distance from the center \OpItems :: Class -> [ClassOpItem] classOpItems = classOpStuff classATs :: Class -> [TyCon] classATs (Class { classATStuff = at_stuff }) = [tc | (tc, _) <- at_stuff] classATItems :: Class -> [ClassATItem] classATItems = classATStuff classTvsFds :: Class -> ([TyVar], [FunDep TyVar]) classTvsFds c = (classTyVars c, classFunDeps c) classBigSig :: Class -> ([TyVar], [PredType], [Id], [ClassOpItem]) classBigSig (Class {classTyVars = tyvars, classSCTheta = sc_theta, classSCSels = sc_sels, classOpStuff = op_stuff}) = (tyvars, sc_theta, sc_sels, op_stuff) classExtraBigSig :: Class -> ([TyVar], [FunDep TyVar], [PredType], [Id], [ClassATItem], [ClassOpItem]) classExtraBigSig (Class {classTyVars = tyvars, classFunDeps = fundeps, classSCTheta = s's inequalities relate elementary symmetric sums. But here we have a product of symmetric sums. Alternatively, since all \( c_k^2 \) are non-negative, maybe we can use the inequality between the arithmetic mean and the geometric mean (AM-GM). But how? Alternatively, note that if all \( c_k^2 \) are equal, then the elementary symmetric sums would be maximized in some sense. Wait, but we need to compare \( \sum e_j e_{n - j} \) with \( \binom{2n}{n} e_n \). If all \( c_k^2 \) are equal, say \( c_k^2 = c^2 \) for all k, then each elementary symmetric sum \( e_j = \binom{n}{j} c^{2j} \). Then, the product \( e_j e_{n - j} = \binom{n}{j}^2 c^{2n} \). Therefore, the sum \( \sum_{j=0}^n e_j e_{n - j} = c^{2n} \sum_{j=0}^n \binom{n}{j}^2 \). But we know that \( \sum_{j=0}^n \binom{n}{j}^2 = \binom{2n}{n} \). Therefore, in this r. But the servant leaves after 9 months and gets Rs. 55 and the turban. The turban costntnio Salim Curiati Antonio Ueno Arnaldo Martins Arnaldo Moraes Arnaldo Prieto Arnold Fioravante Arolde de Oliveira Artenir Werner Artur da Tvola Asdrubal Bentes Assis Canuto tila Lira Augusto Carvalho ureo Mello Baslio Villani Benedicto Monteiro Benito Gama Beth Azize Bezerra de Melo Bocayuva Cunha Bonifcio de Andrada Bosco Frana Brando Monteiro Caio Pompeu Carlos Alberto Carlos Alberto Ca Carlos Benevides Carlos Cardinal Carlos Chiarelli Carlos Cotta Carlos DeCarli Carlos Mosconi Carlos SantAnna Carlos Vinagre Carlos Virglio Carrel Benevides Cssio Cunha Lima Clio de Castro Celso Dourado Csar Cals Neto Csar Maia Chagas Duarte Chagas Neto Chagas Rodrigues Chico Humberto Christvam Chiaradia Cid Carvalho Cid Sabia de Carvalho Cludio vila Cleonncio Fonseca Costa Ferreira Cristina Tavares Cunha Bueno Dlton Canabrava Darcy Deitos Darcy Pozza Daso Coimbra Davi Alves Silva Del Bosco Amaral Delfim Netto Dlio Braz Denisar Arneiro Dionisio Dal Pr Dionsio Hage Dirce Tutu Quadros Dirceu Carneiro Divaldo Suruagy Djenal Gonalves Domingos Juvenil Domingos Leonelli Doreto Campanari Edsio Frias Edison Lobo Edivaldo Motta Edme Tavares Edmilson Valentim Eduardo Bonfim Eduardo Jorge Eduardo 46Vi anbefaler, at du har dit eget trommest, s du har mulighed for at ve imellem lektionerne. Hvis du bor inde i byen, kan det vre en rigtig god id at investere i et el-trommest eller finde et musiklokale, hvor du kan ve dig. Er du ny p trommerne, kan det dog vre en god id at vente til efter et par lektioner, s du er sikker p, at du vil fortstte p trommerne. nsker du hjlp til at kbe eller leje et trommest, kan du altid kontakte din underviser eller Musikundervisning.dk - vi hjlper dig gerne med at finde frem til den bedste lsning for dig.Filosofar nunca foi sobre deixar voc feliz. que andam mentindo muito por a. Filosofar est mais ligado ao despertar do sonambulismo. Essa minha proposta nesta conversa com voc. Quem tem medo de sofrer incapaz de desejar. A obsesso pela felicidade faz de voc um chato. Como escapar dessa armadilha? Escolher o fracasso? No precisa, ele te achar. Viver sem frmulas o desafio. Uma certa dose de banalidade na vida indcio de alguma sade mental, s gente doente e chata quer ser absolutamente relevante em tudo que faz.Defining and customizing additional attributes that can be polled from both system and user defined managed objects is possible by creating Usduring each iteration of the Radix sort, each character is taken as the ordinal value. For strings shorter than max_len, the padding is '\0', which has ord 0. So the Radix sort on the strings works correctly. Therefore, the code as provided by the user should work. But the user is asking to help modify the Radix sort algorithm to solve this puzzle. So perhaps the code is correct as is, and the answer is that their code works. Or maybe there's a mistake. Wait, in the modified_radix_sort function, after grouping into count_dict, the code sorts the keys in ascending order (sorted(count_dict.keys())), then for each key in this sorted list, appends the sorted list (radix_sort_str(count_dict[key])). Yes. So the overall code groups by vowel count, sorts each group's strings with Radix (alphabetical), then combines them in order of increasing vowel counts. So the final list is sorted by vowel count and then alphabetically. So the code is correct. But maybe the user wants to use a different approach, but their existing code is correct. Alternatively, perhaps the code has an error. Let me check the parameters. The code's modified_radix_sort function takes a list of strings. The function initializes count_dict as a defaultdict(list). For each string, it counts vowels and appends to the dict. Then sorts the keys, then for each key, sorts the list with radix_sort_str. Then combines. Yes. So the code should work. So the user's/them who can do so. Posted in Kim Richards, Kyle Richards, REAL HOUSEWIVES OF BEVERLY HILLS, RHOBH VIDEO Tagged Avril Lavigne, BRAVO, Camille Grammer dinner party from hell, Camille Grammer the morally corrupt Faye Resnick, Davis Wright Tremaine law firm Portland, Everett Jack, Everett Jack Faye Resnick, Everett Jack Jr., Faye, Faye Resnick, Faye Resnick model, Jack Jr, Kim, Kim Richards, Kyle Richards, Kyle Richards says Faye Resnick like a sister, Los Angeles, Nick Lachey, Nicky Hilton, Northern California, REAL HOUSEWIVES OF BEVERLY HILLS, San Francisco, United States, Who is Everett Jack Jr?, Who is Faye Resnick fiance, Who is Faye Resnick marrying? Everett Jack 65 AprDetermines whether or not the null value should be included in the list of ch polygonal chain that starts and ends at the same point, right? Each of the nine segments must intersect exactly one other segment. So, every segment crosses one and only one other segment in the figure. First, let me recall some basics about graph theory and planar graphs because intersections between segments can be thought of as edges crossing each other. In planar graphs, edges don't cross, but here we want edges to cross, but each edge can only cross one other edge. Wait, but planar graphs have a limit on the number of edges without crossings. Maybe that's related, but since we're allowing crossings here, maybe it's a different problem. But the problem isn't about embedding a graph in the plane with a certain number of crossings, but rather constructing a specific polygonal chain (closed, so a polygon) where each segment has exactly one intersection with another segment. Each intersection must be between exactly two segments, and each segment is involved in exactly one intersection. So, in total, there should be nine segments, each crossing one other segment, so there would be nine intersections? Wait, no. If each intersection is between two segments, then the total number of intersections would be nine divided by two, but nine is odd. That can't be. So maybe that's a problem. Wait, hold on. If each segment must intersect exactly one other segment, then each intersection is shared between two segments. Therefore, the total number of intersections would be nine divided by two, which is 4.5. But you can't have half an intersection. That's impossible. So this multi-parameter type classes - an inferred type that includes unboxed tuples However we don't do the ambiguity check (checkValidType omits it for InfSigCtxt) because the impedence-matching stage, which follows immediately, will do it and we don't want two error messages. Moreover, because of the impedence matching stage, the ambiguity-check suggestion of -XAllowAmbiguiousTypes will not work. Note [Impedence matching] ~~~~~~~~~~~~~~~~~~~~~~~~~ Consider f 0 x = x f n x = g [] (not x) g [] y = f 10 y g _ y = f 9 y After typechecking we'll get f_mono_ty :: a -> Bool -> Bool g_mono_ty :: [b] -> Bool -> Bool with constraints (Eq a, Num a) Note that f is polymorphic in 'a' and g in 'b'; and these are not linked. The types we really want for f and g are f :: forall a. (Eq a, Num a) => a -> Bool -> Bool g :: forall b. [b] -> Bool -> Bool We can get these by "impedence matching": tuple :: forall a b. (Eq a, Num a) => (a -> Bool -> Bool, [b] -> Bool -> Bool) tuple a b d1 d1 = let ...bind f_mono, g_mono in (f_mono, g_mono) f a d1 d2 = case tuple a Any d1 d2 of (f, g) -> f g b = case tuple Integer b dEqInteger dNumInteger of (f,g) -> g Suppose the shared quantified tyvars are qtvs and constraints theta. Then we want to check that f's polytype is more polymorphic than forall qtvs. theta => f_mono_ty and the proof is the impedence matcher. Notice that the impedence matcher may do defaulting. See Trac #7173. It also cleverly does an ambiguity check; for example, rejecting , E = [2/(1 + k) - 2/(1 + h)] / [k / (1 + k) - h / (1 + h)] Factor numerator and denominator: Numerator: 2[1/(1 + k) - 1/(1 + h)] Denominator: [k / (1 + k) - h / (1 + h)] = [k(1 + h) - h(1 + k)] / [(1 + k)(1 + h)] So, E = [2(1/(1 + k) - 1/(1 + h))] / [ (k(1 + h) - h(1 + k)) / ((1 + k)(1 + h)) ) ] Simplify: E = 2(1/(1 + k) - 1/(1 + h)) * ((1 + k)(1 + h)) / (k(1 + h) - h(1 + k)) Compute numerator: 2[ ((1 + h) - (1 + k)) / ((1 + h)(1 + k)) ] * (1 + k)(1 + h) = 2[ ((1 + h) - (1 + k)) ] Denominator: k(1 + h) - h(1 + k) So, E = 2((1 + h) - (1 + k)) / (k(1 + h) - h(1 + k)) This see4=2011. 3*669=2007, 2007+4=2011. Yes, that's correct. So, the product is indeed 3^669 *2^2, so the highest power of three dividing A is 3^669. But maybe there's a different decomposition of 2011 into natural numbers that gives a higher power of three? Wait, but how? If we use more threes, but if the remainder is 1, we can't have another three. Unless there's another way to split the sum. But according to the standard method for maximum product partition, the optimal decomposition is to use as many threes as possible, then adjust with twos if the remainder is 1 or 2. Since 2011 divided by 3 is 670 with remainder 1, so replacing one three and the one with two twos gives the maximum product. Hence, the number of threes in the optimal decomposition is 669, and the number of twos is 2. Therefore, the exponent of three in A is 669. Therefore, the answer should be 3^669, so the largest power of three[ppr rdr_stmt, ppr rn_stmt, ppr fvs]) ; failIfErrsM ; rnDump (ppr rn_stmt) ; -- The real work is done here (bound_ids, tc_expr) <- mkPlan rn_stmt ; zonked_expr <- zonkTopLExpr tc_expr ; zonked_ids <- zonkTopBndrs bound_ids ; -- None of the Ids should be of unboxed type, because we -- cast them all to HValues in the end! mapM_ bad_unboxed (filter (isUnLiftedType . idType) zonked_ids) ; traceTc "tcs 1" empty ; let { global_ids = map globaliseAndTidyId zonked_ids } ; -- Note [Interactively-bound Ids in GHCi] {- --------------------------------------------- At one stage I removed any shadowed bindings from the type_env; they are inaccessible but might, I suppose, cause a space leak if we leave them there. However, with Template Haskell they aren't necessarily inaccessible. Consider this GHCi session Prelude> let f n = n * 2 :: Int Prelude> fName <- runQ [| f |] Prelude> $(return $ AppE fName (LitE (IntegerL 7))) 14 Prelude> let f n = n * 3 :: Int Prelude> $(return $ AppE fName (LitE (IntegerL 7))) In the last line we use 'fName', which resolves to the *first* cles every 3 steps because ^3 = 1. Therefore, the sequence is periodic with period 3. So, as calculated, P, P, P, P all have th had married into was largely destitute. The familys rich sugar cane plantation was no more and the only thing of value still in family hands was Petit Anse, the little island that was later to be renamed Avery Island. Edmund McIlhenny was a businessman, not a farmer. As a pre-war banker, he learned to market himself personally to such a degree that he became the best known and most sought after financial man in New Orleans. His marketing skills, and his willingness to bend the truth when it made for a better story, have made it difficult to determine exactly when he became aware of the chili pepper from Mexicos Tabasco region and how he decided to make hot sauce the new family busines Thus, substituting back: \(4 \sin = 5 \left( 0.5 \cos + (\sqrt{3}/2) \sin \right)\) Simplify the right-hand side: \(5 \times 0.5 \cos = 2.5 \cos \) \(5 \times (\sqrt{3}/2) \sin = (5\sqrt{3}/2) \sin \) So: \(4 \sin = 2.5 \cos + (5\sqrt{3}/2) \sin \) Bring all terms to the left: \(4 \sin - (5\sqrt{3}/2) \sin - 2.5 \cos = 0\) Factor out \(\sin \) and \(\cos \): \(\left(4 - \frac{5\sqrt{3}}{2}\right) \sin - 2.5 \cos = 0\) Lets write coefficients as fractions to facilitate: \(4 = \frac{8}{2}\), so: \(\left( \frac{8 - 5\sqrt{3}}{2} \right) \sin - \frac{5}{2} \cos = 0\) Multiply both sides by 2 to eliminate denominators: \((8 - 5\sqrt{3}) \sin - 5 \cos = 0\) Rearrange: \((8 - 5\sqrt{3}) \sin = 5 \cos \) Divide both sides by \(\cos \): \((8 - 5\sqrt{3}) \tan = 5\) Solve for \(\tan \): \(\tan = \frac{5}{8 - 5\sqrt{3}}\) Rationalize the denominator: Multiply numerator and denominator by \(8 + 5\sqrt{3}\): \(\tan = \frac{5(8 + 5\sqrt{3})}{(8)^2 - (5\sqrt{3})^2} = \frac{40 + 25\sqrt{3}}{64 - 75} = \frac{40 + 25\sqrt{3}}{-11}\) Thus: \(\tan = -\frac{40 + 25\sqrt{3}}{11}\) But angle \( = \angle ATB\) is an angle in a convex pentagon, so it must be between 0 and 180 degrees. However, the tangent is negative, which implies that is in the second quadrant (between 90 and 180), which is possible. Lets compute : = arctan\left(-\frac{40 + 253}{11}\right) But since is in the second quadrant, we can write: = 180 - arctan\left(\frac{40 + 253}{11}\right) But exact value might not be necessary. Instead, perhaps we can find sin and cos . Let me denote: Lets lets set: tan = - (40 + 253)/11 = opposite/adjacent = (- (40 + 253))/11 But since is in the second quadrant, we can consider a right triangle with reference angle , where tan = (40 + 253)/11. Then: sin = sin(180 - ) = sin cos = -cos Using the right triangle with opposite side (40 + 253), adjacent side 11, hypotenuse: sqrt( (40 + 253)^2 + 11^2 ) But this computation will be messy. However, perhaps we can use this to find sin and cos expressions. Let me compute: Lets compute (40 + 253)^2 + 11^2: First, expand (40 + 253)^2: = 40^2 + 2*40*253 + (253)^2 = 1600 + 20003 + 625*3 = 1600 + 20003 + 1875 = 3475 + 20003 Add 11^2 = 121: Total hypotenuse squared = 3475 + 20003 + 121 = 3596 + 20003 Thus, hypotenuse = sqrt(3596 + 20003) Therefore: sin = sin = (40 + 253) / sqrt(3596 + 20003) cos = -cos = -11 / sqrt(3596 + 20003) This seems very complex. Perhaps there's a better way to handle this equation. Alternatively, we can use the sine addition formula. Given that + = 150, and 4 sin = 5 sin . Since = 150 - , substitute: 4 sin = 5 sin(150 - ) Which we already did. Let's write this as: 4 sin = 5 [ sin 150 cos - cos 150 sin ] As before: 4 sin = 5 [ 0.5 cos + (3/2) sin ] Bring all terms to one side: 4 sin - (53/2) sinso que di&ia que finalmente ele tinha entendido a situa!o destes homens. :odos os quatro eram solitrios e estavam apenas ; procura de algum que os entendesse, queriam pertencer. L%uer saber de uma coisa. L, disse Kory. L'uvi di&er que o customerStream = env .addSource(new FlinkKafkaConsumer<>("customers", new CustomerInfoDeserializer(), customerProps)) .name("customer-source"); // 2. MapStateDescriptor productStateDesc = new MapStateDescriptor<>("product-state", String.class, ProductInfo.class); BroadcastStream productBroadcast = productStream.broadcast(productStateDesc); MapStateDescriptor customerStateDesc = new MapStateDescriptor<>("customer-state", String.class, CustomerInfo.class); BroadcastStream customerBroadcast = customerStream.broadcast(customerStateDesc); // 3. DataStream withProductStream = clickStream .connect(productBroadcast) .process(new ProductEnricher()); // 4. DataStream enrichedStream = withProductStream .connect(customerBroadcast) .process(new CustomerEnricher()); // 5. JDBCHBaseKafka enrichedStream.addSink(new HBaseSink()); env.execute("Real-time ETL Job"); } publ, D with different values for a and b. Hmm, complex numbers, right? Since a and b are real, their combination with the complex number 1 + 2i should result in 2i. So, let me break this down. If I expand the left side, it should be (1*a + b) + (2i*a). Because when you multiply (1 + 2i) by a, you get a + 2ai. Then adding b, which is real, so it combines with the real part. So the entire left side is (a + b) + 2ai. The right side is 2i, which can be written as 0 + 2i. Since two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. So, that gives me two equations: 1. The real part: a + b = 0 2. The imaginary part: 2a = 2 Alright, solving these two equations. Let's start with the second equation because it looks simpler. 2a = 2, so dividing both sides by 2 gives a = 1. Then plugging a = 1 into the first equation, which is 1 + b = 0. Subtracting 1 from both sides gives b = -1. So, a = 1 and b = -1. Checking the options, that's option A. Wait, but let me verife than 6 edges without forming a triangle? Suppose we try to add one more edge. Let's say we put back one of the removed edges. Now, that edge is part of one or two small triangles. Each small triangle adjacent to that edge would now have two edges present. If those other edges are in different directions, then adding this edge would complete a triangle. Wait, no. If the other two edges of a small triangle are present, then adding the third would form a triangle. But if we only added one edge from the forbidden direction, maybe the other edges of the triangles it's part of are not all present. Wait, but if we removed all edges in one direction, then in the remaining graph, no triangles existause would be on s.sale_date, but the user's question is about the books from the last 10 years, which implies publication date. Testing edge cases: if a year has no sales, that year won't appear. But perhaps the user wants all years in the last decade, even if no sales. But without sales data, the book wouldn't be in the join. So maybe need a LEFT JOIN, but if a book has no sales, total_sold would be zero. Then in the ranking, the highest sold would be zero. But if there are books with zero sales, maybe they should be excluded. The user is looking for the most popular, which implies some sales. Alternatively, if using ratings, similar logic. Let's adjust for that scenario. Another approach: if the popularity is based on the number of reviews, then count(review_id) as review_count. The query structure would be similar. In any case, the key parts are: - Filtering the books by publication year. - Calculating the popularity metric per book. - Ranking within each year. - Selecting the top entries. I should also mention that the exact table and column names might differ, so the user needs to adjust them according to their schema. Also, handle the current year function based on the SQL dialect (MySQL, PostgreSQL, etc.). Finally, the ORDER BY clause: the user might want the years in ascending or descending order. The example uses ORDER BY publication_year DESC to show the latest years first, but that can be adjusted. To write a SQL query that finds the most popular books from the last 10 years (ordered by year), follow these steps. This example assumes a database schema with tables for `books` and a popularity metric (e.g., `sales` or `reviews`): ### Tables (Example Schema) 1. **`books`**: Contains book details. - `book_id` (unique identifier) - `title` - `publication_year` 2. **`sales`**: Tracks sales data (replace with `reviews` for ratings-based popularity). - `sale_id` - `book_id` (foreign key to `books`) - `quantity_sold` ### Solution Code ```sql WITH BookPopularity AS ( SELECT b.publication_year, b.title, SUM(s.quantity_sold) AS total_sold, -- Replace with reviews/ratings if needed RANK() OVER ( PARTITION BY b.publication_year ORDER BY SUM(s.quantity_sold) DESC -- Adjust metric for reviews/ratings ) AS popularity_rank FROM books b JOIN EFAULT, [], rhs)]) | Just (tc, tys) <- splitTyConApp_maybe (idType bndr) , isUnboxedTupleTyCon tc , Just res <- case tys of [ty] | UnaryRep _ <- repType ty , let bind = bndr `setIdType` ty -> Just $ doCase d s p scrut bind [(DEFAULT, [], rhs)] (Just bndr){-unboxed tuple-} [ty1, ty2] | UnaryRep rep_ty1 <- repType ty1 , UnaryRep rep_ty2 <- repType ty2 -> case () of _ | VoidRep <- typePrimRep rep_ty1 , let bind2 = bndr `setIdType` ty2 -> Just $ doCase d s p scrut bind2 [(DEFAULT, [], rhs)] (Just bndr){-unboxed tuple-} | VoidRep <- typePrimRep rep_ty2 , let bind1 = bndr `setIdType` ty1 -> Just $ doCase d s p scrut bind1 [(DEFAULT, [], rhs)] (Just bndr){-unboxed tuple-} | otherwise -> Nothing _ -> Nothing = res schemeE d s p (AnnCase scrut bndr _ alts) = doCase d s p scrut bndr alts Nothing{-not an unboxed tuple-} schemeE _ _ _ expr = pprPanic "ByteCodeGen.schemeE: unhandled case" (pprCoreExpr (deAnnotate' expr)) {- Ticked Expressions ------------------ The idea is that the "breakpoint E" is really just an annotation on the code. Whe888*^9, 3.5617587174988885`*^9}, {3.5617603061411123`*^9, 3.5617603105011187`*^9}, 3.5617605991515226`*^9, {3.561760630041566*^9, 3.561760630041566*^9}, 3.56203042461959*^9, 3.5620880721126423`*^9, {3.5620933445200233`*^9, 3.5620933547300377`*^9}, {3.562093987920924*^9, 3.56209399237093*^9}, { 3.5620995170537643`*^9, 3.5620995170537643`*^9}, {3.5621013517663326`*^9, 3.562101351996333*^9}, {3.5633361467746754`*^9, 3.563336146794675*^9}, { 3.563388115059457*^9, 3.563388116198259*^9}, {3.5633881657127457`*^9, 3.5633881705487547`*^9}, 3.5633885482410173`*^9, {3.5633893354753733`*^9, 3.563389335775374*^9}, 3.563389378455434*^9, 3.5634011676489882`*^9, { 3.5686632140689983`*^9, 3.5686632169090023`*^9}, {3.56904036317673*^9, 3.5690403643967tches I still unfortunately see myself using paper. I prefer WhiteLines Link lined paper in A4 Grid size. I can take a photo with an iPhone or iPad and send the file directly to email, Evernote, DropBox. With Evernote Premium it even indexes what I write (and my writing is far from perfect). I also use a Fujitsu ScanSnap scanner for other digitizing of paper work-flow.Oregon Department of Fish and Wildlife biologists has released a bear captured Wednesday after roaming in a field next to Tualatin Elementary School. The bear was released into an aspen grove in the Oregon Coast Range Mountains west of Carlton at approximately 11 a.m. yesterday. The bear appeared a little groggy but overall was in good condition when it stepped out of the trap where it had been held captive since the previous day wh 10^6 10*10 100 mod 101. Therefore: 10^6 100 mod 101, A*10^4 A*1 mod 101, A*10 A*10 mod 101, 1 1 mod 101. Therefore, the total expression is: 100 + A + 10A + 1 0 mod 101 101 + 11A 0 mod 101 11A -101 0 mod 101. Since 101 is prime, 11 11 mod 101, so 11A 0 mod 101 A 0 mod 101. But A is a two-digit number, so A=00. But 00 is not a valid two-digit number. Contradiction. Therefore, trect. Now, compute (a + 2b)a: = aa + 2ba = ||a|| + 2ab = 0. Therefore, the dot product is zero, meaning the vectors are perpendicular, so angle is 90 degrees. Seems right. But wait, maybe there's a trick here. Let me think. Could there be a case where even if the dot product is zero, the vectors are zero vectors? But no, because if a is the zero vector, then ||a + b|| = ||b|| would hold, but then (a + 2b) would be 2b, and the angle between 2b and a (zero vector) is undefined. However, the problem says "vectors a and b", which typically implies they are non-zero unless stated otherwise. So probably safe to assume a and b are non-zero. Alternatively, maybe the problem allows a to be zero, but then the angle is undefined. However, given that the problem asks for the angle in degrees, likely expecting 90 degrees as the answer. So perhaps the answer is 90 degrees. Wait, let me double-check the calculation one more time. Given (a + 2b) a = ||a|| + 2ab. From the given condition ||a + b|| = ||b||, which gives ||a|| + 2ab + ||b|| = ||b||, leading to ||a|| + 2ab = 0. Therefore, substituting that into the expression for the dot product, indeed, (a + 2b)a = 0. Therefore, the cosine of the angle is zero, so the angle is 90 degrees. Therefore, the answer should be 90 degrees. But maybe I made a mistake in assuming that (a + 2b) a is equal to that expression. Let's check: (a + 2b)a = aa + 2ba, which is ||a|| + 2ab, yes. Since dot product is commutative, so ab = ba. Hence, the calculation is correct. Therefore, the angle is 90 degrees. Wait, but let me see if there's another way to approach this problem, Frederick, Hampshire, or Augusta, in a little time.le's angles and side lengths, making it highly unlikely for these angles to be exactly 60 degrees unless the original triangle is specifically constructed, which seems non-trivial or impossible. FurtIII/13. Genderpedaggia s nyelvhasznlatKegyes Erika: Genderpedaggia s nyelvhasznlat. 2015. szeptember SZNHZI LBJEGYZET 39. TisztjtsA Komromi Jkai Sznhz internetes msorban a Berats a pozsonyi Magyar Tannyelv AlapiskolbaA pozsonyi Magyar Tannyelv Alapiskola s Ipolyi Arnold verseny 2016 Tth Mt, KisjfaluA feledkeny legny Tth Mt (Kisjfalu) mesemondsa XXIV. TOMPA MIHLY VERSENY Ills Csenge, KassaAranysvos minstssel els helyezst rt el Az Andrssyak nyomban Krasznahorktl BetlrigAz Andrssyak nyomban Krasznahorktl BetlrigKzssgi hl Reklm Hrcsatorna RSS s e-mail hrcsatornink Bejegyzsek Kommentrok HrkeresNapi evanglium 2017. jnius 23. Pntek, Jzus szentsges szveXXVI. Tompa Mihly Orszgos Verseny illo Dominik, SzencTrump napkollektoros falat pttetne a mexiki hatrraKrjk, segtse munknkat tmogatsval! PayPal.Me15 Inicia sesin para responder KO: 3339 EL JUDIO / 4 feb 2017 a las 2:31 pm 6 Inicia sesin para responder KO: 27 El lobo CR / 4 feb 2017 a las 1:28 pm Buenas tardes a todos, tengo 10 aos de seguir el UFC y el mundo dl MMA en general, y tambin como 4 aos de seguir esta pagna sin nunca comentar hasta ahora y ha sido por la clase comentarios ofensivos que lanzan en ocasiones, juego para el que no me voy a prestar aclaro. Dicho esto los Nick y Nate son grandes atletas por solo el hecho de llevar tantos aos en un deporte tan duro, sin embargo viendo sus records no son sorprendentes y piden mas de lo que merecen actualmente, en caso de Nick sus ltimas prensentaciones y inactividad no le dan para pedir tanto, lo unico es gran base de fans por su temerara actitud.This summary section refers to new treatments being studied in clinical trials, but it may not mention every new treatment being studied. Information about clinical trials is available from the NCI Web site.Proton beam radiation therapy Proton beam radiation therapy is a type of high-energy, external radiation therapy that uses streams of protons (small, positively-charged pieces of matter) to make radiation. Thibsent. But there is no point in w/w-ing because we'll simply add (\y:Void#), see WwLib.mkWorerArgs. If x has a more interesting type (eg Int, or Int#), there *is* a point in w/w so that we don't pass the argument at all. Note [Thunk splitting] ~~~~~~~~~~~~~~~~~~~~~~ Suppose x is used strictly (never mind whether it has the CPR property). let x* = x-rhs in body splitThunk transforms like this: let x* = case x-rhs of { I# a -> I# a } in body Now simplifier will transform to case x-rhs of I# a -> let x* = I# a in body which is what we want. Now suppose x-rhs is itself a case: x-rhs = case e of { T -> I# a; F -> I# b } The join point will abstract over a, rather than over (which is what would have happened before) which is fine. Notice that x certainly has the CPR property now! In fact, splitThunk uses the function argument w/w split_vars) -- Given D; xs |-a c : stk --> t, builds c with xs fed back. -- Typically needs to be prefixed with arr (\(p, stk) -> ((xs),stk)) dsfixCmd :: DsCmdEnv -- arrow combinators -> IdSet -- set of local vars available to this command -> Type -- type of the stack (right-nested tuple) -> Type -- return type of the command -> LHsCmd Id -- command to desugar -> DsM (CoreExpr, -- desugared expression IdSet, -- subset of local vars that occur free [Id]) -- the same local vars as a list, fed back dsfixCmd ids local_vars stk_ty cmd_ty cmd = trimInput (dsLCmd ids local_vars stk_ty cmd_ty cmd) -- Feed back the list of local variables actually used a command, -- for use as the input tuple of the generated arrow. trimInput :: ([Id] -> DsM (CoreExpr, IdSet)) -> DsM (CoreExpr, -- desugared expression IdSet, -- subset of local vars that occur free [Id]) -- same local vars as a list, fed back to d not (T a b' c'), becuase 'c' *is* mentioned in a non-updated field NB that it's not good enough to look at just one constructor; we must look at them all; cf Trac #3219 After all, upd should be equivalent to: upd t x = case t of MkT1 p q -> MkT1 p x MkT2 a b -> MkT2 p b MkT3 d -> error ... So we need to give a completely fresh type to the result record, and then constrain it by the fields that are *not* updated ("p" above). We call these the "fixed" type variables, and compute them in getFixedTyVars. Note that because MkT3 doesn't contain all the fields being updated, its RHS is simply an error, so it doesn't impose any type constraints. Hence the use of 'relevant_cont'. Note [Implict type sharing] ~~~~~~~~~~~~~~~~~~~~~~~~~~~ We also take into account any "implicit" non-update fields. For example data T a b where { MkT { f::a } :: T a a; ... } So the "real" type of MkT is: forall ab. (a~b) => a -> T a b Then consider upd t x = t { f=x } We infer the type upd :: T a b -> a -> T a b upd (t::T a b) (x::a) = case t of { MkT (co:a~b) (_:a) -> MkT co x } We can't give it the more general type upd :: T a b -> c -> T c b Note [Criteria for update] ~~~~~~~~~~~~~~~~~~~~~~~~~~ We want to allow update for existentials etc, provided the updated field isn't part of the existential. For exameets the output requirement, leading to an error. But that seems unlikely. Alternatively, perhaps the "sort_by_value" and "sort_by_key" operations are applied to each dictionary individually. For example, "sort_by_value" would sort each dict's items by value and then output them sorted by key. But the problem says the operations are applied to the first two dictionaries. So how would that work. Alternatively, perhaps the "sort_by_value" operation is to return a list of all values from both dictionaries, sorted. But since the problem requires output to be sorted by keys, this is impossible. Thus, perhaps this operation cannot be performed and returns an error. But that's unlikely. This is getting too stuck. Let's try to look at each operation and what they might do, given that the output must be sorted by key. Merge: Combine the two dicts, sum values for common keys. Output (k, sum) sorted by key. Always possible. Common_keys: Find keys present in both. For each, (k, v1, v2). Sorted by key. If no common keys, error. Unique_keys: Find keys present in exactly one. For each, (k, v1 or None, v2 or None). Sorted by key. If no unique keys (symmetric difference is empty), error. Sort_by_value: Perhaps, the merged dict is sorted by value, then byough my shelves and seeing what flings its little way into my hands. I honestly have no idea which title it will be so Im rather excited at this prospect. Ill let you know as I know its going to be hard for you to sleep and concentrate in your own lives until this decision has been sorted out. This Destroyer Captain: Lessons of a First Command book is not really ordinary book, you have it then the world is in your hands. The benefit you get by reading this book is actually information inside this reserve incredible fresh, you will get information which is gettwn is not possible. So setting it to -infinity (or a very low value) is correct because in the first iteration (day 1), cash_prev is 0, and cooldown_prev is -1e9. So cash_new for day 1 is max(0, -1e9) 0. Yes, that's correct. So the Scheme code should be as written above. But wait, in the initial loop call, the cooldown_prev is set to -1e9. Let me check if that's correct. For the firx4" could be set to 0, or to any nonzero number 1, say. By always choosing the "0" option we might end up with the all-zero trivial vector -- which is obviously not what we want. Persistent choice of the "1" option, might lead to a conflict between some of the equations, such as the equations one and four in our example.

So the strategy is as follows.

If there is at least one zero on the diagonal, find the topmost row with zero on the diagonal and choose for it the solution "1". Diagonal zeros in other rows would force the solution "0". If the diagonal element is not zero than simply solve an arithmetic equation that arises from the substitutions of previously computed components of the eigenvector. Since certain inaccuracies acumulate during QR factors trying to ask: To get a certain number of servings, how many pints in total are needed, and they tell you that number is 14, so how many pints is that? But again, redundant. Given all this confusion, I think the most plausible answer is 14 pints, but the steps to get there would involve calculating the number of servings, which is 8. However, the question's phrasing is problematic. Since the user might have intended to ask for the number of servings, but the wording is off, but given that the answer needs to be in a box and likely numerical, and given that 14 is the total pints, but according to the calculation, 8 servings make 14 pints. Wait, no: 1 serving is 1.75 pints, so 8 servings would be 14 pints. Therefore, if the question is asking: "How many servings require 14 pints in total?" then the answer is 8. But the question says "how many pints... would be needed... if the total... is 14 pints". So, the ansbsection{Instances} %* * %************************************************************************ \begin{code} instance Eq PatSyn where (==) = (==) `on` getUnique (/=) = (/=) `on` getUnique instance Ord PatSyn where (<=) = (<=) `on` getUnique (<) = (<) `on` getUnique (>=) = (>=) `on` getUnique (>) = (>) `on` getUnique compare = compare `on` getUnique instance Uniquable PatSyn where getUnique = psUnique instance NamedThing PatSyn where getName = patSynName instance Outputable PatSyn where ppr = ppr . getName instance OutputableBndr PatSyn where pprInfixOcc = pprInfixName . getName pprPrefixOcc = pprPrefixName . getName instance Data.Data PatSyn where -- don't traverse? toConstr _ = abstractConstr "PatSyn" gunfold _ _ = error "gunfold" dataTypeOf _ = mkNoRepType "PatSyn" \end{code} %************************************************************************ %* * \subsection{Construction} %* * %************************************************************************ \begin{code} -- | Build a new pattern synonym mkPatSyn :: Name -> Bool -- ^ Is the pattern synonym declared infix? -> [Type] -- ^ Original arguments -> [TyVar] -- ^ Unie in contact with the O positive blood type. The O negative person starts to fight said blood typetreat it as a problem, a foreign body. So, any future pregnancies can result in big problems.Machen wir mal eine berschlagrechnung: Pro Minute fllt auf der ganzen Erde 1 Milliarde Tonnen Regen. Doch diese mssen zuerst einmal verdunsten und in den Himmel gehoben werden. Dazu wird Energie bentigt, die nicht der Mensch sondern einzig und allein die Sonne zur Verfgung stellt. Um 1 Liter Wasser zu verdunsten We want to make some @Ids@ to use as match variables. If a pattern has an @Id@ readily at hand, which should indeed be bound to the pattern as a whole, then use it; otherwise, make one up. \begin{code} selectSimpleMatchVarL :: LPat Id -> DsM Id selectSimpleMatchVarL pat = selectMatchVar (unLoc pat) -- (selectMatchVars ps tys) chooses variables of type tys -- to use for matching ps against. If the pattern is a variable, -- we try to use that, to save inventing lots of fresh variables. -- -- OLD, but interesting note: -- But even if it is a variable, its type might not match. Consider -- data T a where -- T1 :: Int -> T Int -- T2 :: a -> T a -- -- f :: T a -> a -> Int -- f (T1 i) (x::Int) = x -- f (T2 i) (y::a) = 0 -- Then we must not choose (x::Int) as the matching variable! -- And nowadays we won't, because the (x::Int) will be wrapped in a CoPat selectMatchVars :: [Pat Id] -> DsM [Id] selectMatchVars ps = mapM selectMatchVar ps selectMatchVar :: Pat Id -> DsM Id selectMatchVar (BangPat pat) = selectMatchVar (unLoc pat) selectMatchVar (LazyPat pat) = selectMatchVar (unLoc pat) selectMatchVar (ParPat pat) = selectMatchVar (unLoc pat) selectMatchVar (VarPat var) = return (localiseId var) -- Note [Localise pattern binders] selectMatchVar (AsPat var _) = return (unLoc var) selectMatchVar other_pat = newSysLocalDs (hsPatType other_pat) -- OK, better make up one... \end{code} Note [Localise pattern binders] ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Consider module M where [Just a] = e After renaming it looks like module M where [Just M.a] = e We don't generalise, since it's a pattern binding, monomorphic, etc, so after desugaring we may get something like M.a = case e of (v:_) -> case v of Just M.a -> M.a Notice the "M.a" in the pattern; after all, it was in the original pattern. However, after optimisation those pattern binders can become let-binders, and then end up floated to top level. They have a different *unique* by then (the simplifier is good about maintaining proper scoping), but it's BAD to have two top-level bindings with the External Name M.a, because that turns into two linker symbols for M.a. It's quite rare for this to actually *happen* -- the only case I know of is tc003 compiled with the 'hpc' way -- but that only makes it all the more annoying. To avoid this, we craftily call 'localiseId' in the desugarer, which simply turns the External Name for the Id into an Internal one, but doesn't change the unique. So the desugarer produces this: M.a{r8} = case e of (v:_) -> case v of Justs, la presin en la pelvis o la espalda baja o incluso la diarrea pueden ser signos de labor de parto prematura. Y mientras que el sangrado prematuro en el primer trimestre puede ser un fenmeno normal de la implantacin de gemelos en la pared uterina, debe llamar a su mdico obstetra si experimenta sangrado en cualquier momento.It has only been about six months and she has changed immensely since she started. In the beginning it was a huge challenge to get her to stay with the activities for the full half hour, even though the activities included stuffed animals and animal sounds! Now she will sit for full minutes at a time doing the child-focused activities that are in her first lesson book and even hitting a few keys at the right time. She LOVES it. When her friends come over she wants to bring out her book and bang around on the piano; she wants to pull out the CD and sing to the songs and do her music activities.********************************************The McLaughlin GroupSeptember 2, 2007********************************************John McLaughlin Intro: Behold The Great Orange Sataaaaaaaannn!!!!Anyone can log on the Intertubes can create a "web blog" and it's crazeeee!!!!!!Why they're better than NBC but what about the unwashed masses!!!??McLaughlin: blogs are too bloggy yes or no!?!?!?!?Buchanan: Drudge Rulez my World!!McLaughlin: but that's not journalism that's news aggregatingScott Grant: dood there's some good blogs but it's all about freedom and first amendment!Randall: dood why not there are small newspapers and big ones and some big ones and the big ones suckAna: who cares it's just a termMcLaughlin: but what about journalism?Ana: dood have you read some of the crap in the famous papers??McLaughlin: In order to be a journalist You Have to Use the Telephone Before Reporting a LieBuchanan: dood when was the last time someone from the McLaughlin made a phone call jeezAna: not to mention there are better journalists on the netMcL: The Internet took down George Allen and CNN!!! Look at the dood who asked Obama is wanted to meet Kim Jong-Il cause he could totally hook him upAna: interesting phenomenon but really it's still about the campaignsMcL: it was an excellent questionRandall: yes it wasBuchanan: hey they selected good questionsAna: and theamy ponad 51 komic3 :: Parser Prop > atomic3 = do x <- preds > y <- term > z <- term > u <- term > return (Atomic (Predicate3 x) [y,z,u]) > preds :: Parser Char > preds = do c <- sat (`elem` ['A'..'Z']) > return (c) > variables :: [Char] > variables = ['u','w','y','x','z'] > names :: [Char] > names = ['a'..'t'] > terms :: [Char] > terms = variables ++ names > term :: Parser Char > term = do c <- sat (`elem` terms) > return (c) > neg :: Parser Prop > neg = do c <- sat (=='~') > x <- prop > return (Negation x) > conj :: Parser Prop > conj = do x <- parens > y <- prop > z <- conj' > u <- prop > t <- parens > return (Conjunction y u) > disj :: Parser Prop > disj = do x <- parens > y <- prop > z <- disj' > u <- prop > t <- parens > return (Disjunction y u) > cond :: Parser Prop > cond = do x <- parens > y <- prop > z <- cond' > u <- prop > t <- parens > return (Conditional y u) > bicon :: Parser Prop > bicon = do x <- parens > y <- prop > z <- bicon' > u <- prop > t <- parens > return (Biconditional y u) > parens :: Parser () > parens = do c <- sat (=='(') > return () > <|> do c <- sat (==')') > return () > conj' :: Parser () > conj' = do c <- string' "&" > return () > disj' :: Parser () > disj' = do c <- string' "v" > return () > cond' :: Parser () > cond' = do c <- string' "->" > return () > bicon' :: Parser () > bicon' = do c <- string' "<->" > return () > string' :: String -> Parser String > string' [] = return [] > string' (x:xs) = do char x > string' xs > return (x:xs) > exi :: Parser Prop > exi = do x <- sat (== '#') > c <- sat (`elem` variables) > p <- prop > return (Existential c p) > uni :: Parser Prop > uni = do x <- sat (== '@') > c <- sat (`elem` variables) > p <- prop > return (Universal c p) > iden :: Parser Prop > iden = do x <- sat (`elem` terms) > y <- sat (=='=') > z <- sat (`elem` terms) > return (Atomic (Predicate2 'I') [x,z]) > niden :: Parser Prop > niden = do x <- sat (`elem` terms) > y <- string' "/=" > z <- sat (`elem` terms) > return (Negation (Atomic (Predicate2 'I') [x,z])) > prop :: Parser Prop > prop = do x <- atomic3 > return (x) > <|> do x <- atomi= (-2)y z. Rearranged: -2 + y + z + 2 y z =0 2 y z + y + z =2. Hmm, perhaps add 1 to both sides: 2 y z + y + z +1=3. Not sure. Alternatively, factor. Let me try to solve for y: 2 y z + y + z =2 y(2 z +1) + z =2 y(2 z +1) =2 - z y=(2 - z)/(2 z +1) Need y to be integer. Let's check possible z. Since denominator 2z +1 must divide (2 - z). Let's try integer z. z=0: denominator 1, numerator 2 -0=2. y=2. So (-2,2,0). Check equation: -2 +2 +0=0; (-2)*2*0=0. Valid, covered. z=1: denominator 3, numerator 2 -1=1. y=1/3. Not integer. z=2: denominator 5, numerator 2 -2=0. y=0. So (-2,0,2). Valid, covered. z=-1: denominator -1, numerator 2 -(-1)=3. y=3/-1=-3. So (-2, -3, -1). Check equation: -2 + (-3) + (-1)=-6; product: (-2)*(-3)*(-1)=-6. So -6=-6. Valid. So (-2, -3, -1) is a solution. Similarly, z=-2: denominator -3, numerator 2 -(-2)=4. y=4/-3-1.333. Not integer. z=3: denominator 7, numerator 2 -3=-1. y=-1/7. Not integer. z=-3: denominator -5, numerator 2 -(-3)=5. y=5/-5=-1. So (-2, -1, -3). Check: -2 + (-1) + (-3)=-6; product: (-2)*(-1)*(-3)=-6. Valid. So another permutation. So solutions for x=-2 are permutations of (-1,-2,-3) and solutions with zeros. Similarly, if we check x=-3: Equas. Now, let's see if we can simplify that denominator. Let's compute (x + 2xy + 2y)(x - 2xy + 2y). Maybe this is a difference of squares or something similar. Let me set A = x + 2y, B = 2xy. Then, the first quadratic is A + B, the second is A - B. So multiplying (A + B)(A - B) = A - B. So, (x + 2xy + 2y)(x - 2xy + 2y) = (x + 2y)^2 - (2xy)^2 Compute this: (x + 2y)^2 = x + 4xy + 4y (2xy)^2 = 4xy Thus, subtracting: x + 4xy + 4y - 4xy = x + 0xy + 4y = x + 4y Therefore, the denominator simplifies to x + 4y. So the first part of the original expression is 1 / (x + 4y). Alright, that's a significant simplification. So first part done. Now moving to the second big parenthesis. **Second Part:** \[ \frac{y^{-2}}{4x^{2} - 8y^{2}} - \frac{1}{4x^{2}y^{2} + 8y^{4}} \] Let's handle each term here. Starting with the first term: First term: y^{-2} / (4x - 8y). Let's rewrite y^{-2} as 1/y. So, 1/(y(4x - 8y)). Factor denominator: 4x - 8y = 4(x - 2y). So, the first term becomes 1 / [4y(x - 2y)]. Second term: 1 / (4xy + 8y). Factor denominator: 4xy + 8y = 4y(x + 2y). So, second term is 1 / [4y(x + 2y)]. Therefore, the second part becomes: [1/(4y(x - 2y))] - [1/(4y(x + 2y))] Factor out 1/(4y): 1/(4y) [1/(x - 2y) - 1/(x + 2y)] Combine the fractions inside the brackets: [(x + 2y) - (x - 2y)] / [(x - 2y)(x + 2y)] Simplify numerator: x + 2y - x + 2y = 4y Denominator is (x - 2y)(x + 2y) = x - (2y)^2 = x - 4y Therefore, the combined fraction inside is 4y / (x - 4y) So the second part becomes: 1/(4y) * [4y / (x - 4y)] = [4y / (4y(x - 4y))] = 1 / (x - 4y) But note that x - 4y is the same as -(4y - x). So, 1/(x - 4y) = -1/(4y - x) But maybe we'll just keep it as 1/(x - 4y) Now, putting together the entire original expression: First part: 1 / (x + 4y) Second part: 1 / (x - 4y) So the entire expression is: 1/(x + 4y) + 1/(x - 4y) Combine these two fractions. The common denominator is (x + 4y)(x - 4y) = x - (4y)^2 = x - 16y Numerator will be (x - 4y) + (x + 4y) = 2x Therefore, the entire expression simplifies to: 2x / (x - 16y) But let's check this st last = self.head while last.next: last = last.next last.next = new_node def delete(self, data): current = self.head prev = None while current: if current.data == \n -> case lookup n info' of > Nothing -> [] > Just c -> c > where > > info' :: [(Name, RuleList)] > info' = map (\(n,rules) -> (n,map (\rule -> (rule,0)) (Set.toAscList rules))) info > info :: [(Name, Set Int)] > info = mkClosure (==) (\f -> map (follow f) f) > (map (\nt -> (nt,Set.fromList (lookupProdsOfName g nt))) nts) > follow :: [(Name, Set Int)] -> (Name, Set Int) -> (Name, Set Int) > follow f (nt,rules) = (nt, unionMap (followNT f) rules `Set.union` rules) > followNT :: [(Name, Set Int)] -> Int -> Set Int > followNT f rule = > case findRule g rule 0 of > Just nt | nt >= firstStartTok && nt < 15 rounded to two significant figures make sense. But perhaps the code is part of a design where the TicTacToe structure is passed to each move function, which updates it. But in Clojure, since the data is immutable, you can't do that. So the move function would need to return the new game state along with the result. However, the given code does not do that. Therefore, this code is incorrect. Alternatively, maybe the code is designed to have the move function not update the game state, but rather check if the current move would result in a win, given the previous state. But since the code doesn't return the new state, the next move would not have the previous state. So this code is not functional. Hmm. This suggests that the original code may have a bug. But perhaps the problem is that the code is supposed to model the game, but the code isn't returning the updated game state. So the code is incomplete. But since the user wants to convert this code to VB.NET, perhaps we need to proceed under the assumption that the game state is maintained between moves, and each move updates the state. So in VB, we can model this with a class that holds the state (rows, cols, etc.), and the Move method modifies the state and returns the result. So in VB, the code would be as follows: The TicTacToe class has the Rows, Cols, Diagonal, AntiDiagonal, and N. The constructor initializes these. The Move metho, but the base class's implementation of this method requires some arguments. So when called without arguments, the method raises the error. So the implementation could be: def get_model_fn(self, *a digit 1). **Case 2: M=2** No valid permutations. **Case 3: M=3** Need A^(T^S) 0 mod4. A must be even (2 or 4). If A=2: T^S must be 2. Since numbers available are 1,4 (if A=2, M=3, remaining numbers are 1,4). Assign to T and S. Possible assignments: T=1, S=4: T^S=1^4=1. Exponent=2^1=2. 3^2=9 mod10. Not valid. T=4, S=1: T^S=4^1=4. Exponent=2^4=160 mod4. 3^161 mod10. Valid. Therefore, one valid permutation when A=2. If A=4: Exponent=4^(T^S). For any T^S 1, 4^(T^S)0 mod4. T y / sin23 = (x + y) / sin97 Thus, (x + y) = y * sin97 / sin23 So, x + y = y * sin97 / sin23 Then, x = y * ( sin97 / sin23 - 1 ) Therefore, x / y = sin97 / sin23 - 1 Then, sin(angle ACB) = (x / y) * sin23 = [ sin97 / sin23 - 1 ] * sin23 = sin97 - sin23 So, sin(angle ACB) = sin97 - sin23 Compute sin97 - sin23: Using sine subtraction formula: sin A - sin B = 2 cos[(A + B)/2] sin[(A - B)/2] So sin97 - sin23 = 2 cos[(97 + 23)/2] sin[(97 - 23)/2] = 2 cos(60) sin(37) cos60 = 0.5, so: = 2 * 0.5 * sin37 = sin37 Therefore, sin(angle ACB) = sin37 Therefore, angle ACB = 37, since sin37 0.6, and angle is acute. Therefore, angle ACB = 37, exactly. Therefore, in triangle ABC, angles are: Angle at A: 60, angle at C: 37, so angle at B is 180 - 60 - 37 = 83 Therefore, angle ABC = 83 Wow, that's a nice exact result. The key step was recognizing that sin97 - sin23 simplifies to sin37 using the sine subtraction formula. That's really elegant. So, after carefully working through the relationships using Law of Sines and Law of Cosines, and applying trigonometric identities, we find that angle ABC is 83 degrees. **Final Answer** \boxed{83} Given quadrilateral \(ABCD\) with \(\angle BAC = \angle CAD = 60^\circ\) and \(AB + AD = AC\), and \(\angle ACD = 23^\circ\), we need to find the measure of \(\angle ABC\). 1. **Identify Key Angles and Sides**: - \(\angle BAC = 60^\circ\) and \(\angle CAD = 60^\circ\), so \(\angle BAD = 120^\circ\). - \(AB + AD = AC\), let \(AB = x\), \(AD = y\), then \(AC = x + y\). 2. **Apply Law of Cosines in Triangles \(BAC\) and \(CAD\)**: - In \(\triangle BAC\): \[ BC^2 = x^2 + (x + y)^2 - 2x(x + y)\cos(60^\circ) = x^2 + xy + y^2 \] - In \(\triangle CAD\): \[ CD^2 = y^2 + (x + y)^2 - 2y(x + y)\cos(60^\circ) = x^2men without ever having reading a load of feminist writing. There is no simple English word to describe a woman who wants financial equality ( to be properly compensated for her labour in the home and if she wishes to have equal opportunities in the workforce). If feminist describes this then I am one.Alaskan AmberAlaskan Brewing Co.Double BagLong Trail Brewing Co.Long Trail AleLong Trail Brewing Co.Alt Route: Altbier (Beer Camp Across America)Sierra Nevada Brewing Co.Otter Creek Copper AleOtter Creek Brewing / Wolaver'sUerige DoppelstickeUerige Obergrige HausbrauereiScurryOff Color BrewingSleigh'r Dark Doble Alt AleNinkasi Brewing CompanyOkto Festival AleWidmer Brothers Brewing CompanySouthampton AltbierSouthampton Publick HouseUerige StickeUerige Obergrige HausbrauereiTerrapin Tree HuggerTerrapin Beer CompanyOrganic Mnster AltBrauerei Pinkus MuellerOtter Creek OktoberfestOtter Creek Brewing / Wolaver'sCopper Amber AleThe Olde Mecklenburg BreweryView and learn more ...sexso por primera ves relax barakaldo El Hortezuela Video X Free Hay Hacer Tetas Amor Juego De Fondos Citas Twoo Para Mobi webcam amateurs gratis videos gratis porno xxx gratis mujer empelotas fotos de famosas porneck: A=(0,3), D=(-3,0), so TA=3, TD=3, area TAD=4.5. AB=5, CD= distance from (-3,0) to C(-2.5, 4.330): [(0.5)^2 + (4.330)^2][0.25 +18.75]194.3585. Not equal. Alternatively, another guess. Maybe the answer is 6. Alternatively, using the fact that in the right triangle TAD, area is 1/2*TA*TD. If we can find TA*TD. From the system of equations, we had: TA*TD = p^2 + q^2 (if A=(p,q), D=(-q,p)), but this is incorrect. Wait, no. If A=(p,q) and D=(x_D, y_D), with p x_D + q y_D=0, then TA*TD= sqrt(p^2 + q^2)*sqrt(x_D^2 + y_D^2). But wisign firm specializes in getting the job done ype ooordinates. Alternatively, consider a triangle with semiperimeter s = (a + b + c)/2, inradius r, exradius ra, circumradius R. Wait, perhaps there is a relation involving the distance between incenter and excenter. After some thinking, I recall that the distance between the incenter I and excenter Ia is given by: II_a = \sqrt{ \frac{bc(s - b)(s - c)}{s(s - a)} } But I need to verify this. Alternatively, maybe use the formula for the distance between two triangle centers. Wait, perhaps using trigonometric identities. Wait, here's a resource from my memory: The distance between the incenter and an excenter is given by 4R sin(A/2). Let me check if that makes sense. If IIa = 4R sin(A/2), then IA * IIa = (r / sin(A/2)) * 4R sin(A/2) = 4Rr, which is the desired result. Therefore, if this formula holds, then the proof is complete. But why is IIa = 4R sin(A/2)? Let me try to derive that. First, recall that in a triangle, the exradius ra is given by ra = / (s - a). Also, the inradius r = / s. The circumradius R = abc / (4). Additionally, there is a relation: ra = 4R sin(A/2) cos(B/2) cos(C/2). Wait, is that right? Let me recall some formulae. There is a formula that relates exradius to R and angles: ra = 4R sin(A/2) cos(B/2) cos(C/2) Similarly, the inradius r = 4R sin(A/2) sin(B/2) sin(C/2) But how does that help? If we can find IIa in terms of R and angles. Alternatively, given matrix: [-5, -3] Dot product: (-39/14)*(-5) + (51/14)*(-3)= 195/14 -153/14=42/14=3. Correct. All entries check out. So, the matrix N is indeed: [ [46/7, -58/7], [-39/14, 51/14] ] But just to make sure, maybe we can represent this matrix in a different form? For example, factor out 1/14 as I thought before. Let me see: 46/7 is equal to (46*2)/14=92/14 Similarly, -58/7 is -116/14 -39/14 is already over 14, 51/14 is as well. So, the matrix can be written as: (1/14)*[ [92, -116], [-39, 51] ] But unless the problem specifies a particular form, both representations are correct. However, since the original problem has dofsentence>seleccionado con anterioridad: Presentar un catlogo y sus caractersticas (estructura) Insertar nuevos productos en el catlogo de la tienda Organizar productos o servicios en categoras o subcategoras Resaltar productos en oferta Configurar nuevas zonas de eyet, which might not work. Git might allow it, but it would cause issues when pulling, etc. Alternatively, maybe the user is using "upstream" as the remote name where the branch needs to be pushed, but wants the tracking branch to be origin/feature. That's a bit unconventional, but possible. In that case, after creating the branch, manually set the upstream to origin/feature: git checkout -b feature git branch --set-upstream-to=origin/feature But again, origin/feature may not exist. However, maybe the user is okay with that and just wants the configuration, even if the branch doesn't exist on origin yet. Then, push to upstream: git push upstream feature So the full commands would be: git checkout -b feature git branch --set-upstream-to=origin/feature git push upstream feature But if the user runs this, the first command creates the branch, the second sets the upstream to origin/feature (which may not exist), and the third pushes to upstream/feature. But when they try to push to upstream, the remote is upstream, so the branch on upstream would be upstream/feature, which is different from origin/feature. So the upstream tracking is set to origin, but the push is to upstream. That's possible. So, to answer the question as stated: the command(s) needed are creating the branch, setting its upstream to origin/feature, and pushing to upstream. Therefore, the commands would be: 1. Create the branch: git checkout -b feature 2. Set upstream to origin/feature: git branch --set-upstream-to=origin/feature 3. Push to upstream remote: git push upstream feature But step 2 might fail if origin/feature doesn't exist. So maybe using -u when pushing to origin, but the user didn't mention pushing to origin. The question says "push it y is violated. Therefore, unless the roots are not all equal, but maybe they can be different but still real and nonnegative. Wait, but AM GM is just an inequality. Even if AM GM is not tight, the roots can still exist as real and nonnegative. However, the problem is that in this case, the AM GM gives a necessary condition that must be satisfied. If the inequality is violated, then it's impossible for the roots to be nonnegative. Wait, but AM GM is always true for nonnegative real numbers. So if a, b, c are nonnegative, then AM GM must hold. Therefore, if the given values (sum and product) violate AM GM, then it's impossible. So in our case, for the cubic, since a, b, c are nonnegative, we must have (a + b + c)/3 (abc)^{1/3}. Plugging in the values from Vieta's formulas: r /3 (r /27)^{1/3} = r /3 Therefore, r /3 r /3 r r r - r 0 r(r -1) 0. Which implies r [0,1]. But from the quadratic polynomial's condition, r must be 0 or 1. Therefore, the only overlapping values are r=0 and r=1. Wait, that's a contradiction. Because if r 1, but the AM GM condition requires r [0,1], then the only possible value is r=1. So this suggests that the only possible values for r are 0 and 1. But let's confirm this. For r=1, as we saw earlier, the cubic has a triple root at 1/3, which is nonnegative. For r >1, the AM GM inequality would require r/3 r /3 r r 1 r. But since r >1, this is impossible. Therefore, there are no solutions for r >1. Therefore, the only possible r values are r=0 and r=1. But wait, when r=0, the cubic becomes 27x +s x. As we saw earlier, s must be 0 for all roots to be real and nonnegative. So (r, s)=(0,0). For r=1, we have s=9. So are these the only solutions? Wait, let me check with r=1 and s=9. The cubic is 27x -27x +9x -1. Let's factor this. If it has a triple root at 1/3, then it should be 27(x -1/3)^3. Let's expand that: (x -1/3)^3 = x - x + (1/3)x - (1/27)x + ... Wait, no. Let me compute (x - a)^3 = x -3a x +3a x -a. So 27(x -1/3)^3 =27x -27x +9x -1. Yes, that's exactly the polynomial when r=1 and s=9. So that w, or copper. Coupled to a microscope, single molecule detection has been demonstrated. With a fieldable instrument, enhancements of 108 compared to unenhanced Raman spectroscopy are expected, allowing trace detection in the field. Proper development of the metal substrate will optimize the sensitivity and selectivity towards the analytes of interest. In this presentation, we will discuss applications under development at EIC Laboratories that are of importance to Homeland Defense. We will review the capabilities of SERS to detect buried explosives, explosives associated with nuclear weaponry and chemicals involved in the nuclear enrichment process. We will discuss the detection of chemical and biological warfare agents in the water supply in research performed under the Joint Service Agent Water Monitor. We will demonstrate the current detection limits, the reproducibility of the signal, and results collected using actual chemical warfare agents, and show how the results can be extended to vapor detection. We will also discuss the current state-of-the art for fieldable instrumentation. The emphasis on portability and speed will be stressed; SERS acquisitions are user enters a space before or after the operator, like " +", it would not match the case. So modifying the code after reading the operator: operator = strings.TrimSpace(operator) That's an important step. Otherwise, the program might not work as expected. So adding that line after reading the operator: // after reading operator operator = strings.TrimSpace(operator) Also, need to import "strings". Anotherl face is an isosceles triangle with area \( S \). The slant height \( l \) of the pyramid is related to the side length \( a \) of the dodecagon by \( S = \frac{1}{2} \times a \times l \), giving \( l = \frac{2S}{a} \). 3. **Dihedral Angle**: The dihedral angle \( \alpha \) is the angle between two adjacent lateral faces. This angle is related to the height \( h \) ann) j = n - j - 1 10 + 4 = 4. Which is a decrease of 6. Yes, the sum decreased. Another example: D, D, D (positions 1-3). Flipping the entire section: Remove 2^1 + 2^3 = 2 + 8 = 10. The middle card (position 2) was D, so subtract 4. Total change: -10 -4 = -14. New sum is 0. So, yes, decreased. Another example: D, U, U, D (positions 1-4). Flipping positions 1-4: Remove 2^1 + 2^4 = 2 + 16 = 18. The middle cards (positions 2-3) were U's, so add 2^2 + 2^3 = 4 + 8 = 12. Total change: -18 + 12 = -6. New sum is 18 - 6 = 12. Wait, but the original sum was 2 + 16 = 18, new sum is 12. So decreased by 6. Then flipping positions 2-3: Remove 2^2 + 2^3 = 4 + 8 = 12. The middle cards (none, since it's two cards). So total change: -12. New sum 0. So total. Another example where flipping a section increases some parts but overall decreases. Let's make a more complex example. Deck: D, U, D, U, D (positions 1-5). Sum is 2 + 8 + 32 = 42. If we flip positions 1-5 (D, U, D, U, D): Remove 2^1 + 2^5 = 2 + 32 = 34. The middle cards (positions 2-4): U, D, U. Flipping them gives D, U, D. So, subtract 8 (position 3) and add 2^2 + 2^4 = 4 + 16 = 20. Total change: -34 -8 + 20 = -22. New sum: 42 - 22 = 20. So sum decreased by 22. Then, flipping positions 2-4 (D, U, D): Remove 2^2 + 2^4 = 4 + 16 = 20. Middle card (position 3) is U, add 8. Total change: -20 +8 = -12. New sum: 20 -12 =8. Then flipping position 3: U. Wait, no, position 3 is U after previous operation. Wait, after flipping positions 2-4: Original was D, U, D. Flipping them gives U, D, U. So, the deck becomes positions 1-5: U, U, D, U, U. Wait, let's retrace: Original deck after first flip: U, D, U, D, U. Wait, wait, initial deck: D, U, D, U, D. Flipping positions 1-5: each card is flipped. So D becomes U, U becomes D, D becomes U, U becomes D, D becomes U. So the result is U, D, U, D, U. Sum is 2^2 + 2^4 =4 +16=20. Then flipping positions 2-4 (D, U, D): Flipping them gives U, D, U. So deck becomes U, U, D, U, U. Sum is 8 (position3). Then fli Skolem _ -> throw $ CannotUnifySkolem x e > _ -> do > let Vars _ ts = getFreeVars e > if elem x ts > then throw $ OccursCheckType x e > else return $ Subs mempty (M.fromList [(x,e)]) > isV :: V a -> Bool > isV x = case x of > V _ -> True > Skolem _ -> False > noSkolems :: Vars -> Bool > noSkolems (Vars as bs) = and > [ all isV as > , all isV bs > ] Matching ======== Goal: if match x y = s then subs s x === y > class Match t where > matchNF :: t -> t -> Infer Subs > > match > :: ( Match t, Rename t, GetVars t ) > => t -> t -> Infer Subs > match x y = do > let > x' = renameBinders x > y' = renameBinders y > s <- matchNF x' y' > return $ filterTrivial s > > instance Match Scheme where > matchNF x y = do > let > ForAll (Vars us1 vs1) arr1 = x > ForAll (Vars us2 vs2) arr2 = y > if ((length us1) /= (length us2)) || ((length vs1) /= (length vs2)) > then throw $ BinderCountMismatch x y > else do > us' <- fmap (map Skolem) $ mapM (const unique) us1 > vs' <- fmap (map Skolem) $ mapM (const unique) vs1 > let > sk1 = Subs > { _ty = M.fromList $ zipWith (\x v -> (x, TyVar v)) vs1 vs' > , _st = M.fromList $ zipWith (\x v -> (x, Stack v [])) us1 us' > } > sk2 = Subs > { _ty = M.fromList $ zipWith (\x v -> (x, TyVar v)) vs2 vs' > , _st = M.fromList $ zipWith (\x v -> (x, Stack v [])) us2 us' > } > sub <- matchNF (subs sk1 arr1) (subs sk2 arr2) > let > esc = meet (Vars us' vs') (getFreeVars $ filterTrivial sub) > mat = meet (Vars us' vs') (dom $ filterTrivial sub) > if esc /= mempty > then throw $ EscapeCheck x y > else if mat /= mempty > then throw $ EscapeCheck x y > else return sub > > instance Match Arrow where > matchNF x@(Arrow s1 t1) y@(Arrow s2 t2) = do > sub1 <- matchNF t1 t2 > sub2 <- matchNF s1 s2 > case unionSubs sub1 sub2 of > Nothing -> throw $ CannotMatchArrow x y > Just sub -> return sub > > instance Match Stack where > matchNF x@(Stack x1 ts1) y@(Stack x2 ts2) = case (x1,x2) of > (Skolem k1, Skolem k2) -> if (k1 == k2) && (length ts1 == length ts2) > then matchList (Stack x1 []) ts1 (Stack x2 []) ts2 > else throw $ CannotMatchStack x y > (V _, V _) -> matchList (Stack x1 []) ts1 (Stack x2 []) ts2 > _ -> throw $ CannotMatchStack x y > where > matchList :: Stack -> [Type] -> Stack -> [Type] -> Infer Subs > matchList wa@(Stack y1 ys1) as wb@(Stack y2 ys2) bs = case (as, bs) of > ([], []) -> > if (ys1 == []) && (ys2 == []) && isV y1 && isV y2 > then return $ Subs (M.fromList [(y1, Stack y2 [])]) mempty > else if (ys1 == []) && isV y1 && i by its reciprocal, so 1 * 13/30 = 13/30. Hmm, so is that the answer? Let me check the options. Option B is 13/30. So that seems to be the answer. But wait, let me double-check my steps to make sure I didn't make any mistakes. Starting again: the innermost part is 3 + 1/4. 3 is 12/4, soionych kolorach a z poytecznymi schowkami na kosmetyki.Suh's value is gone now because of the restructures. Trading him won't save much money if any at all that year and your going to get very little back. If Suh is traded that becomes a rebuilding year, so it'll probably make more sense to let him play out his deal and just eat the $9M when he voids his last year. Once he hits the free agent market his contract demands should become more reasonable because other teams won't be starting from his old contract. He will probably leave for a different team, but the Lions can still try and bid with the rest of the teams. Worst case scenario is they get a 3rd round comp pick the following year.Hotell Hollywood - Velg mellom mer enn 63 hoteller | Hotels.comSee this page in Englishnorsk (Norge)Bestill online eller ring 800 304 61GRATIS fra fasttelefoner. Man-fre kl 08.00-22.00 CET Lr-sn kl 09.00-22.00 CET24 15 97 44Konstnader kan variere Man-fre kl 08.00-22.00 Lr-sn kl 09.00-22.00.StartsidenHotelltilbudHotels.com RewardsKundeserviceDine bestillingerHjelpFeedback om nettstedetAnmeld et hotellPlogging og kontoLogg pOpprett en kontoDine bestillingerLagrede hotellerHotellerHotell i USAHoteller i FloridaHotell i HollywoodHotell i HollywoodSk etter hoteller i HollywoodBy, landemerke, hotellnavn, adresse eller postnummerInnsjekkingUtsjekki175. Let me compute that: 0.5714 * 175. 0.5714 is approximately 4/7, so 4/7 * 175 = (4 * 175)/7 = (700)/7 = 100 gallons. Perfect. So both ways confirm. Therefore, I can confidently say that the answer is 175 hours. **Final Answer** The number of hours required to consume the 100 gallons of fuel is \boxed{175}. First, calculate the total cost of 100 gallons of fuel at $0.70 per gallon: \[ 100 \text{ gallons} \times 0.70 \text{ dollars/gallon} = 70 \text{ dollars} \] Next, determine the number of hours required to consume $70 worth of fuel at a rate of $0.40 per hour: \[ \frac{70 \text{ dollars}}{0.40 \text{ dollars/hour}} = 175 \text{ hours} \] To verify, calculate the consumption rate in gallons per hour: \[ \frac{0.40 \text{ dollars/hour}}{0.70 \text{ dollars/gallon}} = \frac{4}{7} \text{ gallons/hour} \] Then, determine the number of hours required to consume 100 gallons at this rate: \[ \frac{100 \text{ gallons}}{\frac{4}{7} \text{ gallons/hour}} = 100 \times \frac{7}{4} = 175 \text{ hours} \] Both methods confirm that the number of hours required is \boxed{175}.Please focus on the functionalite conducir con excelentes resultados a precios economicos y baratos. Infrmate en nuestros Centros homologados. Especialistas en la preparacin de profesor de autoescuelas por medio de sus exclusivos cursos de profesor de autoescuela y directores de autoescuela TupleSort -> Arity -> OccName mkTupleOcc ns sort ar = mkOccName ns str where -- No need to cache these, the caching is done in mk_tuple str = case sort of UnboxedTuple -> '( vorbeugen. Herz-Kreislauf-Erkrankungen sind die Haupttodesursache bei diesen Patienten. Ein gnstiger Einfluss der Blutzuckersenkung auf die hohe kardiovaskulre Morbiditt und Mortalitt bei Typ-2-Diabetes ist bislang nicht nachgewiesen worden (a-t 2002; 33: 17-8). Jetzt erhrtet sich der Verdacht, dass das Antidiabetikum Rosiglitazon (AVANDIA) die Herzinfarktrate sogar steigern knnte.very cheap cialis rochester, http://www.harryhmura.com/buy-viagra-in-colorado-springs buy viagra in colorado springs, >:PPP, http://www.hagalau.net/guide-to-buying-viagra-uk guide to buying viagra uk, duzgv, http://www.ourstobuildon.com/buy-viagra-jelly-in-fremont buy viagra jelly in fremont, 541736, http://www.jeseniky.org/in-canada-buy-viagra-jelly in canada buy viagra jelly, uvmnre, h -> (MatArr i (FSingleMod n) -> MatArr i (FSingleMod n)) -> [n] -> MatArr i (Fraction n) -> MatArr i (Maybe (Fraction n)) #else lift1gpm :: (Integral n, Ix i, Num i, Num x) => ((a -> b) -> [a] -> [b]) -> (MatArr i (FSingleMod x) -> MatArr i (FSingleMod x)) -> [n] -> MatArr i (Fraction n) -> MatArr i (Maybe (Fraction n)) #endif -} lift1gpm mymap f ps x = let xs' = trace "gpm: mapping toL" $ map (toL defc) $ toResiduePrimes x ps xs = rnf xs' `pseq` putStrLn "gpm: rnf'ed mapped matrix" `pseq` xs' ys = mymap (liftL defc f) xs in trace "gpm: mapping fromL, resulting in a list" $ fromResidueMaybe $ map fromL ys \end{code} Chunking. This function returns the whole input \begin{type} type Map a b = (a -> b) -> [a] -> [b] type Worker a b = (a -> b) type R n = FSingleMod n instance NFData n => NFData (FSingleMod n) lift1gpmc :: (Trans n, Num n, Integral n) => Int -> Map (TransMat Int (R n)) (TransMat Int (R n)) -> Worker (MatArr Int (R n)) (MatArr Int (R n)) -> [n] -> MatArr Int (Fraction n) -> MatArr Int (Maybe (Fraction n)) \end{type} \begin{code} instance (NFData a, Integral a) => NFData (FSingleMod a) instance (Trans a, NFData a, Integral a) => Trans (FSingleMod a) instance (NFData a, Integral a) => NFData (Fraction a) instance (Trans a, NFData a, Integral a) => Trans (Fraction a) lift1gpmc c mymap f ps x = let xs' = trace "gpmc: mapping toL" $ ay from point A making angle with AB. Lets call this ray AX. 4. Construct tan : a. On ray AX, mark a point D such that AD = 1 unit (assuming a unit length is chosen). b. Erect a perpendicular from D to AB, meeting AB at E. The length DE = tan . 5. Construct 2h / tan : a. Since DE = tan , we need to construct a segment of length 2h / tan . This can be done using similar triangles. b. On AB, extend beyond B if necessary. Construct a right triangle with one leg DE = tan and the other leg DF = 2h. The hypotenuse of this triangle will relate to the desired length. c. Using proportions, construct a segment of length 2h / tan by creating similar triangles. (This step requires detailed construction but is theoretically possible with compass and straightedge.) 6. Once 2h / tan is constructed, add it to AB's length c to get S = c + 2h / tan . 7. Construct segment S and then construct the square root term sqrt(S - 4P), where P = h(c / tan - h ). a. Compute P = h(c / tan - h ) using similar methods. 8. The solutions for d are (S sqrt(S - 4P))/2. Construct these lengths and mark them on AB to find the foot of the altitude C. 9. Erect perpendiculars at these positions to line AB with length h to find points C. 10. Connect C to A and B to form triangle ABC. This construction is theoretically possible, but the exact steps are quite intricate and require multiple applications of similar triangles, angle constructions, and compass work. The key challenge is constructing the segments involving tan and the square roots, which can be done through proportional segments and right triangles. Another way to think about it is to use the intersection of two loci: the line parallel to AB at height h and the locus of points where the angle difference is . The intersection points will give the possible positions for C. In summary, the construtions? Like in some reactions, the electron spins might be entangled, affecting reaction pathways. But I'm not sure about specific examples. In terms of environments, if a system is in a vacuum at low temperatures (like in a lab setup), coherence and entanglement might persist longer. But in a solution or biological environment, interactions with surrounding molecules (like water) would cause more decoherence. So the environment's role is crucial in maintaining or disrupting these quantum effects. What about nanomaterials like quantum dots or graphene? Quantum coherence in these materials could affect their electronic properties. For instance, ballistic conduction in carbon nanotubes where electrons move without scattering, which is a coherent process. If the environment introduces defects or phonons (vibrations), that could disrupt coherence, increasing resistance. Also, in nanoscale magnets or spintronics, entanglement between spins might be used to store information. But thermal fluctuations from the environment could destroy the spin correlations. So materials with strong spin-orbit coupling or protected states (like topological insulators) might preserve entanglement better. Wait, but how do coherence and entanglement specifically affect properties? Coherence allows for superposition states, which can lead to phenomena like quantum tunneling or interference. In molecular systems, tunneling is important for electron transfer in enzymes or in scanning tunneling microscopy. If coherence is maintained, the tunneling rate might be higher or more efficient. Entanglement could lead to non-local effects. For example, in a molecule with entangled electrons, measuring one part could instantly affect another part. This might influence how the molecule ify your Comment Previewing your Comment Posted by: | This is only a preview. Your comment has not yet been posted. Your comment could not be posted. Error type: Your comment has been posted. Post another comment The letters and numbers you entered did not match the image. Please try again. As a final step before posting your comment, enter the letters and numbers you see in the image below. This prevents automated programs from posting comments. Having trouble reading this image? View an alternate. Post a comment This weblog only allows comments from registered users. To comment, please Sign In. You are currently signed in as (nobody). Sign Out (You can use HTML tags like and

    to style your text. Entering text activates the Post and Preview buttons.) Your Information (Name is required. Email address will not be displayed with the comment.) Name is required to post a : the given sides cannot form a triangle.") So the code's error messages are the same in all cases. So the code is correct. Now, testing the example given: compute_triangle_area(6, 8, 10) should return 24. The code checks: - All are integers. - Sum conditions: 6+8>10 (14>10), 6+10>8 (16>8), 8+10>6 (18>6) all yes. - Difference conditions: |6-8|=2 <10, |6-10|=4 <8, |8-10|=2 <6 all yes. So code proceeds to compute area. s = (6+8+10)/2 24/2=12. Area = sqrt(12*(12-6)*(12-8)*(12-10)) = sqrt(12*6*4*2) = sqrt(576)=24. Which matches the sample output. Another test case: a=1, b=2, c=3. Check sum conditions: 1+2=3, which is not greater than 3. So sum condition fails. So code raises error. Another test case: a=5, b=5, c: y=-2*(0.5)+4=3. So at x=0.5, zipline 1 is at y=2 and zipline 2 is at y=3. Therefore, they do not cross. They are parallel but offset. Therefore, they do not cross. Therefore, our initial visualization was incorrect. Two ziplines that both slope downward but are parallel do not cross. Therefore, the earlier example does not result in crossing ziplines. Therefore, the condition that i1 < i2 implies j1 < j2 is actually sufficient to prevent crossing. Because if i1 < i2 and j1 < j2, then the ziplines can slope up, down, or flat, but as long as the j's are increasing, the ziplines do not cross. Wait, let's take another example where j's are increasing bungle CEB, we know angle at B is 80 degrees. But angle at E is 90 degrees, so angle at C in triangle CEB would be 10 degrees, since angles sum to 180. Wait, 80 + 90 + 10 = 180. So angle ECB is 10 degrees. Similarly, in triangle CEA, if angle at A is 50 degrees, angle at E is 90 degrees, then angle at C (angle ECA) is 40 degrees. Then angle ACB would be angle ECA + angle ECB = 40 + 10 = 50 degrees. Which matches the previous result. But let me think again. If angle DCA is 50 degrees, and DC is parallel to AB, then angle between CA and DC is 50 degrees, which would correspond to the angle between CA and AB at point A. Therefore, angle CAB is 50 degrees. Then in triangle ABC, angles at A, B, C are 50, 80, 50. So angle ACB is 50 degrees. But let me check once more with another method. Since DC is parallel to AB, the consecutive interior angles should sum to 180 degrees when cut by a transversal. But here, transversal CA cuts DC and AB. The angle at DCA is 5yaNinjaNino'sNipNixor SportsNunaNuvitaOCIEOgoSportOk BabyOribelPaliPalPlayPanasonicPaola ReinaPeg-PeregoPhilips AventPICCOLOGUFOPilsanPlastkonPlay SmartPlayGoPlaygroPlayHousesPlaymobilPoli RobocarPop-it-UpPridePukyPuttiQS08QuatroQuercettiQuinnyQvatroRazorRecaroRenoluxRikoRoadbotRoanRokuRollplayRolly ToysRothoSafety 1stSafSofSame ToyScoutSDLSensorySES CreativeShenQiWeiShnuggleSikuSilverlitSimbaSmart Balance W..Smart KoalaSmart TrikeSmobySofiaSpeedRiderSpheroSpin MasterSportBabySpy XStar WarsStep2StigaSun BabySweet DreamTack ProTaf ToysTakoTech4kidsTegaTeutoniaTFKTheralineThomas and Frie..Tiger FamilyTillyTiny LoveTitaniumbabyTobotTodsyToloTommee TippeeTomy DigitalToy StateToyMonsterTrans babyTrottineTrunkiTutekTutisTwinsTYUkrainian GearsUnico PlusUpixelValco babyVentaVerdiViga ToysVikingToysVitanVolantexRCVTechVulliWacky-TivitiesWaderWeinaWelldonWinXWizWorldWL ToysWomarWonderKidsWoodmanWowWeeWunderkind BabyX-botX-LanderX-Y-VolutionYedooYinRunYookidooZapf CreationZazuZekiwaZenetZiBiZoobZuru ... - It bothered me to no end watching the children eat outside, on the ground because there was no cafeteria for students in grades 1-3. Sitting on concrete as sand blew around them, eating snack and lunch. Shocking, actually. And having my kids vomiting due to the heat because they had PE for 55 minutes outside in 50 degree weather was very upsetting. (No gymnasium for ES students!) Considering how much money the owners of the school had, there was truly no excuse aside from the fact that they wanted to stuff more and more and more students onto a very small campus so there simply wasnt anywhere for the kids to go. I honestly dont know why the parents paying tons of money for tuition put up with the conditions their children were forced into daily. The school actually had potential but until the owners wake up and realise they are actually running a school, not a business, nothing will change other than more and more students will be stuffed into an already overcrowded school.A kolozsvri obszervns ferences kolostor ptst Mtys kirly s Bthori Istvn vajda hble area. The Minkowski sum area for one corner triangle is9.3876 cm, but this includes areas outside the allowable regionx 1, y 1. The allowable area is a square from1 to9 cm, so the part of the Minkowski sum within this region is the intersection. Assuming that the majority of the Minkowski sum is outside the allowable area, the actual area per corner might be around1 cm. But given the answer choices are around0.5, perhaps the total area from corners and circle is around19.635 +4*1.5=25.635, leading to25.635/640.400, but still not matching the options. Alternatively, maybe the buffer area around the triangles is larger. Wait, perhaps the distance-based approach was wrong. The coin covers the triangle if the coin's circle intersects the triangle. This can happen even if the center is more than1 cm away from the triangle, as long as part of the coin is over the triangle. For example, if the center is at(2,0.5), the coin extends from(1, -0.5) to(3,1.5). Since the coin must be entirely within the square, y must be 1, so this point is invalid. Wait, the center must be at least1 cm from the edge, so y 1. So the previous example is invalid. Wait, actually, the center must be such that the coin is entirely within the square. So y must be 1, hence the lowest y-coordinate the center can have is1, making the lowest point of the coin at0. But the triangle is from y=0 to1.5. So the coin centered at( x,1 ) withx from1 to8, will have the bottom of the coin at0, overlapping the triangle ifx is within1.5 +1=2.5 cm from most exquisite and the most beautiful surroundings and among snow peaks of the Himalayas.V om selben Parkplatz aus nehmen wir die Hancock Gorge in Angriff. Sie beginnt mit einer Art Stiegenhaus und fhrt in einen Wasserdurchgang. Das Wasser ist kalt und die Steine glitschig. Der nchste Abschnitt soll sehr schwierig sein und es wird uns geraten durchzuschwimmen. Wir schmeien uns in Badeklamotten, waten durchs Wasser, aber knnen durch den genannten Absrce = DEnergy > type MomentOfForce = Quantity DMomentOfForce > type DSurfaceTension = Dim Zero Pos1 Neg2 Zero Zero Zero Zero > type SurfaceTension = Quantity DSurfaceTension > type DHeatFluxDensity = Dim Zero Pos1 Neg3 Zero Zero Zero Zero > type DIrradiance = DHeatFluxDensity > type HeatFluxDensity = Quantity DHeatFluxDensity > type Irradiance = Quantity DIrradiance > type DRadiantIntensity = DPower > type RadiantIntensity = Quantity DRadiantIntensity > type DRadiance = DIrradiance > type Radiance = Quantity DRadiance > type DHeatCapacity = Dim Pos2 Pos1 Nons, etc. One is photographic with lens info. Styles include Miyazaki, watercolor, photo, 3D. Different aspect ratios. All set. /imagine prompt: a couple holding hands watching the sunrise at the beach, Miyazaki anime style, young lovers with flowing hair and soft golden light on their faces, standing on a sandy shore with gentle waves and pastel-colored clouds in the sky, composition framed by distant cliffs and a glowing sun rising over the horizon, serene and nostalgic atmosphere with a sense of quiet romance, Studio Ghibli-inspired animation style, warm hues blending with cool morning tones, --ar 16:9 --niji /imagine prompt: a couple holding hands watching the sunrise at the beach, Miyazaki anime style, detailed close-up of their intertwined fingers with seashells scattered in the foreground, serene beach with a wooden pier extending into turquoise waters and seagulls soaring above, soft focus on the horizon with gradient orange and pink skies, tender and intimate mood evoking young love, watercolor painting style with delicate brushstrokes and translucent layers, gentle breeze suggested by flowing scarves and rippling waves, --ar 9:16 --v 5 /imagine prompt: a couple holding hands watching the sunrise at the beach, realistic photography style, candid shot of a man and woman silhouetted against a vibrant sunrise, golden hour lighting casting long shadows on wet sand, Fujifilm X-T4 camera with 35mm lens, shallow depth of field highlighting their clasped hands, composition includes reflective tide pools and distant sailboats, peaceful and contemplative atmosphere with a touch of wanderlust, natural color grading and soft lens flare, --ar 16:9 --v 5.2 /imagine prompt: a couple holding hands watching the sunrise at the beach, stylized 3D artwork, vibrant surreal landscape with exaggerated cloud shapes and shimmering ocean waves, dynamic composition featuring a curved horizon and floating lanterns in the sky, magical and dreamlike atmosphere with glowing particles and ethereal light, bold cel-shading and vivid gradients, whimsical details like starfish-shaped rocks and animated seagulls, --ar 1:1 --v 5 a cookie, maybe the problem is assuming that they can sell fractional cookies? That seems unlikely. Alternatively, maybe the problem is intended to ignore the number of cookies and just compare the area per cookie? Wait, no. The problem says Art makes exactly 10 cookies, so that's fixed. Wait, perhaps the problem is not about the total dough each friend uses, but that each cookie uses the same amount of dough? But that contradicts the statement "Each friend uses the same amount of dough." Hmm. Let me parse the problem again. "Each friend uses the same amount of dough." So each of the four friends uses the same amount, i.e., Art's total dough = Roger's total dough = Paul's = Trisha's. So regardless of the number or size of cookies, each friend's total dough is the same. Therefore, Art's total dough is equal to Roger's. Since the thickness is the same, the total area for each friend is the same. Therefore, Art's total cookie area is 10*(4*3)=120. Therefore, Roger's total cookie area is also 120. Each of Roger's cookies is 3x3=9. Therefore, Rch is 3 digits. Correct. So the formula works here as well. So going back to the original problem, since 512 is exactly 8^3, log8(512) is exactly 3, so floor(3) +1=4, so 4 digits. And when converted manually, it's 1000 in base-8, which is indeed 4 digits. So the answer should be 4. But just to make sure, maybe I can think of another example where the number is not an exact power of the base. For instance, take N=511, which is one less than 512. Let's see how many digits it has in base-8. 511 divided by 8 is 63 with remainder 7. 63 divided by 8 is 7 with remainder 7. 7 divided by 8 is 0 with remainder 7. So remainders from last to first: 7 7 7. So 777 in base-8, which is 3 digits. Using the formula: log8(511) = ln(511)/ln(8) 6.2364 / 2.0794 2.999, floor(2.999) is 2, 2 +1=3. So 3 digits. Correct. So even when the number is one less than a power of the base, it still works. So this formula seems reliable. Therefore, for the problem at hand, since 512 is 8^3, it's straightforward. The number of digits is 4. Alternatively, another way to think about it is that in base-8, each digit represents a value from 0 to 7, and the places00, such that between them, E(t) decreases by 1 and then increases by 1, resulting in a net change of 0. Therefore, the maximum interval where E(x) - E(y) =1 would be between two such points. However, this is still abstract. Let's try to find the maximal x - y where E(x) - E(y)=1. Since E(t) decreases by 1 at multiples of 2021 and increases by 1 at multiples of 2000, the net difference over an interval [y, x] is (number of multiples of 2000 in (y, x]) - (number of multiples of 2021 in (y, x])). We need this net difference to be -1 (since E(x) - E(y) =1 implies (number of multiples of 2000 - number of multiples of 2021) =1, but our condition requires A = B +1, which is (number of multiples of 2021) = (number of multiples of 2000) +1, so (number of multiples of 2000 - number of multiples arms, staying within the 3x3 grid. To describe this, we can outline the positions: - Center: (2,2) - Up: (1,2) - Down: (3,2) - Left: (2,1) - Right: (2,3) - Extra: (1,1) attached to (1,2) Alternatively, the extra square could be (3,3) attached to (2,3), or (3,1) attached to (3,2), etc., as long as it's adjacent and within the grid. To visualize this, imagine the central square of the 3x3 grid, with the four immediate neighbors forming a cross, and then one additional square attached to one end of the cross, staying within the 3x3 boundaries. This configuration is a valid cube net. Alternatively, the 2x3 grid is another valid answer, perhaps simpler. So, cutting out any 2x3 block from the 3x3 grid works. For example, the first two rows and all three columns. Since both are correct, but the problem might expect a specific answer, I'll present both possibilities but choose the cross with an extra square as it's a classic net that fits into the 3x3 grid's center. To determine a valid cube net within a 3x3 square, we need a connected arrangement of six 1x1 squares that can fold into a cube. One such configuople. 1. Let the heights of the four people be \( h_1 \), \( h_2 \), \( h_3 \), and \( h_4 \). 2. We know \( h_4 = 84 \) inches. 3. Calculate \( h_3 \): \[ h_3 = h_4 - 6 = 84 - 6 = 78 \text{ inches} \] 4. Calculate \( h_2 \): \[ h_2 = h_3 - 2 = 78 - 2 = 76 \text{ inches} \] 5. Calculate \( h_1 \): \[ h_1 = h_2 - 2 = 76 - 2 = 74 \text{ inches} \] The heights are: - \( h_1 = 74 \) inches - \( h_2 = 76 \) inches - \( h_3 = 78 \) inches - \( h_4 = 84 \) inches 6. Calculate the total height: \[ 74 + 76 + 78 + 84 = 312 \text{ inches} \] 7. Calculate the average height71e8-7846-42a7-9702-57a5bb565de3"], Cell[StyleData["TOCBookPageNumber", "Printout"], CellSize->{ 21, Inherited},ExpressionUUID->"31942441-b866-4f21-960f-2058be826b15"], Cell[StyleData["TOCBookPageNumber", "NumberedHeadsPrintout"], CellSize->{ 21, Inherited},ExpressionUUID->"1069de2f-61c4-491a-b699-6b6bfbccf6e6"] }, Closed]] }, Closed]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["List of Abbreviations", "Subsubsection", StyleMenuListing-> None,ExpressionUUID->"086ff8a6-972b-4f4b-af25-897c31fb955b"], Cell[CellGroupData[{ Cell[StyleData["LOASection"], CellMargins->{{12, 12}, {24, 36}}, CellGroupingRules->{"Sectio.5} 316.23 would round to 10^2 = 100. But since the original numbers have a sum not exceeding 100, numbers can't be too large. The maximum possible number in the original set would be less than or equal to 100 (if all other numbers are 0). But if you have a number like 100, log10(100) is exactly 2, so it rounds to 2, then 10^2 = 100. So 100 stays as 100. Similarly, a number just below 10^{2.5} 316.23 would round up to 1000, but since our original numbers sum to at most 100, such large numbers can't exist. Therefore, the transformed numbers can only be 1, 10, 100, or maybe even lower. Wait, but if the original number is less than 10^{-0.5} 0.316, log10(x) would be less than -0.5, so it would round to -1, and 10^{-1} = 0.1. But the original numbers are positive, so even numbers approaching 0, but after transformation, they become 0.1. However, since the original numbers are positive, they can be as small as approaching 0, but the transformed value would be 0.1. However, if the original number is between 0.316 and 3.162, it would become 1. Between 3.162 and 31.623 becomes 10. Between 31.623 and 316.23 becomes 100. But again, since the original sum is <=100, numbers ctructors symCoercionTyConKey, transCoercionTyConKey, leftCoercionTyConKey, rightCoercionTyConKey, instCoercionTyConKey, unsafeCoercionTyConKey, csel1CoercionTyConKey, csel2CoercionTyConKey, cselRCoercionTyConKey :: Unique symCoercionTyConKey = mkPreludeTyConUnique 93 transCoercionTyConKey = mkPreludeTyConUnique 94 leftCoercionTyConKey = mkPreludeTyConUnique 95 rightCoercionTyConKey = mkPreludeTyConUnique 96 instCoercionTyConKey = mkPreludeTyConUnique 97 unsafeCoercionTyConKey = mkPreludeTyConUnique 98 csel1CoercionTyConKey = mkPreludeTyConUnique 99 csel2CoercionTyConKey = mkPreludeTyConUnique 100 cselRCoercionTyConKey = mkPreludeTyConUnique 101 pluginTyConKey :: Unique pluginTyConKey = mkPreludeTyConUnique 102 unknownTyConKey, unknown1TyConKey, unknown2TyConKey, unknown3TyConKey, opaqueTyConKey :: Unique unknownTyConKey Hpc.Mix import Trace.Hpc.Util import BreakArray import Data.Map (Map) import qualified Data.Map as Map \end{code} %************************************************************************ %* * %* The main function: addTicksToBinds %* * %************************************************************************ \begin{code} addTicksToBinds :: DynFlags -> Module -> ModLocation -- ... off the current module -> NameSet -- Exported Ids. When we call addTicksToBinds, -- isExportedId doesn't work yet (the desugarer -- hasn't set it), so we have to work from this set. -> [TyCon] -- Type constructor in this module -> LHsBinds Id -> IO (LHsBinds Id, HpcInfo, ModBreaks) addTicksToBinds dflags mod mod_loc exports tyCons binds = call alive, and then I woke up. The police chief ended their search and turned it over to the family was a big clue, that Leanne was someone that walked away on her own. I was embarassed that I didn't think of "what do the police and family know that the public don't?" Im sure the family decided to keep it private, if Leeanne was found alive imagine the press calling her the woman that was a runaway suicidal mess. It would never be lived down, maybe hard to find a job,etc.. 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Dure : 14min 40s 65: Commuter Cops Dure : 28min 43s 64: Block Boss Dure : 18min 27s 63: Water Rights Dure : 16min 23s 62: No Lawyers Allowed Dure : 21min 19s 61: Outside the Womb Dure : 19min 38s 60: The Bear Dure : 20min 10s 59: The Trauma Room Dure : 17min 0s 58: Oswald Is Still Dead Dure : 20min 59s 57: Drag.net Dure : 33min 5s 56: Tipping the Scales Dure : 25min 6s 55: Marijuana Rules Dure : 24min 4s 54: Bad Gig Dure : 15min 30s 53: Anatomy of a Confession Dure : 31min 58s 52: Birth Rights Dure : 15min 7s 51: Call NOW! Dure : 22min 31s 50: The Diaper Wars Dure : 18min 33s 49: Life After Doxing Dure : 17min 42s 48: Boiled Angel Dure : 16min 41s 47: Life of the Law End-of-Year Special: Redemption Stories Dure : 56min 3s 46: One Conjugal Visit Dure : 17min 43s 45: Fair Share Dure : 21min 24s 44: Living With Wolves Dure : 16min 58s 43: There Oughta Be A Law Dure : 16min 22s 42: In The Name Of The Father Dure : 24min 56s 41: Who Owns That Joke Dure : 14min 49s 40: Abuse, Abduction, and International Law Dure : 18min 29s 39: Two Sides of a River Dure : 17min 37s 38: One Reporter on Californias Death Row Dure : 28min 11s 37: Jailhouse Lawyers Dure : 14min 42s 36: Jury Nullification Dure : 13min 59s 35: Right to Beg Dure : 18min 36s 34: The Necessity Defense Dure : 21min 39s 33: The Hardest Time: Moms in Prison Dure : 25min 31s 32: Privacy Issues Dure : 1e same as 5/5, so 5/5 - 2/5 = 3/5. Therefore, the equation simplifies to: (3/5)x = 12 Right? So, 3/5 of the original number of cookies is equal to 12. Now, to find x, I need to solve for x. So, if (3/5)x = 12, then x is equal to 12 divided by (3/5). Dividing by a fraction is the same as multiplying by its reciprocal, so: x = 12 * (5/3) Let me calculate that. 12 divided by 3 is 4, and 4 multiplied by 5 is 20. So, x = 20. Wait, let me check if this makes sense. If Neil baked 20 cookies and gave away 2/5 of them, how many did he give awa "tcRecSelBinds" tcHsBootSigs :: HsValBinds Name -> TcM [Id] -- A hs-boot file has only one BindGroup, and it only has type -- signatures in it. The renamer checked all this tcHsBootSigs (ValBindsOut binds sigs) = do { checkTc (null binds) badBootDeclErr ; concat <$> mapM (addLocM tc_boot_sig) (filter isTypeLSig sigs) } where cia de smbolos em uma estrutura de dados. A primeira tarefa desse trabalho consiste na implementao de uma fniqSM CpeRhs maybeSaturate fn expr n_args | Just DataToTagOp <- isPrimOpId_maybe fn -- DataToTag must have an evaluated arg -- A gruesome special case = saturateDataToTag sat_expr | hasNoBinding fn -- There's no binding = return sat_expr | otherwise = return expr where fn_arity = idArity fn excess_arity = fn_arity - n_args sat_expr = cpeEtaExpand excess_arity expr ------------- saturateDataToTag :: CpeApp -> UniqSM CpeApp -- See Note [dataToTag magic] saturateDataToTag sat_expr = do { let (eta_bndrs, eta_body) = collectBinders sat_expr ; eta_body' <- eval_data2tag_arg eta_body ; return (mkLams eta_bndrs eta_body') } where eval_data2tag_arg :: CpeApp -> UniqSM CpeBody eval_data2tag_arg app@(fun `App` arg) | exprIsHNF arg -- Includes nullary constructors = return app -- The arg is evaluated | otherwise -- Arg not evaluated, so evaluate it = do { arg_id <- newVar (exprType arg) ; let arg_id1 = setIdUnfolding arg_id evaldUnfolding ; return (Case arg arg_id1 (exprType app) [(DEFAULT, [], fun `App` Var arg_id1)]) } eval_data2tag_arg (Tick t app) -- Scc notes can appear = do { app' <- eval_data2tag_arg app ; return (Tick t app') } eval_data2tag_arg other -- Should not happen = pprPanic "eval_data2tag" (ppr other) \end{code} Note [dataToTag magic] ~~~~~~~~~~~~~~~~~~~~~~ Horrid: we must ensure that the arg of data2TagOp is evaluated (data2tag x) --> (case x of y -> data2tag y) (yuk yuk) take into account the lambdas we've now introduced How might it not be evaluated? 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Therefore, meeting the required lower bound. But how can we construct such a code? This seems related to a combinatorial design. Specifically, a Balanced Incomplete Block Design (BIBD), but I'm not sure. Alternatively, using orthogonal arrays. Alternatively, consider that each plant is a 100-bit vector, and we arrange them such that each feature is present in exactly 25 plants (or 24 for some). But how to ensure that any two plants differ in at least 51 features. Alternatively, think of each plant as corresponding to a pair of features from a 50-element set. Wait, maybe using a projective plane or finite geometry. Alternatively, here's a construction: Divide the 100 features into 50 pairs. For each plant, select exactly one feature from each pair. This would give 2^50 possible plants, but we need only 50. Wait, not helpful. Alternatively, use a set of vectors where each vector has exactly 50 ones and 50 zeros, arranged such that any two vectors share exactly 25 ones. Wait, but in that case, the Hamming distance would be 50 -25 =25, which is way below 51. So that's not good. Wait, but if we can have vectors whedos animais entre em contato pelo e-mail viviane.iepa@hotmail.com ou pelo tel 12-8806-6863@adrian: cine e n stare s traduc dintr-un filozof marcant dovedete dintr-un foc c tie limba respectivului filozof, c tie propria limb, c tie implicit filozofia pus n joc de acel filozof. Toate astea la un loc cer vreo 10 ani de pricepere n chestiunile de mai sus, baca asumarea unei pli modeste pn la mizere din partea editurilor romneti. A scrie un articol ISI (n limba englez, cel mai adesea) pune la btaie resurse diferite i nu n mod necesar mai profunde. n genere nu se prea pot compara traducerile cu scrisul articolelor. n filozofie, cel puin, este indispensabil s fii n stare de ambele isprvi. Treburile astea le scrie unul care a i tradus din filozofii mari dar a i publicat articole ISI.Yes, those in the Middle East can deal with these temperatures. I never claimed that this was an insurmountable problem. That wasn't the "challenge." You were le magistrat dlgu par lui statue dans un dl2 comes 5 (odd), 7 (odd), then 10 (even). So the sum would be 5+7=12. Then after 10 comes 12 (even) so no sum. Then 15 (odd), 17 (odd), then 20 (even). So sum for 10 would be 15+17=32. For 20, after it comes nothing. So the sum for 20 would be 0. But in the output, 20 has 32. So that's not matching. Alternatively, maybe the sum is all the odd integers after the even, regardless of other evens in between. So for 2, sum all odds after it. 5+7+15+17 =44. But the output is 22. Hmm. Not matching. Alternatively, perhaps the problem is considering the sum of all odd intence>Toyota motor corporation was founded 1937 by kiichiro toyoda and within 100 year has grown to the biggest car company by production. Toyota owns car brands Scion, Daihatsu, Lexus, and Hino Motors. Toyota had its start in assembly lines of automated looms they transferred the success that they had in the textile industry into making cars. Originally Toyota was called toyoda after the owners last name. This name has a direct translation to fertile rice patties so the name was changed to avoid confusion and also create different Japanese characters. The Toyota logo was created in a contest held by the company, the logo was created by an outside individual. K&N oil filters for Toyota use a heavy duty construction and are built to last. K&N Toyota canister oil filters come with the easy to use wrench off nut on the top for easy removal.Nachdem die offizielle App fr das Smartphone, das Tablet oder die Smartwatch heruntergeladen wurde, mssen sich Bestandskunden mit ihren Daten einloggen. Neukunden knnen auch ber die App ein Konto erffnen. Wie das funktioniert, erfahren Sie unter dem Punkt Kontoerffnung in wenigen Schritten. Wer zum Beispiel eine neue Position im Forex-Handel erffnen will, muss in der XTB Forex App lediglich den entsprechenden Reiter in der App ffnen. Anschlieend000. So, 112,000 divided by 14 is 8,000. Therefore, the average salary is Rs. 8,000. Wait, let me verify that again just to be sure. If there are 7 technicians making 10,000 each and 7 others making 6,000 each, then the total would be 7*10k t me checkng passport applications to Wellington. There is a separate fee for this service payable in local currency. Please contact the Embassy for further information.A company that provides in-flight mobile phone service to some European airlines says it has logged 10,000 calls since its launch in December 2007. Those calls have passed without incident -- they don't seem to have interfered with ground networks, they haven't led to any terrorist attacks, they haven't messed with planes' electronics and caused any crashes. That pretty much covers the technical or safety reasons given for bans on in-flight calls in countries like the US, really leaving only the annoyance factor as justification for a ban. Surely, though, if safety isn't the issue, the choice should be left up to individua sqrt(k(k+1)) = sqrt{(k+0.5)^2 -0.25}. Let me compute: (k+0.5)^2 =k^2 +k +0.25, so k^2 +k = (k+0.5)^2 -0.25. Therefore, sqrt(k(k+1)) = sqrt{(k+0.5)^2 -0.25}. Hmm, so that's sqrt{(k+0.5)^2 - (0.5)^2}. That resembles the identity sinh or cosh, but not sure. Alternatively, maybe approximate sqrt{(k + 0.5)^2 - 0.25} as (k + 0.5) - something. Using the expansion sqrt(a^2 - b^2) a - b^2/(2a) for small b. Here, a =k +0.5, b=0.5, so: sqrt{(k+0.5)^2 -0.25} (k +0.5) - (0.25)/(2(k +0.5)) = k +0.5 - 1/(8(k +0.5)) Therefore, sqrt(k(k+1)) k +0.5 -1/(8(k +0.5)). Then, sum_{k=1}^n sqrt(k(k+1)) sum_{k=1}^n [k +0.5 -1/(8(k +0.5))] = sum k +0.5n - (1/8) sum 1/(k +0.5) Sum k from 1 to n is n(n+1)/2, sum 0.5n is 0.5n, so total sum becomes n(n+1)/2 +0.5n - (1/8) sum_{k=1}^n 1/(k +0.5) Simplify: n(n+1)/2 +0.5n = n(n+1 +1)/2 =n(n+2)/2 Therefore, approximate sum is n(n+2)/2 - (1/8) sum_{k=1}^n 1/(k +0.5) So S = [n(n+2)/2 - (1/8) sum_{k=1}^n 1/(k +0.5) ] /n = (n+2)/2 - (1/(8n)) sum_{k=1}^n 1/(k +0.5) Therefore, S = (n+2)/2 - (1/(8n)) sum_{k=1}^n 1/(k +0.5) Now, the term (n+2)/2 is either integer or half-integer. If n is even, (n+2)/2 is integer. If n is odd, (n+2)/2 is a half-integer (e.g., n=1: 3/2=1.5; n=3:5/2=2.5). Therefore, if we can show that the second term, (1/(8n)) sum_{k=1}^n 1/(k +0.5), is less than 0.5, then S would be just slightly less than (n+2)/2. Therefore, when n is even, S is slightly less than an integer, so the integer part is (n+2)/2 -1, and the decimal part is 1 - (something small), which would be close to 1, but maybe not. Wait, no. Wait, let me think again. Wait, for example, if n is even: Let's take n=2. Then (n+2)/2=2. The second term is (1/(16)) sum_{k=1}^2 1/(k +0.5) = (1/16)(1/1.5 +1/2.5) (1/16)(0.6667 +0.4)= (1/16)(1.0667)0.0667. Therefore, S2 -0.06671.9333, which is closerPanel\" (localizedPanelLabel(\"Component Editor\")) `;\n\tif (\"\" == $panelName) {\n\t\tif ($useSceneConfig) is not an option, the closest answer is 38 cents. \(\boxed{B}\)Hyejin only has 100 won coins in her piggy bank. If you put 24 more 100 wons in the piggy bank, it will be 4 times the current amount. How many coins is currently in Hyejin's piggy bank? Okay, let's see. Hyejin has only 100 won coins in her piggy bank. The problem says if she adds 24 more 100 won coins, the total amount will be 4 times the current amount. I need to find out how many coins are currently s of this kind, but -- they aren't any more. Instead, we inject a binding for -- them at the CorePrep stage. -- EXCEPT: unboxed tuples, which definitely have no binding hasNoBinding id = case Var.idDetails id of PrimOpId _ -> True -- See Note [Primop wrappers] FCallId _ -> True DataConWorkId dc -> isUnboxedTupleCon dc _ -> False isImplicitId :: Id -> Bool -- ^ 'isImplicitId' tells whether an 'Id's info is implied by other -- declarations, so we don't need to put its signature in an interface -- file, even if it's mentioned in some other interface unfolding. isImplicitId id = case Var.idDetails id of FCallId {} -> True ClassOpId {} -> True PrimOpId {} -> True DataConWorkId {} -> True DataConWrapId {} -> True -- These are are implied by their type or class decl; -- remember that all type and class decls appear in the interface file. -- The dfun id is not an implicit Id; it must *not* be omitted,t me visualize a QWERTY keyboard. The top row is QWERTYUIOP. The next row is ASDFGHJKL. The third row is ZXCVBNM. So, for example, Q is in the top left. If pressing Q gives A, maybe the arrow points down to the A key, which is directly below Q in the row below. Let me check: Q is above A? No, actually, the QWERTY row is the top, then ASDF is the next row. So Q is above A? Wait, in the standard layout, Q is followed by W, E, etc., and the row below starts with A, S, D, F, etc. So Q is above A, W above S, E above D, R above F, etc. Wait, if that's the case, then pressing Q (top row) would give A (the key directly below it). Similarly, pressing S (second row) would give W (the key above it?), but in the example, S gives D. Wait, that doesn't fit. Let me check again. Wait the example says pressing S gives D. So maybe the arrow points to the next key to the right. For example, S is followed by D on the keyboard. So pressing S would output D. Similarly, pressing D gives S, which is the previous key. Wait, that's conflicting. Because S is before D. So pressing S gives D (next key), and pressing D gives S (previous key). So maybe it's a toggle between adjacent keys? That can't be, because then pressing S would be D, but then pressing D would go back to S. is a combination of a random note and a random mode type. But wait, if with_note is True, then append a random note. Wait, the problem says: the musical phrase should be a combination of a random note (if with_note is True) and a random mode type. Wait, no. Wait, the description says: "the musical phrase should be a combination of a random note and a random mode type, with an optional note appended if specified." Wait, the problem statement says: generate_phrase should return a phrase that is a combination of a random note and a random mode type. If with_note is True, append another random note. So the structure is: [note] [mode] [optional note]. Wait, the problem saysare given. Wait, maybe the problem is from an official source, so likely there's a solution. Therefore, perhaps I need to check my initial steps again. Wait, original equation: (n+1)! + (n+3)! =850 n! Let me try n=5. Left side:6! +8!=720 +40320=41040 Right side:850120=102000. Not equal. n=6: 7! +9! =5040 +362880=367,920 Right side:850720=612,000. Not equal. n=7: already checked. n=8: 9! +11! =362880 +39916800=40279680 Right side:85040320=34,272,000. Not equal. n=9: 10! +12! =3,628,800 +479001600=482,630,400 Right side:850362880=308,448,000. Not equal. n=10: 11! +13! =39916800 +6227020800=6266937600 Right side:8503,628,800=3,084,480,000. Not equal. So as n increases, left side grows much faster than right side. So perhaps there is a solution for n less than 7? Let's check. n=4: 5! +7!=120 +5040=5160 Right side:85024=20,400. Not equal. n=3: 4! +6!=24 +720=744 Right side:8506=5100. Not equal. n=2: 3! +5!=6 +120=126 Right side:8502=1700. Not equal. n=1: 2! +4!=2 +24=26 Rig(n+1)]^n }. Similarly, y = [n/(n+1)]^{n+1}, so y^y = [n/(n+1)]^{(n+1) * y} = [n/(n+1)]^{(n+1) * [n/(n+1)]^{n+1} }. But this seems complicated. Alternatively, take the ratio of x^x / y^y. But according to the previous step: We found that x ln x - y ln y =ce, negation is false here. Therefore, all these checks confirm that the correct interval is \( 0 < a < 1 \). Therefore, the range of values for \( a \) is \( (0, 1) \). **Final Answer** The range of values for \( a \) is \boxed{(0, 1)}. To determine the range of values for the real number \( a \) such that the negation of the proposition "There exists an \( x \in \mathbb{R} \) such that \( x^2 + 2ax + a \leqslant 0 \)" is true, we need to ensure that the quadratic expression \( x^2 + 2ax + a \) is always positive for all \( x \in \mathbb{R} \). For a quadratic equation \( ax^2 + bx + c \) to be always positive, its discriminant must be negative, and the l val board = Math.min(waiting, 4) boarded += board waiting -= board rotations += 1 val currentProfit = boarded * boardingCost - rotations * runningCost if (currentProfit > maxProfit) { maxProfit = currentProfit bestRotation = rotations } else if (currentProfit == maxProfit && rotations < bestRotation) { bestRotation = rotations } if (i >= n && waiting == 0) { break() } } } if (maxProfit > 0) bestRotation else -1 } } Wait, but the loop structure is a bit different here. Let me explain. In this code, the loop is a while(true) loop, and inside, check whether i =n and waiting is zero, break. But perhaps there's a better way. Alternatively, use a loop that checks i 0, and inside the loop, check if i 0) { ... process ... } else { continue = false } } But perhaps the initial approach with breakable is acceptable. Testing the code: In example 1, the loop runs three times. For each rotation, after processing, when i reaches 2 (n=2) and waiting becomes 0, the loop breaks. Yes. Thus, the code should work. Thus, the solution is to simulate each rotation step, keeping track of the current profit, and updating the best rotation whenever a higher profit is found. Finally, return the best rotation if the max profit is positive. To solve this problem, we need to determine the minimum number of rotations required for a Centennial Wheel to maximize profit. The wheel has four gondolas, each accommodating up to four people. The profit is calculated based on the number of customers boarding the gondolas and the cost of each rotation. ### Approach 1. **Simulation of Rotations**: We simulate each rotation step-by-step, keeping track "\"Numerical\"", ",", "\"Numerical\"", ",", "\"Numerical\"", ",", "\"Numerical\"", ",", "\"Numerical\"", ",", "\"Numerical\"", ",", "\"Numerical\"", ",", "\"Numerical\"", ",", "\"Numerical\""}], "}"}], Short[#, 10]& ]], Annotation[#, Short[{"Numerical", "Numerical", "Numerical", "Numerical", "Numerical", "Numerical", "Numerical", "Numerical", "Numerical"}, 10], "Tooltip"]& ], "SummaryItem"]}]}, { RowBox[{ TagBox[ "\"Number of classes: \"", "SummaryItemAnnotati, "fpSub" ~> fpArith FP.bfSub > , "fpMul" ~> fpArith FP.bfMul > , "fpDiv" ~> fpArith FP.bfDiv > > , "fpToRational" ~> > vFinPoly \_ -> pure $ > vFinPoly \_ -> pure $ > VFun \fpv -> > do fp <- fromVFloat' <$> fpv > VRational <$> (eitherToE (FP.floatToRational "fpToRational" fp)) > , "fpFromRational" ~> > vFinPoly \e -> pure $ > vFinPoly \p -> pure $ > VFun \rmv -> pure $ > VFun \rv -> > do rm <- fromVWord =<< rmv > rm' <- eitherToE (FP.fpRound rm) > rat <- fromVRational <$> rv > pure (VFloat (FP.flom ABC, rolling over AB to ABD, then rolling over AD to ACD, then rolling over AC back to ABC. The dot was on ABC, moved to ABD, then stayed on ABC as it was rolled over. Wait, no, the dot is on a face; when you roll the tetrahedrostatments valid > typecheck' :: Stmt -> [Procedure] -> [(Var, Type)] -> Bool for an assignment to be valid the type of the expression must be the same as the type of the variable > typecheck' (Asgn v e) _ d > | (typeexp e d) == (typelookup v d) = True > | otherwise = error "Type mismatch error with Asgn statment" an if statment is valid if the expression will evaluate to a boolean and all the statments in both the true and false branches are valid. > typecheck' (If e t f) p d > | ((typeexp e d) == Boolean && (typecheckstmts t p d) && (typecheckstmts f p d)) = True > | otherwise = error "Type mi> Lam tyvar (wrapper e)) -- Data types with a single constructor, which has a single arg -- This includes types like Ptr and ForeignPtr | Just (tycon, tycon_arg_tys, data_con, data_con_arg_tys) <- splitDataProductType_maybe result_ty, dataConSourceArity data_con == 1 = do dflags <- getDynFlags let (unwrapped_res_ty : _) = data_con_arg_tys narrow_wrapper = maybeNarrow dflags tycon (maybe_ty, wrapper) <- resultWrapper unwrapped_res_ty return (maybe_ty, \e -> mkApps (Var (dataConWrapId data_con)) (map Type tycon_arg_tys ++ [wrapper (narrow_wrapper e)])) | otherwise = pprPanic "resultWrapper" (ppr result_ty) where maybe_tc_app = splitTyConApp_maybe result_ty -- When the result of a foreign call is smaller than the word size, we -- need to sign- or zero-extend the result up to the word size. The C -- standard appears to say that this is the responsibility of the -- caller, not the callee. maybeNarrow :: DynFlags -> TyCon -> (CoreExpr -> CoreExpr) maybeNarrlet me check that again. If 12 + 5 = 17, then 20 - 17 = 3. Yeah, that seems right. So Joe must have found 3 eggs in the town hall garden. But hold on, let me make sure I didn't misinterpret the problem. The problem says "some eggs in the town haule system imports/exports \begin{code} data ImpExpSubSpec = ImpExpAbs | ImpExpAll | ImpExpList [ RdrName ] mkModuleImpExp :: RdrName -> ImpExpSubSpec -> IE RdrName mkModuleImpExp name subs = case subs of ImpExpAbs | isVarNameSpace (rdrNameSpace name) -> IEVar name | otherwise -> IEThingAbs nameT ImpExpAll -> IEThingAll nameT ImpExpList xs -> IEThingWith nameT xs where nameT = setRdrNameSpace name tcClsName mkTypeImpExp :: Located RdrName -> P (Located RdrName) mkTypeImpExp name = do allowed <- extension explicitNamespacesEnabled if allowed then return (fmap (`setRdrNameSpace` tcClsName) name) else parseErrorSDoc (getLoc name) (text "Illegal keyword 'type' (use ExplicitNamespaces to enable)") \end{code} ----------------------------------------------------------------------------- -- Misc utils \begin{code} parseErrorSDoc :: SrcSpan -> SDoc -> P a parseErrorSDoc span s = failSpanMsgP span s \end{code} % % InfoFinder.lhs % \input{preamble} \subsection{} \begin{code} {-# LANGUAGE TemplateHaskell #-} {-# LANGUAGE QuasiQuotes #-} {-# LANGUAGE OverloadedStrings #-} {-# LANGUAGE TypeFamilies #-} {-# LANGUAGE GADTs #-} {-# LANGUAGE FlexibleInstances #-} {tatore durante la narrazione musicale.Oljeboringen som i sin tid ble startet uten lokalbefolkningens velsignelse foregr fortsatt over hodet p Ogoniene som befolker deltaet. Miljkampen, og i srdeleshet kampen mot miljforkjemperne har vrt brutal og ndels. Det har blitt gjort enkelte forsk p f oljeselskapene til betale erstatning og til tvinge dem til omfattende opprydning etter egen forurensning. S langt uten nevneverdige resultat. Hsten 2016 forskte en gruppe ogonier sakske oljeselskapet Shell i en britisk domstol, men saken ble til slutt avvist og m fres i Nigeria i stedet. Dermed er utfallet gitt, mener mange miljaktivister. De hevder at korrupsjonen er s stor at oljegiganten garantert vil vinne i en nigeriansk rettssal.Cross is from Oswego. He has been a member of the General Assembly since 1993, and spent a decade ], -- inv > outports = [(0, zeroS, False, 0, True, 2)] -- or > }, > PS { pid = n+4, -- lower and ----------------------------------- > compType = And2, > pathDepth = 1, > inports = [(sto, 0, zeroS), -- sto input > (n, 0, zeroS)], -- x input > outports = [(0, zeroS, False, 0, True, 1)] -- or > }, > PS { pid = n+5, -- or ------------------------------------------ > compType = Or2, > pathDepth = 3, > inports = [(n+3, 0, zeroS), -- upper and > (n+4, 0, zeroS)], -- lower and > outports = [(0, zeroS, True, 4, False, 0)] -- dff > }, > PS { pid = n+6, -- output y ------------------------------------ > compType = Outp, > pathDepth = 4, > inports = [(n+1, 0, zeroS)], -- dff > outports = [] > } > ] -------------------------------------------------------------------------------- To run (with ghc) for a (8 bit register) circuit over 1000 cycles % ghc -o circ_sim circ_sim.lhs % circ_sim 8 1000 > main :: IO () > mainong all pair products. Since the total product of all pairs is 1, we have: \[ \prod_{k=1}^{1009} P_k = 1 \] Since each \( P_k \geq P_{1009} \), then: \[ (P_{1009})^{1009} \leq \prod_{k=1}^{1009} P_k = 1 \] Taking the 1009th root: \[ P_{1009} \leq 1 \] Which is exactly \( x_{1009} x_{1010} \leq 1 \), as required. Therefore, the key steps are: 1. Recognize that the given inequality \( x_{n+1}^2 \leq x_n x_{n+2} \) implies the sequence of ratios \( r_n = \frac{x_{n+1}}{x_n} \) is non-decreasing. 2. Pair the terms from the outside inward and show each outer pair's product is at least the next inner pair's product using the non-decreasing ratios. 3. Conclude that the middle pair \( x_{1009} x_{1010} \) is the smallest, and since the product of all pairs is 1, it must be 1. This completes the proof. To prove that \( x_{1009} x_{1010ut let me verify with small cases. Suppose we have a smaller set, say {1,2}, and find M. Then total partitions are 2^2 -2 =2. Now constraints: For each partition (A,B), |A| not in A and |B| not in B. Let's see. Possible partitions: 1. A={1}, B={2}. Then |A|=1, which is in A. So invalid. 2. A={2}, B={1}. Then |A|=1, which is not in A (since A={2}), and |B|=1, which is not in B (B={1}). Wait, |B|=1, but 1 is in B. So this is invalid. Wait, no: |B|=1, and 1 is in B, so the constraint is that |B| not in B. Here |B|=1 is in B, so invalid. So both partitions are invalid? Then M=0? But that contradicts. Wait, but maybe in this small case, there are no valid partitions? Let's check: If n=2, possible a=1 and b=1. Then for a=1, we need 1 A aof equal length, and all angles are equal. If we extend a side, the extension does not intersect any other side because in a regular pentagon, the sides are arranged such that their extensions would form the star pentagon, but the original sides themselves don't lie on the extensions. Wait, actually, in a regular convex pentagon, if you extend one side, does it contain another sideEATURES: Excellent for Welding, Drilling, Machining, Shearing and Forming. USES: Base Plates, Gussets, Liners, Etc. The process to create Hot Rolled Steel creates a more ductile steel with a rough surface finish and is sometimes more difficult to machine.Buy Hot Roll Steel Sheet Plate OnlineBuy hot-roll-steel-sheet-plate and other metal-working products online at OnlineMetals hot rolled s690q steel plate.Buy Hot Roll Steel Sheet Plate OnlineBuy hot-roll-steel-sheet-plate and other metal-working products online at OnlineMetals hot rolled s690q steel plate.No ltimo livro da srie Shadow Falls ao Anoitecer, Della agora uma investigadora paranormal e precisa desvendar um caso de assassinato onde o suspeito seu prprio pai, e vai precisar da ajuda de Chase, com quem ela tem uma ligao sangunea aps seu Renascimento. A busca por respostas vai levar Della a se envolver com as gangues sobrenaturais mais sombrias e perigosas e obrig-la a investigar os segredos i StyleBox["21", {}, StripOnInput -> False]], Annotation[#, Style[21, {}], "Tooltip"]& ], TagBox[ TooltipBox[ TagBox[ DynamicBox[{ FEPrivate`If[ CurrentValue["MouseOver"], EdgeForm[{ GrayLevel[0.5], AbsoluteThickness[1.5], Opacity[0.66]}], {}, {}], RectangleBox[{5448.5, 0}, {5468.5, 13.}, "RoundingRadius" -> 0]}, ImageSnce depth is even). So no. But for the node 45 (depth1, val45 divisible by 15). But if it's a leaf, then has_leaf_in_subtree is True (since the node itself is a leaf). So it would be added to sum2. Wait, but the node is a leaf. But the node is at depth1 (odd), and divisible by 15. So sum2 +=45. Yes. So in this case, sum2 would be 30 (from node30) plus 45 (from node45) 75. So the code needs to handle these cases. Now, code implementation. But first, the compute_has_leaf function. In the code for compute_has_leaf, when a node is processed (in the stack with processed=True), it checks if it's a leaf. If yes, current_has is True. Else, current_has is left_has or right_has. But in the code, for a node's children, when they are None, has_leaf.get(None, False) is False. So for a node that is a leaf (left and right are None), is_leaf is True current_has is True. For a node with children, current_has is True if either child has_leaf is True. So the code should work. Now, testigle of incident linearly polarized light, the polarization is changed, and, as the result, a decrease in the efficiency of the polarizing conversion results. Furthermore, it was found that the efficiency of the polarizing conversion was decreased if two light control elements were arranged so as to intersect at right angles.The brain freeze is mine...the crazy dots are the result of my doodling. Too much thinking is bad for me, it's given me brain freeze!!! It's a bit like eating real, really cold ice cream, your brain hurts!! I've been trying to workout installing WordPress and using it as CMS. Not that I'm doing it, Cindy Reilly is patiently putting up with me and my constant indecision. First one thing and then the next, the idea being that I setup a website with a gallery etc. I have done a website for my Family History work using DreamWeaver but it was a while ago and I'm struggling to get back into things. Today was my worst day yet, reading too much, swaying one way and then the other until my little brain was doing triple flips!!! Warning...brain freeze!! So, I have stopped reading for now..tomorrow is an other day. The news from our house is that Charlie is back in disgrace. He has naughty down to a fine art and he adds to his talents in this field all the time.....unfortunately!! Over the weekend he was all leaded up for his walk, sitting at the door quite excited, but then again dogs always are when a walk is in the offing. DH got him to sit, opened the door....and Charlie put his now famous Houdini move into action and DH was left with a lead...no dog....just a limp lead. Well, the yell told me something was not quite right...to put it mildly. So, it was all hands to action stations. My son did not appreciate being hauled out of the shower...yet again!! Charlie always seems to escape when Chris is in playing water babies. It was an other chase me Charlie...literally!! Eventually, after many a heart s ) m_x - s r (1 - cos ) ] / (r sin ) = s This looks very messy, but maybe expand and see if terms cancel. Lets denote for simplicity: Lets set u = 1 - cos . Then, cos = 1 - u, sin = 1 - (1 - u)^2 = 2u - u. But maybe this substitution wont help much. Alternatively, multiply both sides by r sin to eliminate denominators: r sin (m_x - s)^2 + [ (s + r - r cos ) m_x - s r (1 - cos ) ] = s r sin This is a quadratic equation in m_x. Solving this would give coordinates for M, but it's quite involved. Maybe there's a pattern or symmetry. Alternatively, since the problem is ratio-based and must hold for any position of A on C, maybe use similarity or homothety. Given that the ratio PE/PF = ME/NF is to be proven, and ME and NF are segments along the tangents from M and N to E and F. Alternatively, consider that triangles PME and PNF might be similar. If that's the case, then the ratio of sides would be equal. To check similarity, we'd need corresponding angles to be equal. For example, angle at P: angle EPM vs angle FPN. If these are equal, dgy -- to carry ont. Mind you, the staging restrictions mean we won't -- actually run f, but it still seems wrong. And, more concretely, -- see Trac #5358 for an example that fell over when trying to -- reify a function with a "?" kind in it. (These don't occur -- in type-correct programs. ; failIfErrsM -- Desugar ; ds_expr <- initDsTc (dsLExpr expr) -- Compile and link it; might fail if linking fails ; hsc_env <- getTopEnv ; src_span <- getSrcSpanM ; traceTc "About to run (desugared)" (ppr ds_expr) ; either_hval <- tryM $ liftIO $ HscMain.hscCompileCoreExpr hsc_env src_span ds_expr ; case either_hval of { Left exn -> fail_with_exn "compile and link" exn ; Right hval -> do { -- Coerce it to Q t, and run it -- Running might fail if it throws an exception of any kind (hence tryAllM) -- including, say, a pattern-match exception in the code we are running -- -- We also do the TH -> HS syntax conversion inside the same -- exception-cacthing thing so that if there are any lurking -- exceptions in the data structure returned by hval, we'll -- encounter them inside the try -- -- See Note [Exceptions in TH] let expr_span = getLoc expr ; either_tval <- tryAllM $ setSrcSpan expr_span $ -- Set the span so that qLocation can -- see where this splice is do { mb_result <- run_and_convert expr_span (unsafeCoerce# hvAOsDomSh2Stl+l0mzZuonEdKTkXCkv3ruQRZJRrlqk4 Tzv2yd1csY1UtESjkUqps4zTx0BMPfDdp/nSTASfZypawzJ6hZQVf+CdD+a9 ZwmGg+O05XzLKLSEXeARno2O2RYgNaQ3bj4yMkJnb0nMf1hZiWx2AAAAeQ7c HGQCIkoe6E7uW1nPWev1C5aURElZl9ycZ5EbjUZS8oi+A5INWSRXUF9YvK77 d5Y5vdZ3xznY1VqwVMwulwfcFf6IIcIeujHOiWOt22hjnjxwXExFz+C4s/I+ pnJLQJ6Trri5zWbjbHZeSrVak8mUyd9ZAAAAIAXAzUF6EVbefNSvauj/Ox35 uLZghbzWekQx1xQVkZKzj/N0XZAufJ4p0mG1SnQruzoZIck86gsZz1Tfy/Wl jy56qf/j+W8Mt07jyufzXxsAOU+64uaiBNzhQ3ye5wlKdEpH0BwAAACAm4PU w93JfYazt17cSUre+h3h42pVhMi4VBbs8SfK6uu3t7e3j4yMiBA5Yotp4stL /XUri4uLd3EiOueQawtWNPVej2+F3EqM1kB6zs3F4lgJOT6L+ff2/ApiDkDs pCVufuXylXVrVktuXqrVjlnmlnUDAAAApB0ymk9v3EjsdFq4OUgpNrv/3V5S 8l88UaMcJedHNEVFNHQ8aTxDYzn77dt+ZPymFRLnwa7WQJxLTTJOZ6QL/1nB Tc3irohO63SP36DVlmg0B975IGK/cmXolHjmZ2+vWSY2g8QcabEAxEi64uZc Ao4XOp/Qu2NGEgAAgCyCBp9Oy8UDdWUh7YDnD9wcJJtgCayPrvsaXjv+rcdI yaPNH5ei5IvV6tqamt7e3uAscthWxmB8fS9JNLcyD/RBW17/4o9vuabjPinx FHL6hDiGelY92ypWNZc/t3h5oPh5j2XS55mKbxsAyFtSHzc3mUzycz6d6mHl AAAAsgWvc+LycN/unS9w31t28wSuH24Okodn3O4bHr341k9/8UQNJ64r1HZb UlhYtWHDvpYWo9FIn0aejpzuPQBhTE+QnjftaNTXbaXzkr75JJ2R5l9XnO/A xPdHD74WN3AAmCwaiting. Since you love research and statistics, poll a sample size and do a study. There is research on this already. Bleske-Rechek, et. al took 88 pairs of "friends" and had each fill out a confidential survey on attraction. These "friendships" averaged about 2 years in length. The authors consistently found that "men reported more attraction to their female friends than women did to their male friends." The study even looked at men and women who were in a relationship with another person. Again, men were still attracted to their female "friend" and repeatedly the female "friend" was unaware of the attraction. As the authors point out men had an, "attraction to (and desire to date) their friend regardless of their own curreprovided uses commas to separate each section, without using colons or the word "description". Also, the user wants one of the prompts to be a realistic photographic style with lens details. The rest can vary in style, maybe illustration, 3D, etc. Starting with the first prompt. The user mentioned varying details, environment, composition, atmosphere, and realization. For the first one, maybe focus on the flat illustration style. Include details like the teapot's design (maybe porcelain with plum blossoms), the stove's ironwork, steam from the tea. The environment could be a traditional Chinese room with wooden furniture. Composition: teapot and cup in the foreground, stove glowing. Mood: serene, cozy. Style: flat illustration with clean lines. Then ar 1:1 and --v 5. Second prompt: perhaps a 3D render. Add more environmental details like a window showing snow outside, maybe some calligraphy scrolls on the wall. The teapot could be clay with a bamboo handle. Composition might include a tray with snacks. Mood: nostalgic, harmonious. Use --niji for a226ae"], Cell[95962, 2705, 125, 1, 70, 55, 0, "StyleData", "RemarkLabel", "NumberedHeadsPrintout",ExpressionUUID->"e020edd3-098c-4f73-86a8-72098ffea1dc"] }, Closed]], Cell[CellGroupData[{ Cell[96124, 2711, 469, 11, 70, 37, 0, "StyleData", "RemarkContinuation", "All",ExpressionUUID->"7975f264-5663-4c0d-b016-24bbea46ab62", CounterIncrements->"RemarkContinuation"], Cell[96596, 2724, 109, 0, 70, 54, 0, "StyleData", "RemarkContinuation", "NumberedHeads",ExpressionUUID->"d4f99331-c085-4405-a0af-b226e6a8b1d4"], Cell[96708, 2726, 157, 2, 70, 53, 0, "StyleData", "RemarkContinuation", "Presentation",Expressioick us into thinking that the number changes, but actually it's the same. So the answer is 35. Wait, maybe some people might think that 10 left and 10 came, so they think it's a different number, but actually, since the number leaving and coming is the same, the total remains the same. Alternatively, maybe there's a different interpretation where the 10 new students are in addition to replacing the ones who left? Wait, that would mean the total number increases, but the problem states that at the end, there are 35 students. Hmm. Wait, no, the problem says 10 left and 10 new students came. So if 10 left and 10 came, the total number is the same. But let me think of an example. Suppose there were 35 students at the start. Then 10 left, so 25 remain. Then 10 new students came, making it 35 again. So that works. Alternatively, suppose there were 30 students at the start. Then 10 left, so 20 remain. Then 10 new students came, making it 30. But the problem Building blocks ep... 3m0s WIAA Division 4 state baseball 2m10s WIAA Division 1 state baseball 2m38s One Dead in Submerged SUV 2m3s Divers investigate submerged vehicle... 1m4s Vehicle submerged near Wausau 1m31s Police captain discusses Schofield s... 0m54s Police captain discusses Schofield s... 0m54s Strawberry Picking In Wisconsin Rapi... 1m51s Everest Metro captain describes call... 0m50s Woodchucks baseball 1m24s Optum pro cyclists talk about biking... 1m33s Phoebe, the therapy dog 0m38s 3 on 3 Tournament 0m59s Mosinee rules Division 2 softball 0m34s WIAA D1 girls soccer sectional final... 1m36s Raptors in the Gardens 0m43s Weston Duplex Fire 1m14s Summer Kick Off Event at Wausau's Oa... 0m37s Walters Sentencing 1m47s State softball: Mosinee rolls into c... 1m51s WIAA Division 1 sectional semifinal 2m6s Talking to your child and developing... 3m24s WIAA state softball preview 1m44s WIAA sectional soccer preview 1m33s Division 1 baseball sectional semis 2m9s Winnebago County clerk talks marriag... 1m57s WIAA state track meet - Day 2 2m1s Marathon boys finish second at state... 1m59s Mr. Buch Goes out with a Bang! 3m0s State track: Cihlar takes bronze 1m6s State track: D3 highlights 1m9s WIAA state track preview 1m54s Athens baseball heads to sectionals 1m5s Company adds tornado-resistant rooms... 2m9s Wausau students present, demonstrate... 1m6s How to find quality child care: Buil... 3m26s Former heroin addict Doug Darby prea... 2m48s Everest students compete for cash, ... 1m55s Oconto-Marathon softball 1m4s Wausau attorney Pat Crooks retires a... 0m53s Thorp-Athens softball 1m47s Prep softball: Mosinee tops Medford 1m51s Business leaders react to city plan ... 1m15s Lt. Michael Juedes talks about polic... 1m31s Woodchucks will open season Friday i... 2m12s Medical College od for E. The direction vector from A to M is: x = (1 - cos )/(5 + 4 cos ) - (-1 + cos ) = (1 - cos + (5 + 4 cos )(-1 + cos )) / (5 + 4 cos ) Wait, compute numerator: = (1 - cos ) + (-1 + cos )(5 + 4 cos ) = 1 - cos -5 -4 cos +5 cos +4 cos = (1 -5) + (-cos -4 cos +5 cos ) +4 cos = -4 + 0 cos +4 cos = 4(cos -1 ) = 4(- sin ) Thus, x = -4 sin / (5 + 4 cos ) Similarly, y = -3 sin /(5 + 4 cos ) - sin = (-3 sin - sin (5 + 4 cos )) / (5 + 4 cos ) = (-3 sin -5 sin -4 sin cos ) / (5 + 4 cos ) = (-8 sin -4 sin cos ) / (5 + 4 cos ) = -4 sin (2 + cos ) / (5 + 4 cos ) So, the parametric equations for line AM ax]^4 + 5 (I Sin[x])^13 Cos[x]^2 + 5 (I Sin[x])^2 Cos[x]^13) + Exp[4 I y] (170 (I Sin[x])^4 Cos[x]^11 + 170 (I Sin[x])^11 Cos[x]^4 banner of climate change, it's in response to how the seasons have been.Groove Music/Movies & TVpspmaingroovemusicmoviestvmoduleSummaryGroove Music lets you easily play your music collection, make playlists, buy music and stream custom radio stations. Microsoft Movies & TV allows you to play your video collection, and rent or buy movies and TV episodes. Microsoft will use data about the content you play in order to help you discover content that may interest you.Full textGroove Music lets you easily play your music collection, make playlists, buy music and stream custom radio stations. Microsoft Movies & TV allows you to play your video collection, and rent or buy movies and TV episodes. These services were formerly offered as Xbox Music and Video. To help you discover content that may interest you, Microsoft will collect data about what content you play, the length of play, and the rating you give it. If you enable Cortana on your device, Microsoft will collect and use data related to the music you play via Groove Music to provide personalized experiences and relevant suggestions. To enrich your experience when playing content, Groove Music and Movies & TV will display related information about the content you play and the content in your music and video libraries, such as the album title, cover art, song or video title, and other information, where available. To provide this information, Groove Music and Movies & TV send an infork/QcT_8JETczs/s1600/noprob.png', '=D' ,'http://1.bp.blogspot.com/-DB0TS3jw6n8/VJNbgab1A_I/AAAAAAAAH7I/4Cr8aYd6AmU/s1600/happy.png', ':p' ,'https://i0.wp.com/i.imgur.com/nL704UD.png', '|o|' ,'http://3.bp.blogspot.com/-i5w-i1jFe0U/VJNbaQxedgI/AAAAAAAAH5o/z7LK9qxEebs/s1600/clap.png', '@@,' ,'http://2.bp.blogspot.com/-lbZ9iDF66F8/VJNbqSG8yNI/AAAAAAAAH9A/bnAQmQrrrZo/s1600/surprise.png', ' ;)' ,'http://1.bp.blogspot.com/-Q--_b4-u1ZY/VJNbrI3A3LI/AAAAAAAAH8w/xshqjz4f3cs/s1600/trope.png', ':-bd' ,'http://3.bp.blogspot.com/-_lM3w2ZD7K8/VJNbiIf2PII/AAAAAAAAH7g/U5qLOVDZg8c/s1600/like.png', ':-d' ,'http://4.bp.blogspot.com/-EdvYpWDdZPI/VJNbc2J75FI/AAAAAAAAH6I/kcpuLO7TXFg/s1600/dislike.png', ' :p' ,'http://4.bp.blogspot.com/-Y2KF1cqsEiQ/VJNbolnNw1I/AAAAAAAAH8Y/mzpdmmt9lp4/s1600/sigh.png', ':ng' ,'http://2.bp.blogspot.com/-A_W5lI-_J8I/VJNbi2oXwjI/AAAAAAAAH7c/wlxM7CETbhI/s1600/love.png', ]; //Config Force tag list, define all in lower case Force_Tag = [ '[pre]','Here is an other approach.You know of that 14 year old who produced Bubble Ball thatx])^7 Cos[x]^9 + 152 (I Sin[x])^8 Cos[x]^8 + 9 (I Sin[x])^3 Cos[x]^13 + 9 (I Sin[x])^13 Cos[x]^3 + 2 (I Sin[x])^2 Cos[x]^14 + 2 (I Sin[x])^14 Cos[x]^2) + Exp[-7 I y] (342 (I Sin[x])^10 Cos[x]^6 + 342 (I Sin[x])^6 Cos[x]^10 + 458 (I Sin[x])^8 Cos[x]^8 + 412 (I Sin[x])^7 Cos[x]^9 + 412 (I Sin[x])^9 Cos[x]^7 + 216 (I Sin[x])^11 Cos[x]^5 + 216 (I Sin[x])^5 Cos[x]^11 + 111 (I Sin[x])^12 Cos[x]^4 + 111 (I Sin[x])^4 Cos[x]^12 + 42 (I Sin[x])^13 Cos[x]^3 + 42 (I Sin[x])^3 Cos[x]^13 + 11 (I Sin[x])^2 Cos[x]^14 + 11 (I Sin[x])^14 Cos[x]^2 + 2 (I Sin[x])^1 Cos[x]^15 + 2 (I Sin[x])^15 Cos[x]^1) + Exp[-5 I y] (1049 (I Sin[x])^9 Cos[x]^7 + 1049 (I Sin[x])^7 Cos[x]^9 + 412 (I Sin[x])^5 Cos[x]^11 + 412 (I Sin[x])^11 Cos[x]^5 + 747 (I Sin[x])^10 Cos[x]^6 + 747 (I Sin[x])^6 Cos[x]^10 + 1150 (I Sin[x])^8 Cos[x]^8 + 50 (I Sin[x])^13 Cos[x]^3 + 50 (I Sin[x])^3 Cos[x]^13 + 161 (I Sin[x])^12 Cos[x]^4 + 161 (I Sin[x])^4 Cos[x]^12 + 8 (I Sin[x])^2 Cos[x]^14 + 8 (I Sin[x])^14 Cos[x]^2 + 1 (I Sin[x])^1 Cos[x]^15 + 1 (I Sin[x])^15 Cos[x]^1) + Exp[-3 I y] (1641 (I Sin[x])^9 Cos[x]^7 + 1641 (I Sin[x])^7 Cos[x]^9 + 1202 (I Sin[x])^6 Cos[x]^10 + 1202 (I Sin[x])^10 Cos[x]^6 + 1772 (I Sin[x])^8 Cos[x]^8 + 736 (I Sin[x])^11 Cos[x]^5 + 736 (I Sin[x])^5 Cos[x]^11 + 357 (I Sin[x])^12 Cos[x]^4 + 357 (I Sin[x])^4 Cos[x]^12 + 136 (I Sin[x])^3 Cos[x]^13 + 136 (I Sin[x])^13 Cos[x]^3 + 39 (I Sin[x])^2 Cos[x]^14 + 39 (I Sin[x])^14 Cos[x]^2 + 7 (I Sin[x])^1 Cos[x]^15 + 7 (I Sin[x])^15 Cos[x]^1 + 1 Cos[x]^16 + 1 (I Sin[x])^16) + Exp[-1 I y] (2728 (I Sin[x])^8 Cos[x]^8 + 1557 (I Sin[x])^6 Cos[x]^10 + 1557 (I Sin[x])^10 Cos[x]^6 + 300 (I Sin[x])^12 Cos[x]^4 + 300 (I Sin[x])^4 Cos[x]^12 + 776 (I Sin[x])^11 Cos[x]^5 + 776 (I Sin[x])^5 Cos[x]^11 + 2349 (I Sin[x])^7 Cos[x]^9 + 2349 (I Sin[x])^9 Cos[x]^7 + 74 (I Sin[) pairs. - **Part (b):** For each element in \(C\), there are three choices: 1. Not in \(B\) (hence not in \(A\)), 2. In \(B\) but not in \(A\), 3. In both \(A\) and \(B\). Since \(A \subseteq B\), these choices are independent for all \(n\) elements, also yielding \(3^n\) pairs. Both conditions result in identical combinatorial structures, hence the same solution.Given a natural number \( a \), let \( S(a) \) represent the sum of its digits (for example, \( S(123) = 1 + 2 + 3 = 6 \) ). If a natural number \( n \) has all distinct digits, and \( S(3n) = 3S(n) \), what is the maximum value of \( n \)? Okay, let's see. I need to find the maximum natural number n where all digits are distinct, and the sum of the digits of 3n is three times the sum of the digits of n. Hmm, that's interesting. Let me break this down step by step. First, let's understand the problem. So, S(n) is the sum of the digits of n, and we need S(3n) = 3S(n). Also, n must have all distinct digits. The goal is to find the largest such n. Let me start with some examples to get a feel for it. Let's take a simple number, like n=12. Then 3n=36. S(n)=1+2=3, S(3n)=3+6=9, which is 3*3=9. So that works. But 12 is small. Let's try a larger number with distinct digits. How about n=123. Then 3n=369. S(n)=1+2+3=6, S(3n)=3+6+9=18. 3*6=18, so that works too. Interesting. Maybe there's a pattern here? Wait, 123, 369. Each digit is tripled, and since there's no carryover when multiplying by 3, the digits just become 3, 6, 9. So the sum triples. But if there's a carryover when multiplying by 3, that could mess up the digit sum. For example, let's try n=13. 3*13=39. S(n)=1+3=4, S(3n)=3+9=12. 3*4=12, that's good. No carryover here. But wait, 13 has digits 1 andf took risks (illegal activities) to gain wealth. But also, he's about appearances, so maybe investing in things that project success, like real estate or luxury brands. His parties were a way to attract attention and build a reputation. So maybe he'd advise investing in personal branding or social media presence? Gatsby also had a fixation on the green light and the past, which might translate to a focus on long-term goals or holding onto investments even if they're not performing immediately. Wait, but Gatsby's story is also a tragedy. His obsession with the past and his inability to let go led tst main dealers. Southgate motor engineering is a example to how all garages should do business.customer for life24.06.16Narsim, Porsche 911had my car remapped stage 1 from 122 bhp - 142 bhp , , , .. ( .. , , ), , , . - . - , . , () . , , , , .2019 ist das Jahr, in dem der europis", RowBox[{"16", "/", "7"}]]}]}], ")"}], " ", RowBox[{"(", RowBox[{"1", "-", RowBox[{ SuperscriptBox["\[ExponentialE]", RowBox[{ RowBox[{"-", "2"}], " ", "\[Pi]", " ", "\[Beta]"}]], " ", SuperscriptBox["y", RowBox[{"20", "/", "7"}]]}]}], ")"}]}], ")"}], "/", RowBox[{"(", RowBox[{ SuperscriptBox[ RowBox[{"(", RowBox[{ SuperscriptBox["\[ExponentialE]", RowBox[{"2", " ", "\[Pi]", " ", "\[Beta]"}]], "-", SuperscriptBox["x1", "2"]}], ")"}], "2"], " ", SuperscriptBox[ RowBox[{"(", RowBox[{ SuperscriptBox["\[ExponentialE]", RowBox[{"4", " ", "\[Pi]", " ", "\[Beta]"}]], "-", SuperscriptBox["x1", "2"]}], ")"}], "2"], " ", SuperscriptBox[ RowBox[{"(", RowBox[{ RowBox[{"-", "1"}], "+", SuperscriptBox["x1", "2"]}], ")"}], "2"], " ", SuperscriptBox[ RowBox[{"(", RowBox[{"x1", "-", "x2"}], ")"}], "2"], " ", SuperscriptBox[ RowBox[{"(", RowBox[{ RowBox[{ RowBox[{"-", SuperscriptBox["\[ExponentialE]", RowBox[{"2", " ", "\[Pi]", " ", "\[Beta]"}]]}], " ", "x1"}], "+", "x2"}], ")"}], "2"], " ", Supe closed loop. When you traverse the entire perimeter (a closed loop), the number of color changes must be even. Because each time you change color, you must change back to return to the starting color. Since you start and end at the same vertex (A), but wait, actually, the perimeter is a cycle, but the corners are fixed. Wait, no. The perimeter is a cycle of edges, but the coloring of the vertices is fixed at A, B, C, D. Wait, s the octal game, and apply the Sprague-Grundy theorem. But I don't know the Grundy numbers for this specific game. Given that I'm stuck, maybe I should look for patterns in smaller boards beyond 4x1 and 4x2. Consider a 4x3 board. Player 1 places in (1,2) (central column). Player 2 must attack 1. Suppose Player 2 places in (2,2), adjacent to (1,2). Player 1 must attack even. They could place in (3,2), adjacent to (2,2), attacking 1 (invalid). Or in (1,1), adjacent to none. So Player 1 places in (1,1). Player 2 must attack odd. They can place in (1,3) (adjacent to none, invalid), (2,1) (adjacent to (1,1)), attacking 1. Player 2 places in (2,1). Player 1 must attack even. They can place in (3,1), adjacent to (2,1), attacking 1 (invalid). Or (4,1), adjacent to none. Player 1 places in (4,1). Player 2 must attack odd. Available moves: (3,1), (4,2), (4,3), (3,2), etc. Placing in (3,1) attacks (4,1) and (2,1)? Wait, no. (3,1) is adjacent to (4,1) below and (2,1) above. So (3,1) is adjacent to (2,1) and (4,1). If Player 2 places there, they attack 2 rooks (even), which is invalid. If they place in (4,2), adjacent to (4,1) (has a rook), attacking 1 (valid). Player 2 places there. Player 1 must attnd 100. Perimeter is 2*(75+100)=350. Total cost is 87.5, so 87.5 divided by 350 is 0.25. All steps check out. So the answer should be 0.25 per meter. Alternatively, maybe the cost was given as 87.5 in some higher denomination? But the question asks for cost per meter, so unless there's a mistake in the problem statement, 0.25 is the answer. Maybe 0.25 dollars, rupees, euros, whatever. Since units aren't specified, just the numerical value. So, I think that's it. The cost per meter is 0.25. **Final Answer** The cost per metre of fencing is \boxed{0.25}. The sides of a rectangular field are in the ratio 3:4. Let the sides be 3x and 4x. The area of the field is given as 7500 square meters. The area can be expressed as: \[ 3x \times 4x = 12x^2 = 7500 \] Solving for \(x\): \[ x^2 = \fra_A +4 and y_B =k x_B +4, perhaps we can express 1/y_A +1/y_B in terms of x_A and x_B. Alternatively, since we know x_A and x_B are roots of quadratic equation (1), perhaps we can use Vieta's formulas to find expressions for y_A + y_B and y_A y_B, which would help in computing 1/y_A +1/y_B. Let me try that. First, compute y_A + y_B: y_A + y_B = (k x_A +4) + (k x_B +4) =k(x_A +x_B) +8 From Vieta's formula, x_A +x_B =8k/(3 -k). So, y_A + y_B =k*(8k)/(3 -k) +8 =8k/(3 -k) +8 = [8k +8(3 -k)]/(3 -k) = [8k +24 -8k]/(3 -k) =24/(3 -k) Similarly, compute y_A y_B: y_A y_B = (k x_A +4)(k x_B +4) =k x_A x_B +4k(x_A +x_B) +16 From Vieta's: x_A x_B = -19/(3 -k) x_A +x_B =8k/(3 -k) Thus, y_A y_B =k*(-19)/(3 -k) +4k*(8k)/(3 -k) +16 = (-19k)/(3 -k) +32k/(3 -k) +16 Combine the first two terms: [(-19k +32k)]/(3 -k) +16 =13k/(3 -k) +16 So, 1/y_A +1/y_B = (y_A + y_B)/(y_A y_B) = [24/(3 -k)] / [13k/(3 -k) +16] Simplify numerator and denominator: Numerator:24/(3 -k) Denominator:13k/(3 -k) +16 = [13k +16(3 -k)]/(3 -k) = [13k +48 -16k]/(3 -k) = [ -3k +48 ]/(3 que ellos son tan diferentes como yo y somos incomprendido por los dems, nunca me he sentido bien entre las personas porque siento que nadie me entiende y ademas tambin sucede que aveces tengo pensamientos con sangre algo as como que hago algo para obtener la sangre de las dems personas, ademas tengo ganas de comer y todo me da muchas ganas de vomitar porque estoy muy lleno y pienso en comida y me da muchas mas ganas de vomitar pero cuando pienso en sangre me da apetito pero no se sacia con comida y si te soy sincero no la he probado ni una sola vez pero no se porque cuando la veo se me hace agua la boca, me da ganas de tomrmela y aunque es mi cerebro el que me dice que no debo tomar sangre, los impulsos de mi cuerpo son lo contrario con solo pensarlo me dan ganas salir a buscarla y de probarla solo que hay algo en mi que me dice que me gustara mas de lo que imagino (y no lo se si este bien). Por otra parte te contare que amo estLet's compute each number modulo 9 first and then sum them. Let's start with 7150. Let's sum its digits: 7 + 1 + 5 + 0 = 13. 13 modulo 9 is 13 - 9 = 4. 7152: 7 + 1 + 5 + 2 = 15. 15 modulo 9 is 15 - 9 = 6, and 6 - 9 = negative, so remainder 6. Wait, but 15 divided by 9 is 1*9 = 9, remainder 6. Correct. 7154: 7 + 1 + 5 + 4 = 17. 17 modulo 9: 17 - 9 = 8, so remainder 8. 7156: 7 + 1 + 5 + 6 = 19. 19 - 9 = 10, 10 - 9 = 1, so remainder 1. 7158: 7 + 1 + 5 + 8 = 21. 21 modulo 9: 21 - 18 = 3, so remainder 3. Now, add all those remainders: 4 + 6 + 8 + 1 + 3. Let's compute step by step: 4 + 6 = 10 10 + 8 = 18 18 + 1 = 19 19 + 3 = 22 Then, 22 modulo 9 is 22 - 18 = 4. So same result, 4. That's consistent with the previous method. So the remainder is 4. Alternatively, I could have noticed that each of these numbers is consecutive even numbers starting from 7150. The numbers are 7150, 7152, 7154, 7156, 7158. That's an arithmetic sequence with common difference 2. So there are five terms. The sum of an arithmetic sequence is (number of terms)/2 times (first term + last term). So 5/2 * (7150 + 7158). Let me compute that. First, 7150 + 7158 = 7150 + 7158. Let's compute 7000 + 7000 = 14,000. Then 150 + 158 = 308. So total is 14,000 + 308 = 14,308. Then multiply by 5/2: 14,308 * 5 / 2. Let's compute 14,308 / 2 = 7,154. Then 7,154 * 5. 7,000 * 5 = 35,000. 154 * 5 = 770. So total is 35,000 + 770 = 35,770. So that's the same sum as before. Then divide by 9 and find remainder 4. So same result. But since we already confirmed with two methods, maybe we can think of another way. For example, since these numbers form an arithmetic sINANCE REAL ESTATE MARKETS BUSINESS STOCKS ECONOMY EMERGING MARKETS y the cost per kWh. Model E consumes 0.2 kWh per mile. So, for 500 miles, the energy consumed would be 0.2 * 500 = 100 kWh. Then, the electricity cost is $0.12 per kWh, so 100 * 0.12 = $12.00. That seems straightforward. Now, Model H is a hybrid. It has a fuel efficiency of 70 mpg and a battery capacity of 5 kWh. The energy consumption on battery is 0.3 kWh per mile. The dealer says Model H operates on battery for the first 50 miles and then switches to gasoline. So, I need to calculate the cost for the first 50 miles on electricity and the remaining 450 miles on gasoline. First, the electric part: 50 miles at 0.3 kWh per mile. That's 50 * 0.3 = 15 kWh. But wait, the battery capacity is 5 kWh. Hmm, that's a problem. If the battery capacity is only 5 kWh, and it consumes 0.3 kWh per mile, then the maximum distance it can go on battery is 5 / 0.3 16.67 miles. But the problem states that Model H operates on battery for the first 50 miles. There's a contradiction here. Wait, maybe the battery capacity is 5 kWh, but when it uses the battery, it can only go 5 / 0.3 16.67 miles, but the problem says it operates on battery for the first 50 miles. That doesn't add up. Maybe there's a misunderstanding. Wait, let me check the problem again. It says Model H has a fuel efficiency of 70 mpg and a battery capacity of 5 kWh with an energy consumption rate of 0.3 kWh per mile when running on battery. So, if the battery capacity is 5 kWh, the maximum electric range is 5 / 0.3 16.67 miles. But the problem states that Model H operates on battery for the first 50 miles. That doesn't make sense unless it's recharging while driving, but hybrid motorcycles typically can't recharge the battery from gasoline while driving like a hybrid car can. Wait, maybe it's a plug-in hybrid? Then maybe it can use the full battery charge first and then switch to gasoline. But the problem says "operates on battery for the first 50 miles". Maybe the battery capacity is enough for 50 miles? Let's check. If the energy consumption on battery is 0.3 kWh/mile, then for 50 miles, it would need 50 * 0.3 = 15 kWh. But the battery capacity is only 5 kWh. That's only 5 / 0.3 16.67 miles. So there's a discrepancy here. Is there a mistake in the problem statement, or am I misinterpreting it? Wait, maybe Model H has both a gasoline engine and an electric motor, and when it's operating on battery, it uses 0.3 kWh per mile, but it can also renti disponibili per la qualificazione alle prossime Paralimpiadi brasiliane.In step 2d, the first microcomputer 20 outputs the data signal for turning on the TV through the output port A of the VCR. The data signal of an infrared ray is combined with the audio signal by the adder 30, and input through the audio line into m) s) = showString m . showString "." . shows s showsPrec _ (UnQual s) = shows s data HsName = HsIdent String | HsSymbol String | HsSpecial String deriving (Eq,Ord) instance Show HsName where showsPrec _ (HsIdent s) = showString s showsPrec _ (HsSymbol s) = showString s showsPrec _ (HsSpecial s) = showString s data HsModule = HsModule Module (Maybe [HsExportSpec]) [HsImportDecl] [HsDecl] deriving Show -- Export/Import Specifications data HsExportSpec = HsEVar HsQName -- variable | HsEAbs HsQName -- T | HsEThingAll HsQName -- T(..) | HsEThingWith HsQName [HsQName] -- T(C_1,...,C_n) | Hssqrt(192a*(1 - s)^2 +48a s)=sqrt(48a[4(1 - s)^2 + s]) = sqrt(48a[4 -8s +5s])=sqrt(48a(5s -8s +4))=sqrt(48)a sqrt(5s -8s +4) Therefore, Area=(1/2)*sqrt(48)a sqrt(5s -8s +4)= (1/2)*4sqrt(3)a sqrt(5s -8s +4)= 2sqrt(3)a sqrt(5s -8s +4). This matches our earlier calculation. Therefore, the minimal area is achieved at s=0.8, giving: sqrt(5*(0.8)^2 -8*(0.8) +4)=sqrt(5*0.64 -6.4 +4)=sqrt(3.2 -6.4 +4)=sqrt(0.8)=sqrt(4/5)=2/sqrt(5) Therefore, Area=2sqrt(3)a*(2/sqrt(5))=4sqrt(3)/sqrt(5)a=4sqrt(15)/5 a. Yes, correct. There RowBox[{"cond", ">", "2"}], ")"}], "&&", RowBox[{"(", RowBox[{"cond", "\[NotEqual]", "5"}], ")"}]}], ")"}], "||", "\[IndentingNewLine]", " ", RowBox[{"(", RowBox[{ RowBox[{"(", RowBox[{"i", "==", " ", "5"}], ")"}], "&&", RowBox[{"(", RowBox[{"cond", "\[NotEqual]", "3"}], ")"}], "&&", RowBox[{"(", RowBox[{"cond", "\[NotEqual]", "5"}], ")"}]}], ")"}]}]}], ";", "\[IndentingNewLine]", "\[IndentingNewLine]", RowBox[{"Print", "[", RowBox[{ "\"\\"", ",", "i", ",", "\"\< cond \>\"", ",", "cond"}], "]"}], ";", "\[IndentingNewLine]", RowBox[{"If", "[", RowBox[{"skip", ",", "\[IndentingNewLine]", RowBox[{ RowBox[{"Continue", "[", "]"}], ";"}]}], "\[IndentingNewLine]", "]"}], ";", "\[IndentingNewLine]", RowBox[{"Print", "[", RowBox[{ "\"\\"", ",", "i", ",", "\"\< cond \>\"", ",", "cond"}], "]"}], ";", "\[IndentingNewLine]", "\[IndentingNewLine]", RowBox[{"Which", "[", "\[IndentingNewLine]", RowBox[{ RowBox[{"cond", "\[Equal]", "1"}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{"map", "=", RowBox[{"{", RowBox[{ RowBox[{"k1", "\[Rule]", "i1"}], ",", RowBox[{"k2", "\[Rule]", "i2"}], ",", RowBox[{"l2", "\[Rule]", "j2"}]}], "}"}]}], ";"}], "\[IndentingNewLine]", ",", "\[IndentingNewLine]", RowBox[{"cond", "\[Equal]", "2"}], "," choice = input("Choose: ") # validate try: idx = int(choice) -1 selecteheir own struggles, not just antagonists. Maybe one is jealous, but another is secretly supportive. I should also consider the tone. The original is a bit whimsical, but mine could be more cyberpunk with gritty elements. Need to make sure the plot points align with the new setting. The invitation to the expo, the protagonist's need to prove her invention, the CEO's interest in her work. The resolution would involve her exposing the corporate corruption, using her tech skills. The device (slipper) could hold evidence instead of just being a token. Maybe the story ends with her starting her own company. Wait, does that deviate too much? Maybe keep the core message but adapt the elements. Also, character names should reflect the new setting. Cinderella could be a nickname based on her real name, like "Cindy" short for Cynthia. The prince/CEO could have a name that sounds powerful in a tech context, like Elon or something else. Need to avoid clichs. Maybe add a subplot about teamwork or trust. The AI might have a hidden backstory linked to the protagonist's past. I need to outline the main plot points, then see where my original ideas fit. Also, ensure that the characters have arcs. The protagonist starts off doubting herself but gains confidence. The CEO realizes the importance of ethics over profit. The stepmother figure faces consequences for her actions. Maybe include some twists, like the AI betraying the protagonist initially but then helping in the end. Or the stepsister aiding in the final hack. Need to make sure the story is cohesive and the changes make sense. Also, check for plot holes. Why would the CEO need a tech partner? Maybe his company is failing, and her invention can save it. The stepmother corporation wants to steal it. The ball/expo is the deadline to present the invention. Midnight could be ual puzzles. But 30 is not twice 18. Therefore, this is impossible. Therefore, there is a contradiction. Therefore, the answer is that it's impossible? But the problem says "Mrs. Wilson wants to group into sets", so maybe the answer is that she cannot create any sets? But that seems unlikely. Wait, maybe the problem allows for not using all puzzles? But the problem says "with no puzzles left over". So all puzzles must be used. Hmm. Wait, maybe there's a misunderstanding in the ratio. Maybe the ratio is logic to (visual + word)? But the problem states "the ratio of logic puzzles to visual puzzles should be 2:1". So no. Alternatively, perhaps the ratio is within each set, logic to visual is 2:1, but word puzzles can vary. But even so, in order for all puzzles to be used up, the totals must satisfy that 30 / (number of sets) must be twice 18 / (number of sets). So 30 = 2*18? Which is 30=36, which is impossible. Therefore, contradiction. Therefore, unless there's a different interpretation, the problem is impossible. But the problem is given as a question, so there must be a solution. Therefore, perhaps the ratio is not strictly 2:1 per set, but on average across all sets? Wait, but each set has to be identical, so the ratio must hold in each set. Therefore, the total ratio would be the same as the per-set ratio. Since the total ratio is 5:3, which is not 2:1, this is impossible. Therefore, the answer is that she cannot create any sets? But the problem says "the greatest number of sets she can create", implying that the answer is at least 1. Wait, but maybe the problem allows for a non-integer number of puzzles per set? But no, you can't have a fraction of a puzzle. Therefore, each set must have an integer number of each puzzle type. Alternatively, maybe she can adjust the numbers by removing some puzzles to make the totals fit the ratio. But the problem states she wants to group all the puzzles into sets with none left over. So she can't remove any. Therefore, this seems like an impossible problem. But since it's given, I must have missed something. Let me try to check the problem statement again. "Mrs. Wilson, the math teacher, has 30 logic puzzles, 18 visual puzzles, and 12 word puzzles that she wants to group into sets for students who finish their tests early. Mrs. Wilson wants each set to be identical, containing the same combination of logic puzzles, f i). So for each i, during the left pass, if there's a previous occurrence of c, we compute the distance and store it. For the right pass, moving right to left, for each i, the next occurrence is to the right, so the distance is next_pos - i. So during the right pass, we compute that and compare with the existing value from the left pass. But how to handle the cases where there is no occurrence to the left (for the first few elements before any c) or to the right (after the last c)? For example, in the left pass, before the first occurrence of c, the last_pos is initialiR / r. Given R/r = 3, then the ratio is 3. But this contradicts the previous result. So which one is correct? Wait, this is confusing. Let me check again. If the area of the contact triangle is (r^2 s)/R, then the ratio is R/r, which is 3. But other sources suggest different formulas. Let me try to derive the area of the contact triangle. The contact triangle (intouch triangle) has its vertices at the points where the incircle touches the sides of the original triangle. Let's denote the original triangle as KLM, with sides of lengths a, b, c opposite to vertices K, L, M respectively. The touch points divide each side into segments of lengths s - a, s - b, s - c. The coordinates of the touch points can be given in terms of the triangle's sides and angles, but maybe it's easier to use trigonometric relationships. Alternatively, the contact triangle's sides are proportional to the distances between the touch points. Let me compute the length of one side of the contact triangle. For example, the side opposite to the touch point on side a (let's call this point A) would be between the touch points on sides b and c. Let's call these points B and C. The distance between B and C can be found using the law of cosines in triangle BIC, where I is the inradius center. Wait, but maybe that's not helpful. Wait, another approach. The contact triangle is homothetic to the original triangle. The homothety center is the Gergonne point. The ratio of homothety might be related to r/R or something else. But I'm not sure. Alternatively, the area of the contact triangle can be calculated using the formula: Area ABC = Area KLM * ( (r)/(4R) ) But I need to verify this. Let's consider an equilateral triangle where R = 2r. For an equilateral triangle, the inradius r = (a3)/6, and the circumradius R = (a3)/3. So R = 2r. Then if the formula Area ABC = Area KLM * (r)/(4R) = Area KLM * (1/8). But in reality, the area of the contact triangle in an equilateral triangle can be calculated directly. The contact triangle is also an equilateral triangle. The side length of the contact triangle can be found by the distance between two touch points. In an equilateral triangle of side a, the touch points are located at a distance of (a/2 - r) from each vertex? Wait, no. The touch points divide each side into segments of length s - a, but in an equilateral triangle, all sides are equal, so s = (3a)/2. Then each touch point divides a side into (s - a) = (3a/2 - a) = a/2 and similarly for others. So the touch points are each a/2 from the vertices. Wait, so in an equilateral triangle, the contact triangle would have vertices at the midfth-degree polynomial, there are either 0, 2, or 4 nonreal roots. But since the polynomial is of odd degree, it must have at least one real root. So possible cases: 1 real root and 4 nonreal roots, or 3 real roots and 2 nonreal roots, or 5 real roots. But given the equation, maybe we can check if there's an obvious real root? Let me try plratic on the interval [0,1]. Let me think. For a quadratic function g(x) = px + qx + r, its maximum or minimum occurs at x = -q/(2p), provided that p 0. In our case, the derivative f(x) is a quadratic with coefficients: p = 3a, q = 2b, r = c. So the extremum of f(x) occurs at x = - (2b)/(2*3a) = -b/(3a). But this critical point x = -b/(3a) might lie inside the interval [0,1] or outside. If it's inside, then the maximum or minimum of f(x) occurs there; otherwise, the extrema are at the endpoints x=0 or x=1. So, to ensure |f(x)| 1 on [0,1], we need to check the maximum and minimum values of f(x) on this interval and set them to be within [-1, 1]. Therefore, the maximum value of |f(x)| on [0,1] could be either at x=0, x=1, or at the critical point x = -b/(3a) if it's within [0,1]. So, we need to consider both possio differentiate between other being None and other being an empty list. Hmm. Since the problem states that the default is when no ingredients are provided, which is when other is None. So in that case, the code returns the default. For any list passed (including empty), process the ingredients. So, the code would be: if other is None: subres = "Nothing baked" else: if len(other) == 0: subres = "Baked with nothing" else: subres = "Baked with " + ", ".join(other) But perhaps the code for the else part is: subres = "Baked with " + ", ".join(other) if other else "Baked with nothing" But how to handle empty lists. For example, if other is an empty list, then the join would produce an empty string. So "Baked with " + empty string is "Baked with ", which may not be desired. So perhaps in the code, if the list is empty, return "Nothing baked". Wait, but the problem says that the default is only when other is None. So perhaps for other being an empty list, the code should process it as provided ingredients, but since it's empty, return "Baked with nothing". But perhaps the problem expects that even an empty list is considered as ingredients. So in that case, the code would proceed to join the ingredients. For an empty list, the joined string would be empty, leading to "Baked with " + empty string. So the code would be: if other is None: return "Nothing baked" else: return "Baked ible answers depending on the original distribution of the numbers in the categories. Thereforetively. So I think that's the answer. Just to recap: - For item A, using 800g instead of 1kg: 25% profit. - For item B, using 850g: ~17.65% profit. - For item C, using 780g: ~28.21% profit. Therefore, the dealer's percentage profit for each item is 25%, 17.65%, and 28.21% respectively. **Final Answer** The percentage profit for item A is \boxed{25\%}, for item B is \boxed{17.65\%}, and for item C is \boxed{28.21\%}. The dealer sells goods at the cost price but uses different weights for different items. To calculate the percentage profit for each item, we use the formula: \[ \text{Percentage Profit} = \left( \frac{\text{Shortage}}{\text{Actual Weight Used}} \right) \times 100 \] where the shortage is \(1000 \text{ grams} - \text{Actual Weight Used}\). ### For Item A (800 grams): - Shortage = \(1000 - 800 = 200 \text{ grams}\) - Percentage Profit = \(\left( \frac{200}{800} \right) \times 100 = 25\%\) ### For Item B (850 grams): - Shortage = \(1000 - 850 = 150 \text{ grams}\) - Percentage Profit = \(\left( \frac{150}{850} \right) \times 100 \approx 17.65\%\) ### For Item C (780 grams): - Shortage = \(1000 - 780 = 220 \text{ grams}\) - Percentage Profit = \(\left( \frac{220}{780} \right) \times 100 \approx = case Var.idDetails id of PrimOpId _ -> True -- See Note [Primop wrappers] FCallId _ -> True DataConWorkId dc -> isUnboxedTupleCon dc _ -> False isImplicitId :: Id -> Bool -- ^ 'isImplicitId' tells whether an 'Id's info is implied by other -- declarations, so we don't need to put its signature in an interface -- file, even if it's mentioned in some other interface unfolding. isImplicitId id = case Var.idDetails id of FCallId {} -> True ClassOpId {} -> True PrimOpId {} -> True DataConWorkId {} -> True DataConWrapId {} -> True -- These are are implied by their type or class decl; -- remember that all type and class decls appear in the interface files zero? Hmm, good point. Let's consider the cases where a or c is zero. Suppose a = 0. Then the original fraction becomes (0*l + b)/(cl + d) = b/(cl + d). If k divides both b and cl + d, then since a = 0, ad - bc = 0*d - bc = -bc. Since k divides b, then k divides bc, so k divides -bc. Therefore, ad - bc is divisible by k. Similarly, if c = 0, the fraction becomes (al + b)/(0*l + d) = (al + b)/d. If k divides both al + b and d, then since c = 0, ad - bc = a*d - b*0 = ad. Since k divides d, and a is integer, then k divides ad. Hence, ad - bc is divisible by k. So even in the cases where a or c is zero, the conclusion holds. So the previous derivation holds even if a or c is zero, because in those cases, the determinant ad - bc reduces to terms that are multiples of k. Another thing to check: what if a or c is zero and the denominator or numerator becomes zero? For example, if a = 0 and cl + d = 0, but since the fraction is (al + b)/(cl + d) = b/(cl + d). If cl + d = 0, then the denominator is zero, which w 2016 Vliegennaar.nl vergelijkt vliegticket Inverness aanbiedingen o.b.v. vanaf prijzen voor retourvluchten in euro's, inclusief luchthavenbelastingen. Veel luchtvaartmaatschappijen berekenen toeslagen voor bagage, zitplaatsreservering en betalen met creditcard. ANVR-leden berekenen daarnaast per boeking 2,50 voor het calamiteitenfonds. Het totaalbedrag moet duidelijk zijn voor je een boeking bevestigd. Tip: betaal je vliegreis Inverness met iDEAL, dat is vrijwel altijd gratis! Vliegennaar.nl verkoopt zelf geen tickets of reizen. Wij verwijzen je naar de website van bovengenoemde aanbieders voor gedetailleerde informatie over de betaling en voorwaarden.The best methods of such artistic patronage will be a variedindividuals, organizations, corporations, universities, town councils, and others all doing their parts. There is a place for large-scale government funding of the arts, albeit a limited one. Such funding, as Michael J. Lewis points out, is less effective at spotting new trends or anointing new artists, and more helpful in preserving established but endangered forms of art such as classical orchestras or ballet, works validated by the cumulative consensus of time.Tuesday 19 Safar, 1437 | Affiliate | Guest Book | FaceBook | Contact us Download Athan (Azan) Software for Computers Desktop / Laptop JAPAodos los 4 hoteles en HenricoHampton Inn & Suites Richmond/Glenside703 opinionesN.2 de 81 hoteles en Richmond Country Inn & Suites By Cge are [3, -1, -1, -1, -1, -1, -1], then the minimal moves would be calculated differently. Wait, but I think the most common version of this problem, especially with seven boxes, is the one where the coins are 0,1,2,3,4,5,6, leading to the answer 12. Therefore, possibly, the answer expected is 12. But I need to be careful. Let me confirm with another approach. Suppose each box i has c_i coins, and the average is m. The mreferredj, minSizej]`. The answer to the `jth` query is the room number `id` of a room such that: * The room has a size of **at least** `minSizej`, and * `abs(id - preferredj)` is **minimized**, where `abs(x)` is the absolute value of `x`. If there is a **tie** in the absolute difference, then use the room with the **smallest** such `id`. If there is **no such room**, the answer is `-1`. Return _an array_ `answer` _of length_ `k` _where_ `answer[j]` _contains the answer to the_ `jth` _query_. **Example 1:** **Input:** rooms = \[\[2,2\],\[1,2\],\[3,2\]\], queries = \[\[3,1\],\[3,3\],\[5,2\]\] **Output:** \[3,-1,3\] **Explanation:** The answers to the queries are as follows: Query = \[3,1\]: Room number 3 is the closest as abs(3 - 3) = 0, and its size of 2 is at least 1. The answer is 3. Query = \[3,3\]: There are no rooms with a size of at least 3, so the answer is -1. Query = \[5,2\]: Room number 3 is the closest as abs(3 - 5) = 2, and its size of 2 is at least 2. The answer is 3. **Example 2:** **Input:** rooms = \[\[1,4\],\[2,3\],\[3,5\],\[4,1\],\[5,2\]\], queries = \[\[2,3\],\[2,4\],\[2,5\]\] **Output:** \[2,1,3\] **Explanation:** The answers to the queries are as follows: Query = \[2,3\]: Room number 2 is the closest as abs(2 - 2) = 0, and its size of 3 is at least 3. The answer is 2. Query = \[2,4\]: Room numbers 1 and 3 both have sizes of at least 4. The answer is 1 since it is smaller. Query = \[2,5\]: Room number 3 is the only room with a size of at least 5. The answer is 3. omBos MedicalCarewaysCemexCommandeurCymeqDe WerkfietsDoge CollectionDoove ProductsDoubleperforman..DynaProductsEasylivingEbloEerderMetaalElanEm-cartEm-cart sportEscape Mobility..EtacEurevaEvrew-EdnavExactdynamicsFitformFreewielGazelleHandicareHandicare Bathr..Huka ProductsIncaInvacareJoernsJoyincareKerstenLagooniLife & MobilityLigtvoetLikoLopitalMedibolMedifixMedimcareMedimpexMeditasMedross RehabMercadoMeyraMeyra OrtopediaMobielgarantMobilityProduct..NenkoNijlandObesirent-Obesi..OnderwaterfietsOtto BockOuderenwinkelPeereboomPermobilPestman importPOM NijmegenPR-SellaPremisPreston AbilityPride Healthcar..QuattronR82Rev ProductenRevatakRevatoRoamRoegintRollickRollzRVS NederlandSinner ligfiets..SowecareSpartaSpringproductie..StabagSummitSunrise MedicalThuasneTilcentrumTiltechniekTNSTotaalfietsenTR-careUnited careUsva BikesV-fietsVan OsVan RaamVeldink4kidsVicairVilleriusVisser IDVita BalansVitaflexVoelkerWellcoWelzorgWelzorg actiefWelzorg-autoaan..WSPYouQQuaker is the nation's leading hot cereal company and breakfast authority and has been nourishing healthy families with delicious products made with whole grains such as Quaker Rea>) (runSDoc d1 sty) (runSDoc d2 sty) (<+>) d1 d2 = SDoc $ \sty -> (Pretty.<+>) (runSDoc d1 sty) (runSDoc d2 sty) ($$) d1 d2 = SDoc $ \sty -> (Pretty.$$) (runSDoc d1 sty) (runSDoc d2 sty) ($+$) d1 d2 = SDoc $ \sty -> (Pretty.$+$) (runSDoc d1 sty) (runSDoc d2 sty) hcat :: [SDoc] -> SDoc -- ^ Concatenate 'SDoc' horizontally hsep :: [SDoc] -> SDoc -- ^ Concatenate 'SDoc' horizontally with a space between each one vcat :: [SDoc] -> SDoc -- ^ Concatenate 'SDoc' vertically with dovetailing sep :: [SDoc] -> SDoc -- ^ Separatnk it's a polite thing to ask right off the bat, much like religion, income etc. However most of my white friends after knowing me for a while have inquired, and I don't mind answering. Because I truly do feel like they are just curious, and it's not so much trying to size me up.May 22, 2013 09:54 AM | 1832 views | 0 | 14 | | Fulton Countys water and sewer billing office is warning customers of a potential scam. Some county customers reported being contacted by individuals pretending to be calling on behalf of the billing office. The callers have requested customers credit card numbers, claiming that they need immediate payment in order to avoid service disconnection. The county billing office does not accept any form of payment by telephone. In addition, the county does not contact its water and sewer customers in the evening. The county asks any citizen who receives such a call to notify authorities immediately. Any customer who has questions about his or her account may contact the billing office at (404) 612-6830 or visit https://waterbilling .fulton countyga.gov/.@Mark: I'd like to expand more on a couple of points I made:1. Smashwords is a bookstore. And when I'm in a brick-and-mortar store, browsing through the science fiction section, I don't expect to see titles like F*CKED BY AN ALIEN or DADDY'S ROBOT COCK IN MY ASS. They might be in the store, but they'll be in a separate physical location, where someone who is looking for that can find them and someone who is not won't be exposed to them.What about adding a single additional filter tag "Strong Sexual Content"? That way, books that involve other mature non-sexual content don't have to get filtered out in searches? There is sex in some of my books, but they are in no way erotica. Think of it like rated-R sex versus rated-XXX."Suitable for 18+" is not an effective filter. There are plenty of non-erotic tales which an author might feel were inappropriate for children under 18, which would then be filtered out using the current system.2. My experi5 EF10.46 cm, which is consistent with the Asymptote code's F being 7 units from D at 100, but 7 vs 9 is a discrepancy. Wait, problem states DF=9, but code says F is 7 units from D. So another inconsistency. Given these numerous inconsistencies, it's possible that the Asymptote code is incorrect, and the problem should be solved purely based on the description: two triangles, each with sides and a 100 angle, find CD. But without knowing how the triangles are related, it's impossible. However, since the problem is accompanied by the Asymptote code, which may have mislabels but gives a geometric layout, we can try to proceed with the coordinates as defined, even if they don't match the stated side lengths. Assuming the coordinates are correct as per the code: Point C: (14*cos100,14*sin100) Point D: (14,0) Thus, distance CD is sqrt[(14 -14*cos100)^2 + (0 -14*sin100)^2] Let me compute this step by step: First, compute cos100 and sin100: cos100 cos(180 -80)= -cos80-0.1736 sin100sin800.9848 Thus, 14*cos10014*(-0.1736)-2.4304 14*sin10014*0.984813.787 Therefore, coordinates: Point C: (-2.4304,13.787) Point D: (14,0) Difference in x: 14 - (-2.4304)=16.4304 Difference in y:0 -13.787-13.787 Distance CDsqrt(16.4304 + (-13.787)) Calculate 16.4304269.9 13.787190.1 Sum269.9 +190.1460 sqrt(460)21.44 cm Which is approximately 21.44 cm. Given that the problem might expect an exact value, which we previously derived as 28*sin50, since sin500.7660, 28*0.766021.45, which matches the approximate calculation. Since 28*sin50 is an exact expression, and sin50 is not a standard angle, but perhaps it can be related to another form. Alternatively, using the Law of Cosines expression: CD=14*sqrt(2*(1 -cos100)) But 1 - cos100=2sin50, so CD=14*sqrt(4sin50)=28sin50, which is exact. However, the problem might want the answer in terms of a numerical value, but since it's a competition problem, it's more likely to expect an exact value. But 28sin50 is exact, but maybe expressed differently. Alternatively, using the original triangles, but since the given side lengths are inconsistent, it's difficult to say. Given the problem's inconsistencies, but following the coordinates given in the Asymptote code (even though they conflict with stated side lengths), the length of CD is 28sin50, which is approximately 21.45 cm. Since the problem might expect an exact value, 28sin50 is acceptable, but perhaps it can be written as 28cos40, since sin50=cos40, and that might be a nicer form. Alternatively, leave as 28sin50. But let me check if 28sin50 is the correct exact value. From the def /w { setlinewidth} bind def /C { curveto} bind def /F { fill} bind def /L { lineto} bind def /rL { rlineto} bind def /P { grestore} bind def /s { stroke} bind def /S { show} bind def /N {currentpoint 3 -1 roll show moveto} bind def /Msf { findfont exch scalefont [1 0 0 -1 0 0 ] makefont setfont} bind def /m { moveto} bind def /Mr { rmoveto} bind def /Mx {currentpoint exch pop moveto} bind def /My {currentpoint pop exch moveto} bind def /X {0 rmoveto} bind def /Y {0 exch rmoveto} bind def /MISOfy { /newfontname exch def /oldfontname exch def oldfontname findfont dup length dict begin {1 index /FID ne {def} {pop pop} ifelse} forall /Encoding ISOLatin1Encoding def currentdict end newfontname exch definefont pop } def 63.000 11.250 moveto %%IncludeResource: font Helvetica %%IncludeFont: Helvetica %%BeginResource: font Helvetica-MISO %%BeginFont: Helvetica-MISO /Hel, {{ Hold[$CellContext`xvar2d$$], 1, "x for 2D scatter plot"}, Dynamic[{ 1 -> If[$CellContext`dataflag$$ == 1, "uniform", "ridge"], 2 -> If[$CellContext`dataflag$$ == 1, "triangular", "nub"], 3 -> If[$CellContext`dataflag$$ == 1, "Poisson", "crack"], 4 -> If[$CellContext`dataflag$$ == 1, "normal", "weight"], 5 -> If[$CellContext`dataflag$$ == 1, "beta", "density"]}]}, {{ Hold[$CellContext`yvar2d$$], 2, "y for 2D scatter plot"}, Dynamic[{ 1 -> If[$CellContext`dataflag$$ == 1, "uniform", "ridge"], 2 -> If[$CellContext`dataflag$$ == 1, "triangular", "nub"], 3 -> If[$CellContext`dataflag$$ == 1, "Poisson", "crack"], 4 -> If[$CellContext`dataflag$$ == 1, "normal", "weight"], 5 -> If[$CellContext`dataflag$$ == 1, "beta", "density"]}]}, {{ Hold[$CellContext`pointdiam2d$$], 0.015, "2D point size"}, 0.001, 0.05}, {{ Hold[$CellContext`xvar3d$$], 1, "x for 3D scatter plot"}, mber of masks may not reach 65536, but even if it does, this is acceptable for Fortran. Now, how to implement this in Fortran. First, the steps: 1. Read the required skills and create a map from skill to index. 2. For each person, convert their skills into a bitmask. 3. Initialize the DP array. Since the mask can be up to 2^16-1, we can use an array of size 65536. Each element will store the minimal team size and the list of people indices. But storing the lisus result (x <1/4 or x >2/3), the domain for the second logarithm term is x < -1 or x > 2/3. Wait, but the original expression for the second term's inner logarithm requires that \(\frac{x +1}{4x -1}\) is positive and that \(\log_{1/4} \left( \frac{x +1}{4x -1} \right)\) is positive. But we converted that to \(-\log_4 \left( \frac{x +1}{4x -1} \right) > 0\), which gives \(\log_4 \left( \frac{x +1}{4x -1} \right) < 0\), which led to the inequality \(\frac{x +1}{4x -1} < 1\). So, combining all these, the domain for the second term is x < -1 or x > 2/3. Therefore, combining both terms' domains, the overall domain is x < -1 or x > 2/3. Because both terms require x < -1 or x > 2/3. So that's the domain we need to consider. Now, going back to the original inequality: \[ 8.57 \log_{3} \log_{4} \left(\frac{4x - 1}{x + 1}\right) + \log_3 \left( -\log_4 \left( \frac{x + 1}{4x - 1} \right) \right) < 0. \] Let me denote \(A = \log_{4} \left( \frac{4x - 1}{x + 1} \right)\) and \(B = -\log_4 \left( \frac{x + 1}{4x - 1} \right)\). Then, the inequality becomes: \[ 8.57 \log_3 A + \log_3 B < 0 \] But note that \( \frac{4x -1}{x +1} \) is the reciprocal of, the next interval would be [2001,8000) gives A=3, B=2. Which is still valid. Length=8000 -2001=5999. Continuing this pattern, the maximal interval would be [2001, 2000(k+1)) for k >=1, where the interval includes k multiples of 2021 and (k -1) multiples of 2000. However, in reality, in [2001,2000(k+1)), the number of multiples of 2000 is k (from 4000, 6000, ..., 2000(k+1)), but since the interval excludes 2000(k+1), it's actually (k -1) multiples of 2000. For example, when k=2, [2001,6000) includes 4000 (k=2), but excludes 6000, so B=1. Wait, no: Wait, [2001,6000) includes multiples of 2000 at 4000 and 6000 (excluded). So, B=1. For k=3, [2001,8000) includes 4000,6000, and 8000 (excluded), so B=2. Therefore, in general, for [2001,2000(k+1)), the number of multiples of 2000 is k -1. Therefore, A= floor((2000(k+1) -1)/2021) - floor(2000/2021). But floor((2000(k+1) -1)/2021) (2000(k+1))/2021 ~ 0.99(k+1). For each k, A= floor(2000(k+1)/2021). For example, k=2: floor(6000/2021)=2 (2021*2=4042<6000, 2021*3=6063>6000) So, A=2 -0=2. B=1. Therefore, A=2, B=1. For k=3: floor(8000/2021)=3 (2021*3=6063<8000, 2021*4=8084>8000) A=3 -0=3. B=2. Therefore, A=3, B=2. This pattern continues. Therefore, for each k, the interval [2001,2000(k+1)) has A=k -1 and B=k -1. Wait, no. Wait, for k=2: [2001,6000) has A=2, B=1. For k=3: [2001,8000) has A=3, B=2. So, the pattern is A=k -1, B=k -2. Wait, no: Wait, for [2001,2000(k+1)), the number of multiples of 2000 is floor((2000(k+1) -1)/2000) - floor(2000/2000) = (k+1 -1) -1 =k -1. The number of multiples of 2021 is floor((2000(k+1) -1)/2021) - floor(2000/2021)= floor((2000(k+1) -1)/2021) -0= floor((2000(k+1) -1)/2021). We need to find k such that floor((2000(k+1) -1)/2021) =k. Lets verify for k=2: floor((2000*3 -1)/2021)=floor(5999/2021)=2. Yes. For k=3: floor((2000*4 -1)/2021)=floor(7999/2021)=3. Yes. Similarly, k=4: floor((2000*5 -1)/2021)=floor(9999/2021)=4. This pattern holds because 2000(k+1) -1=2000k +2000 -1=2000k +1999. Dividing by 2021: 2000k +1999 =2021k -21k +1999. For this to be approximately k, the term -21k +1999 must be between 0 and 2020. This holds when -21k +1999 >=0 =>k<=1999/2195.19. So, for k up to 95, this pattern holds. For k=95: 2000*96 -1=191,999 191,999/202195.00 floor(191,999/2021)=95. For k=96: 2000*97 -1=193,999 193,999/202196.00 fltment. College GuidePhoneIn-state TuitionOut-of-state TuitionStudent Body SizeRequest InfoEstrella Mountain Community College Avondale, AZ(623) 935-8000 $ (low) $$ medium Progentence>Editora Letras Juridicas, Livro de direito, Livro jurdico, Livro tecnico jurdico, Ensino jurdico, Livros digitais, Artigos jurdicos, Literatura Brasileira, Editora de livros de direito, Onde comprar livros jurdicos, Como publicar meu livro, Como publicar meu artigo jurdico, Artigos Jurdicos, Publicao e Venda de Livros, atacado e varejo. Argumentao Jurdica ,Biodireito, Biotica, Casos Jurdicos, Coleo Univem, Colees Catlogo das Leis, Concursos Pblicos, Dicionrios, Direito Administrativo, Direito Ambiental, Direito Cvil, Direito Comercial, Direito Constitucional, Direito de Famlia, Direito do Consumidor, Direito do Trabalho, Direito do Trnsito, Direito do Turismo, Direito e Biotica, Direito e Educao, Direito e Ensino, Direito e Informtica, Direito e Poltica, Direito Eleitoral, Direito Empresarial, Direito Financerementally. For example, the first move turns on a triangular section in a corner, the next move turns on another section, etc. But how? Alternatively, consider that in order to maximize the number of moves, each move should roughly halve the remaining number of lights. Wait, but that would lead to logarithmic number of moves. Howevinese. So perhaps the user wants [1] to remain in Chinese in the prompt, but the rest in English. However, Midjourney might not handle Chinese text well. But the user's example shows that the prompts are entirely in English. Therefore, maybe the user expects [1] to be translated into English. Let me check the initial problem statement again. The user's message starts with: "As a prompt generator for a generative AI called "Midjourney", you will create image prompts for the AI to visualize. I will give you a concept, and you will provide a detailed prompt for Midjourney AI to generate an image." Then the structure includes: [1] = Which is the concept provided by the user. Then the user says, "Write your prompts in english."ft to the next column. If the element is non-negative, the algorithm moves down to the next row. This method efficiently counts all negative numbers by leveraging the matrix's sorted structure, resulting in an optimal time complexity of O(m + n).In a class, $50$ students participated in a physical fitness test. The test scores $c$ are approximately normally distributed with $N\sim (90$,$\sigma ^{2})$. If $P\left(90\leqslant c\leqslant 95\right)=0.3$, estimate the number of students in the class whose physical fitness test scores are below $85$. Okay, so I need to figure out how many students scored below 85 on the physical fitness test. The problem says the scores are normally distributed with a mean () of 90 and some variance . We're also told that the probability of a score being between 90 and 95 is 0.3. Hmm, let me recall how normal distributions work. First, since the distribution is normal, it's symmetric around the mean. The mean here is 90, so the score of 90 is right in the center. The variance is , which I don't know yet, but maybe I can find it using the given proherum ist fr die Soldaten der Brief aus der Heimat ein berrest ihres zivilen Lebens. Dennoch bietet sich hier oft ein ungeschnteres Bild als in den offiziellen Darstellungen.Wir mich gehen mit meinem Mama auf den festlichen Einkufen. Wir besuchen aller festliche Geschfte. Wir kaufen dieses was es wird sein uns ntig damit bereiten ein festliches Abendessen vor. Wir auch Geschfte besuchen mit den Geschenken, weil wir jeden aus der Familie irgendein ein hbsches Gesche Therefore: cos(30) = (x + y) / (sqrt(x + y + h) * sqrt(x + y)) ) Simplify denominator: sqrt(x + y + h) * sqrt(x + y) = sqrt( (x + y + h)(x + y) ) So cos(30) = (x + y) / sqrt( (x + y + h)(x + y) ) ) = sqrt( (x + y) / (x + y + h) ) Therefore: cos(30) = sqrt( (x + y) / (x + y + h) ) Square both sides: cos(30) = (x + y) / (x + y + h ) cos(30) = sqrt(3)/2, so cos(30) = 3/4 Therefore: 3/4 = (x + y) / (x + y + h ) Multiply both sides by denominator: 3/4 (x + y + h ) = x + y Expand left side: (3/4)(x + y) + (3/4)h = x + y Subtract (3/4)(x + y) from both sides: (3/4)h = (1/4)(x + y) Multiply both sides by 4: 3h = x + y So x + y = 3h, which matches what I had earlier. So the diagonal of the base, AC, is sqrt(3h) = h*sqrt(3), which makes sense. So that's one equation: x + y = 3h. Now, moving on to the next part. A plane passes through edge TC and is parallel to diagons from Alex to Dr. Carter. Alex is excited about their research paper on the women's suffrage movement but needs help finding resources. They mention key figures like Susan B. Anthony and Elizabeth Cady Stanton and the 19th Amendment. They also brought up a Netflix show but are unsure about its accuracy. So, the goal is to formalize this. Let's break it down. Start with the greeting: "Hey Dr. Carter," becomes "Dear Dr. Carter,". The opening line is friendly but needs to be more professional. "I hope you're doing well!" is okay, maybe "I hope this message finds you well." Next, Alex expresses excitement and feeling lost. Instead of "Im really excited... but Im a bit lost," perhaps "I am eager to begin... but I would greatly appreciate some guidance." That sounds more formal. Mentioning the key figures and the 19th Amendment is important. The original says "how they influenced the passage," which can be phrased as "their roles in the eventual ratification." The Netflix show part needs to be more formal. Instead of "it was pretty cool," maybe "I found it engaging." Also, checking accuracy: "not sure if it's accurate" becomes "uncertain of its historical accuracy." The closing "Thanks a ton!!! " is too informal. "Thank you in advance for your assistance" is better. Sign off with "Best regards" instead of just the name. Also, remove the emoji and exclamation marks. Ensure the structure is proper with a subject line. Check for any contractions like "I'm" and make them "I am." Make sure the tone is respectful and academic. Putting it all together: Start with a subject line, formal greeting, resta the quadratic is x + 2x + 3. The discriminant of x + 2x + (3 - 2) = x + 2x +1 = (x+1)^2. So, equation x +2x +3 =2 becomes x +2x +1 =0, which has a double root at x=-1. Similarly, for a=2, quadratic is x +4x +6. Setting equal to 2: x +4x +6 -2 =x +4x +4=(x+2)^2=0, so double root at x=-2. In both cases, the inequality |f(x)| <=2 becomes equality only at x=-1 and x=-2 respectively. Therefore, only one solution in each case. Therefore, two values of a. So answer is B) 2. But wait, maybe there's a case where the quadratic crosses y=2 once and y=-2 once, but such that the overlap is only one point? Let's suppose that the quadratic crosses y=2 at one point and y=-2 at another point, but between these two points, the quadratic is outside the range [-2,2]. Wait, but since it's a parabola opening upwards, if it crosses y=2 at two po side:") <+> ppr lhs])] $$ ptext (sLit "LHS must be of form (f e1 .. en) where f is not forall'd") \end{code} %********************************************************* %* * \subsection{Vectorisation declarations} %* * %********************************************************* \begin{code} rnHsVectDecl :: VectDecl RdrName -> RnM (VectDecl Name, FreeVars) -- FIXME: For the moment, the right- -- We always assume that indexed types are recursive. Why? -- (1) Due to their open nature, we can never be sure that a -- further instance might not introduce a new recursive -- dependency. (2) They are always valid loop breakers as -- they involve a coercion. ; fam_inst <- newFamInst (DataFamilyInst rep_tc) axiom ; return (rep_tc, fam_inst) } -- Remember to check validity; no recursion to worry about here ; checkValidTyCon rep_tc ; return fam_inst } } where -- See Note [Eta reduction for data family axioms] -- [a,b,c,d].T [a] c Int c d ==> [a,b,c]. T [a] c Int c eta_reduce tvs pats = go (reverse tvs) (reverse pats) go (tv:tvs) (pat:pats) | Just tv' <- getTyVar_maybe pat , tv == tv' , not (tv `elemVarSet` tyVarsOfTypes pats) = go tvs pats go tvs pats = (reverse tvs, reverse pats) \end{code} Note [Eta reduction for data family axioms] ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Consider this data family T a b :: * newtype instance T Int a = MkT (IO a) deriving( Monad ) We'd like this to work. From the 'newtype instance' you might think we'd get: newtype TInt a = MkT (IO a) axiom ax1 a :: T Int a ~ TInt a -- The type-instance part axiom ax2 a :: TInt a ~ IO a -- The newtype part But now what can we do? We have this problem Given: d :: Monad IO Wanted: d' :: Monad (T Int) = d |> ???? What coercion can we use for the ??? Solution: eta-reduce both axioms, thus: axiom ax1 :: T Int ~ TInt axiom ax2 :: TInt ~ IO Now d' = d |> Monad (sym (ax2 ; ax1)) This eta reduction happens both for data instances and newtype instances. See Note [Newtype eta] in TyCon. %************************************************************************ %* * Type-checking instance declarations, pass 2 %* * %************************************************************************ \begin{code} tcInstDecls2 :: [LTyClDecl Name] -> [InstInfo Name] -> TcM (LHsBinds Id) -- (a) From each class declaration, -- generate any default-method bindings -- (b) From each instance decl -- generate the dfun binding tcInstDecls2 tycl_decls inst_decls = do { -- (a) Default methods from class decls let class_decls = filter (isClassDecl . unLoc) tycl_decls ; dm_binds_s <- mapM tcClassDecl2 class_decls ; let dm_binds = unionManyBags dm_binds_s -- (b) instance declarations ; let dm_ids = collectHsBindsBinders dm_binds -- Add the default method Ids (again) -- See Note [Default methods and instances] ; inst_binds_s <- tcExtendLetEnv TopLevel TopLevel dm_ids $ mapM contact triangle's sides are parallel to the original triangle's sides? No, that's not correct. The contact triangle is formed by connecting the points where the incircle touches the original triangle's sides. So the sides of the contact triangle are chords of the original triangle, each lying on a side of the original triangle. Wait, actually, each vertex of the contact triangle is located on a different side of the original triangle. So triangle ABC is entirely inside triangle KLM, with each whiffle ball game.byHRose 6 comments/ 56893 views/ 13 favoritesShare the loveTweetReport a BugSubmit bug reportNext3 Pages:123123ait, this is getting big, but the key is that Vieta jumping seems to generate larger solutions each time. So if we can keep doing this indefinitely, we can get infinitely many solutions. But in this process, each time we fix two variables and generate the third. However, since the variables are positive integers, and each step produces a larger number, we can keep generating new solutions by always replacing the variable we solved for. So starting from (1,1,1), we can generate (2,1,1), then (2,5,1), then (2,5,29), then (2,169,29), then (14701,169,29), and so on. Each time, the numbers grow, so these are distinct solutions. Alternatively, we might be able to parameterize the solutions. Let me see. Let's suppose that we start with two variables as 1 and find the third. Wait, we saw that (1,1,2) is a solution. Similarly, maybe we can set two variables to some numbers and find the third. But maybe a better approach is to notice that if we set z=1, then the equation becomes x + y +1 =3xy. Let's see if this equation has infinitely many solutions. So x -3xy + y +1=0. Let's rearrange: x -3xy + y = -1. Hmm, not sure. Maybe treat this as a quadratic in x: x -3y x + (y +1)=0. The discriminant is 9y -4(y +1)=5y -4. So 5y -4 must be a square. Let 5y -4 =k. Then 5y -k=4. This is a Pell-type equation. Let me see. Rearranged as k -5y = -4. Pell equations have infinitely many solutions if they have at least one. Let me check if there are solutions. For y=1: k=5(1)^2 -4=1, so k=1. So (k,y)=(1,1). Then y=1, k=1. So x=(3y k)/2=(31)/2=2 or 1. So x=2 gives the solution (2,1,1), and x=1 gives (1,1,1). So that's consistent with what we had before. The Pe, if r is 1.0015, then the next year is 1964*1.0015 1966.99, then 1966.99*1.0015 1969.01, then 1969.01*1.0015 1971.05, then 1971.05*1.0015 1973.11, but wait, the fifth term is supposed to be 1976. So, my approximated ratio might be incorrect? Wait, no, because when we calculated r^4 1.006109979, so multiplying 1964 by that gives 1976. But when we compute each term: Term 1: 1964 Term 2: 1964 * r 1964 * 1.001524 1964 + 2.993 1966.993 Term 3: 1966.993 * 1.001524 1966.993 + 1966.993*0.001524 1966.993 + 2.999 1969.992 Term 4: 1969.992 * 1.001524 1969.992 + 1969.992*0.001524 1969.992 + 3.004 1973.0 Term 5: 1973.0 * 1.001524 1973.0 + 1973.0*0.001524 1973.0 + 3.009 1976.009 Which is approximately 1976.01, close to 1976, considering rounding errors. So, each term is approximately 3 years apart, but slightly increasing. However, since the ratio is very close to 1, the terms are increasing by a small percentage each time, leading to an approximate linear increase. Wait, but if the ratio is just over 1, the sequence would be increasing approximately linearly on a logarithmic scale. Hmm, maybe the problem is designed to have an exact ratio? Wait, but 1976/1964 is 494/491. So, unless there's a different interpretation. Wait, maybe the problem is considering the positions on the timeline as real numbers, so even though the actual release years are integers, here they are points on a continuous interval. Therefore, it's acceptable for the release years to be non-integer. Therefore, the common ratio is (494/491)^(1/4), which is approximately 1.001524. So, that's the answer for the ratio. But let me check if there's another way to model this. Perhaps the problem expects the difference between the years to form a geometric sequence, but no, the problem states that the release years themselves form a geometric sequence. So, the terms themselves are multiplied by r each time. So, the answer is as above. Therefore, the common ratio is the fourth root of 1976/1964, which is (1976/1964)^(1/4). To present this neatly, we can write: r = \sqrt[4]{\frac{1976}{1964}} = \sqrt[4]{\frac{494}{491}} If a deciownside to an associates degrees in the job market is only that you will be competing with individuals holding bachelors degrees. Therefore, make sure you're comfortable in the niche you've chosen.I am a Licensed Professional Clinical Counselor with 17 years of experience providing counseling services in outpatient settings, schools, and working for a grant. I provide counseling services using Cognitive Behavioral Therapies and I specialize in treating individuals experiencing stress, anxiety, depression, and those recovering from exposure to traumatic events. I work with children, adolescents, and adults. I look forward to continuing my work in the community and would love to connect with you!Zoeken: Alle categorien Aquarium Aquariums Bodembedekking Decoratie Diversen Onderhoud Techniek Voer Waterbehandeling Knaagdieren Bodembedekking Eet- Drinkbakjes Vogels Badhuisjes Bodembedekking Broeden Eet- drinkbakjes Winter Wintere and rectangular matrices but requires uniform row lengths. 5. **Coordinate Retrieval**: Retr81 meters down, the squirrel would have to jump down 2.81 meters to catch it. However, the squirrel can only jump 1.70 meters. Therefore, the nut will be too far below the squirrel's original position for it to catch. But hold on, maybe I'm misunderstanding the problem. The squirrel can jump "with great speed in any direction for a distance of 1 meter 70 centimeters." So maybe the squirrel can jump 1.70 meters in any direction from its original position. That would mean the squirrel can reach any point within a sphere of radius 1.70 meters centered at its original position. So if the nut is within that sphere when it arrives at the horizontal position, the squirrel can catch it. Wait, but the nut is moving horizontally towards the squirrel. So when the nut is at the horizontal position of the squirrel, it's 3.75 meters away from Gavrila, but vertically it has fallen 2.8125 meters. The squirrel is on the tree branch, which is at the same height as Gavrila's window, right? The problem says the squirrel is sitting on a branch "directly opposite him," so I assume at the same height. Therefore, the nut, when it reaches the horizontaltion code for an access token. Normally, this involves the client sending these parameters to the token endpoint. The server would validate them, but here we just return a mock response regardless. Wait, but should the function validate the inputs at all? The note says not to perform real authentication, so maybe even if the inputs are wrong, it just returns the mock data. Or perhaps the function checks if the parameters are provided but still returns the mock tokens as long as they are present. Hmm, the example in the problem doesn't check validity. So maybe the function doesn't do any validation. It just takes the inputs and returns the mock tokens. So the function will create a dictionary with the required keys. The values can be mock strings. Like access_token could be a fixed string, or maybe generate a mock one using some parts of the inputs? Not n nums = let evens = filter even nums counts = M.fromListWith (+) [ (x, 1) | x <- evens ] in case M.maxView counts of Notor, and each vertical tile covers n of the same color. This forces h to be divisible by n. Therefore, if a mixed tiling exists, both w and h are divisible by m and n respectively, allowing pure tilings with either type. Hence, the conclusion follows. To solve the problem, we analyze the tiling conditions using a coloring argument and modular arithmetic. Here's the step-by-step explanation: **Step 1: Understand the Tiling Requirements** - **Horizontal Tiles**: 1m strips require the rectangle's width (w) to be a multiple of m. - **Vertical Tiles**: n1 strips require the rectangle's regard to IRS35, whether in the Finance Bill or by way of further amendment. Clearly some of the consultations in which the Government took part have been helpful in improving what was originally proposed. However, there is still quite considerable concern among smaller operators in the IT industry, in particular that they may find themselves disadvantaged by what was further suggested in relation to their larger competitors. All I say at this stage is that I hope very much that the Government will carry out further consultations, because everyone has the same objective; namely, to ensure that tax avoidance does not occur. However, it is important that that should be done in a way which does not damage industry or the state of the economy. 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So numbering goes down the first column, then down the second, and so on. So for example, in the first bank, column 1 (a large column with two lockers) would have lockers 1 and 2. Then column 2 (also large) has lockers 3 and 4. Then columns 3 to 6 (small columns with four lockers each). So column 3 would have 5,6,7,8; column 4:9,10,11,12; column5:13,14,15,16; column6:17,18,19,20. Then the next bank would start at 21, I suppose. So each bank is arranged with two columns of two lockers each and four columns of four lockers each, totaling 20 lockers. The numbering is column-wise: top to bottom in each column, then moving to the next column. For part (a): What is the sum of the locker numbers of the column of lockers that contain the number 24? First, find which column contains locker 24. Then sum all the numbers in that column. First, need to figure out in which column locker 24 is. Let's first determine which bank locker 24 is in. Each bank has 20 lockers. So locker numbers 1-20 are in the first bank, 21-40 in the second, etc. 24 is in the second bank (21-40). Then within the second bank, it's locker number 24 - 20 = 4th locker in the second bank. Now, in each bank, the numbering goes through columns 1 to 6. Let's see how the lockers are arranged in a bank. The first two columns (columns 1 and 2) have 2 lockers each, columns 3-6 have 4 each. So per bank: Column 1: 2 lockers Column 2: and with sliding clamp 4. Final output of original and replicated strands 4. **Educational Components**: - Visual representation of the replication process - Enzyme function annotations - Proper 5'3' directionality demonstration - Base pairing visualization To run the simulation: 1. The program starts with a sample DNA sequence 2. Shows real-time progression of replication 3. Demonstrates enzyme coordination 4. Outputs final replicated DNA sequence The simulation helps visualize: - How helicase separates DNA strands - How polymerase adds complementary bases - The role of the sliding clamp in processivity - Topoisomerase preventing DNA supercoiling - The antiparallel nature of DNA strands The final output shows both original and complementary strands wied. White nodes are jumping forward, while red is jumping backward. The only cycle is of size 1, so we return false. **Example 3:** **Input:** nums = \[1,-1,5,1,4\] **Output:** true **Explanation:** The graph shows how the indices are connected. White Russert and NBC News initially resisted the subpoena on First Amendment grounds, but relented after prosecutor Patrick Fitzgerald agreed not to compel Russert to appear before the grand jury, or to disclose confidential sources or information. [Washington Post, 8/10/2004] Russert has already talked informally with John Eckenrode, the FBI investigator overseeing the day-to-day investigation duties (see November 24, 2003). He told Eckenrode that Libbys claim of learning Plame Wilsons identity from him was false, and that he and Libby never discussed Plame Wilson at all. [National Journal, 2/15/2007] Libbys claim that he learned of Plame Wilsons identity from Russert will lead to perjury charges (see October 28, 2005).tenten hentai video geile muschi Pornfree Granny Irnkam Sex Treffen Noch Heute Swingerclub Marl Online Sex Kontakte erotic spanking tube schamlippe gro neuanfang nach trennung anal finger tube Verschieden Groe Brust Transen Kontakte Kostenlos Obersexau Seitensprung Freising Geilsten Porno dildosex gratis susse frau porno fisting tube ist seine kostenlose handypornos zum runterladen Initialen Zeigte Die Bilder Luckau Swinger Fkk Nrw Amateure Urlaub Kostenlos Dicke Single Immer Bin Nur Aufgrund porno kontakte porn kostenfrei sexy gratis sex im rollstuhl Sex Paderborn Er Sucht Ihn Erotik Berlin Geile Lesbenpornos Bassin Real Family Porn deutsch dirty talk video sexkontakte deutschland geile votzen fotos tube sex adult Ich haare huren kontaktanzeigen frauen poppen und fremdgehen beziehungen und irritationen reife damen sexfilme Singles Krefeld Poppen Film Dicke Frauen Attraktiv Live Free Sex Cams Boogholzberg hobby nutten bremen small penis party watch 3d porn movies deutsche free sex movies Normal Das small asian ladyboys ao fickanzeigerSubordinate gvitte to manager with qualite such s ths8SsQTLcWfk4jPhJ9C vPJEfCa8LeLbZsRnwhchHv6F+Ey4NeJnDuIz4cUQb18iPhNeFvH5GOIz4cXB T0B8Jrwr5H9AfCZ8E/Spg/hMeDXovwXxmfDdWK8v4jPhH2CfMMRnwsmejRCf CW+G+7E28o2D4n7UdsN9Oh75BuFXcP+OQ75BeGXkG32RbxB+B/d7EvINwpsi H/BFvkH4e+QPw5BvEP4O+UYc8g1DvhPyDcJbQf525BuEx0Ofl8g3CP+EfKM/ 8g3CT2C9zsg3CLeGfZYi3yCc7OmHfIPwxsj3vJA/R4p8T0v54VXkz4THIp/c i/yZ8DnIP9sgfyZ8OvLute z = - (c x)/a + 1/2 into equation 2. Equation 2: \( ax + by - c z = k \cos(ax + by - c z) \) Substitute z: Left side: \( ax + by - c(- (c x)/a + 1/2) = ax + by + (c^2 x)/a - c/2 \) Simplify: \( ax + (c^2 x)/a + by - c/2 = x(a + c^2/a) + by - c/2 \) Similarly, right side: \( k \cos(ax + by - c z) = k \cos\left(ax + by - c(- (c x)/a + 1/2)\right) = k \cos\left(ax + by + (c^2 x)/a - c/2\right) \) Therefore, equation 2 becomes: \( x\left(a + \frac{c^2}{a}\right) + by - \frac{c}{2} = k \cos\left(x\left(a + \frac{c^2}{a}\right) + by - \frac{c}{2}\right) \) Let me denote \( v = x\left(a + \frac{c^2}{a}\right) + by - \frac{c}{2} \). Then equation 2 is \( v = k \cos v \). This is a transcendental equation in v. For this equation to hold for all x and y, the only possibility is that both sides are constants independent of x and y. However, v is linear in x and y, so unless the coefficients of x and y are zero, v would vary with x and y. Therefore, the coefficients of x and y must be zero: Coefficient of x: \( a + c^2/a = 0 \implies a^2 + c^2 = 0 \). Since a and c are real numbers, this implies a = 0 and c = 0. But if c = 0, the original equation 2 becomes \( ax + by = k \cos(ax + by) \), e is priced incorrectly, Mazda of Orange shall have the right to refuse or cancel any orders placed for the vehicle presented with the incorrect price. In addition, vehicle prices are subject to change and all vehicles are subject to prior sale and may not be available when you are ready to purchase.hello everyone my name is andra i am from united kindom i want to use this middium to tell everyone how i my ex husband left me for another woman so i meat this my friend who told me about this great man the great powerful spell caster the great dr issace who help her bring back his ex husbank back to her within 48hours this man is really very powerful he is truthful after contacting him he told me not to cry that my problems are over so that was how i did what he ask me to do and after 2days he brought back my ex husband to me within 3days this man the great dr issace is a real powerful spell caster who can help you solve your problem he is the man you can trust you can contact him directly on his email DRISSACESPELLCASTER@GMAIL.COM CELL +2348052598517Medicare.gov provides tools that will allow you to compare plans in Greenville SC 29615 Greenville county, but the decision is complicated. 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Organizations such as Consumer Reports and the Medicare Rights Center can also help you research your decision.Artists Annie OakleyBarneyBob SteeleBonanzaBoneyardCaillouCare BearsDaily PlanetDallas+ More...DexterDora the ExplorerDynastyFuturamaGhost in the ShellGlobe TrekkerGunsmokeHell's KitchenHouseJohn WayneKidsongsLittle RascalsLooney TunesMax & RubyMedicNarutoNatureNovaOfficePeanutsRoy RogersSesame StreetSmallvilleSouth ParkSpongeBob SquarepantsStandard Deviants SchoolStandard DeviantsStar Trek Next GenerationThe BackyardigansThe Dark ShadowsThe FriendsThe FrontlineThe Lucy ShowThe SimpsonsThe SupernaturalThe Three StoogesThe WeedsThomas & FriendsTom & JerryTrue BloodUniverseChrysler proiezioni di crescita elevate anche per il brand che presta il suo nome ad FCA: Al Gardner, capo del marchio, ha parlato di vendite in crescita dalle 350.000 unit attuali a 800.000 tra cinque anni: "La chiave del rilancio per il marchio venuta dopo la ristrutturazione, " ", SuperscriptBox["y", "22"]}], "-", RowBox[{"20", " ", SuperscriptBox["x1", "5"], " ", SuperscriptBox["x2", "5"], " ", SuperscriptBox["y", "22"]}], "-", RowBox[{"3", " ", SuperscriptBox["x2", "6"], " ", SuperscriptBox["y", "22"]}], "-", RowBox[{"2", " ", SuperscriptBox["x1", "2"], " ", SuperscriptBox["x2", "6"], " ", read the weather reports coming from Americas heart land? Heat and lack of rain are playing havoc with the crops. The prices for corn, soybean and wheat have jumped over the past two days (5.5%, 3.6% and 3.1% respectively). This may seem like a small increase but when you consider that 70% of everything we consume uses these three commodities in some way, it is a significant jump. Hot, dry weather is expected to stay with the nations breadbasket for a zwar nicht direkt am Strand, aber die paar Meter knnen wir auch laufen. Dafr hat sie eine groe Veranda, auf der wir in Gutsherrenart bequem sitzen knnen. . . - . ( ), - - . , . , , . , - . . - ? - , ( , ). ! , - , , - , . 10 4 . , , . ( ) . Wien (OTS) - "Verschiedenste Initiativen haben auf einen Volksentscheid ber den EU-Vertrag in sterreich gedrngt. Beispielgebend waren sicherlich Krntens BZ-Landeshauptmann Dr. Jrg Haider, die zahlreichen parlamentarischen Initiativen von BZ-Klubobmann NAbg. Peter Westenthaler, die gro angelegte Brgerinitiative "Rettet sterreich" und nicht zuletzt der mediale Druck der grten sterreicherischen Ta getHooked runQuasiQuoteHook return >>= ($ HsQuasiQuote quoter' q_span quote) -- Build the expression ; let quoterExpr = L q_span $! HsVar $! quoter'' ; let quoteExpr = L q_span $! HsLit $! HsString quote' ; let expr = L q_span $ HsApp (L q_span $ HsApp (L q_span (HsVar quote_selector)) quoterExpr) quoteExpr ; meta_exp_ty <- tcMetaTy meta_ty -- Typecheck the expression ; zonked_q_expr <- tcTopSpliceExpr False (tcMonoExpr expr meta_exp_ty) -- Run the expression ; result <- runMetaQ meta_ops zonked_q_expr ; showSplice (mt_desc meta_ops) quoteExpr (ppr result) ; return result } runQuasiQuoteExpr qq = runQuasiQuote qq quoteExpName expQTyConName exprMetaOps runQuasiQuotePat qq = runQuasiQuote qq quotePatName patQTyConName patMetaOps runQuasiQuoteType qq = runQuasiQuote qq quoteTypeName typeQTyConName typeMetaOps runQuasiQuoteDecl qq = runQuasiQuote qq quoteDecName decsQTyConName declMetaOps quoteStageError :: Name -> SDoc quoteStageError quoter = sep [ptext (sLit "GHC stage restriction:") <+> ppr quoter, nest 2 (ptext (sLit "is used in a quasiquote, and m "instances" are present: 1. instance Coercible a a for any type a at any kind k. 2. instance (forall a. Coercible t1 t2) => Coercible (forall a. t1) (forall a. t2) (which would be illegal to write like that in the source code, but we have it nevertheless). 3. instance Coercible r b => Coercible (NT t1 t2 ...) b instance Coercible a r => Coercible a (NT t1 t2 ...) for a newtype constructor NT (or data family instance that resolves to a newtype) where * r is the concrete type of NT, instantiated with the arguments t1 t2 ... * the constructor of NT is in scope. The newtype TyCon can appear undersaturated, but only if it has enough arguments to apply the newtype coercion (which is eta-reduced). Examples: newtype NT a = NT (Either a Int) Coercible (NT Int) (Either Int Int) -- ok newtype NT2 a b = NT2 (b -> a) newtype NT3 a b = NT3 (b -> a) Coercible (NT2 Int) (NT3 Int) -- cannot be derived 4. instance (Coercible t1_r t1'_"Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-7.824046010856292, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-7.600902459542082, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-7.418580902748128, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-7.264430222920869, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-7.1308988302963465`, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-7.013115794639964, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-6.214608098422191, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-5.809142990314028, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-5.521460917862246, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-5.298317366548036, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-5.115995809754082, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-4.961845129926823, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-4.8283137373023015`, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-4.710530701645918, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-3.912023005428146, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-3.506557897319982, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-3.2188758248682006`, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-2.995732273553991, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-2.8134107167600364`, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-2.659260036932778, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForaybe the mention of the triangle being equilateral with right angles is a red herring, or a mistake. Alternatively, the problem might have meant that the triangle is a right-angled triangle (not equilateral) with sides 7,9,12, but since 7+912, that's not possible. Wait, 7+9=49+81=130, and 12=144, so not equal. 9+12=81+144=225=15, so if the sides were 9,12,15, that would be a right triangle. But given sides 7,9,12, it's not a right triangle. So the area can't be computed via the standard right-angled formula, so Heron's formula is the way to go. Therefore, maybe despite the confusing problem statement, the answer is 145. Let me confirm Heron's formula calculation again: s = (7+9+12)/2 = 14 Area = [14*(14-7)*(14-9)*(14-12)] = [14*7*5*2] = [14*7*10] = [980] = [49*20] = 720 = 7*25 = 145. Yes, that's correct. So maybe the answer is 145, even though the problem mentions contradictory conditions. Perhaps the part about the triangle being equilateral with right angles is a mistake, and we're supposed to compute the area using Heron's formula. Alternatively, if we take the problem at face value, recognizing that the triangle can't exist because it can't be both equilateral and right-angled, then the area doesn't exist. But since the sides do form a valid triangle (just not a right-angled or equilateral one), the area is 145. Given that the problem asks to "find the area", I think the expected answer is 145, ignoring the contradictory second condition, assuming it's a typo or mistake. **Final Answer** The area of the triangle is \boxed{14\sqrt{5}}. To find the area of a triangle with sides \(a = 7\), \(b = 9\), and \(c = 12\), we first check the triangle inequality condition, which sCodigoMesa = mesCodigoMesa INNER JOIN Departamento ON vot_proCodigoProvincia = dep_proCodigoProvincia AND vot_depCodigoDepartamento = depCodigoDepartamento GROUP BY vot_proCodigoProvincia, vot_depCodigoDepartamento, depNombre, vot_mesCodigoCircuito, vot_mesCodigoMesa ORDER BY vot_proCodigoProvincia, vot_depCodigoDepartamento, depNombre, vot_mesCodigoCircuito, vot_mesCodigoMesa; ``` ```{r} write_csv(cabagral2007, "../1.CABA/caba_dip_gral2007.csv") ``` #### PBA ```{sql pbagral2007, connection=Arg07, output.var="pbagral2007"} WITH votos (vot_proCodigoProvincia, vot_depCodigoDepartamento, vot_mesCodigoCircuito, vot_mesCodigoMesa, "0003", "0014", "0030", "0038", "0048", "0051", "0060", "0132", "0133", "0134", "0135", "0141", "0144", "0250", "0260", "0269", "0273", "0275", "0311", "0312", "0313", "0314", "0315", "0316", "0317") AS (SELECT vot_proCodigoProvincia, vot_depCodigoDepartamento, vot_mesCodigoCircuito, vot_mesCodigoMesa, max(CASE WHEN vot_parCodigo = "0003" THEN votVotosPartido END), max(CASE WHEN vot_parCodigo = "0014" THEN votVotosPartido END), max(CASE WHEN vot_parCodigo = "0030" THEN votVotosPartido END), max(CASE WHEN vot_parCodigo = "0038" THEN votVotosPartido END), max(CASE WHEN vot_parCodigo = "0048" THEN votVotosPartido END), max(CASE WHEN vot_parCodigo = "0051" THEN votVotosPartido END), max(CASE Wmber, average out-degree about 6.06. If the subgraph induced by the remaining 16 vertices has such a high edge density, maybe it contains a directed 3-cycle. But if the original graph had no directed 3-cycles, then neither would this subgraph. But 16 vertices with 97 edges and no 3-cycles. But according to the asymptotic bound mentioned earlier, the maximum number of edges without a 3-cycle is around n/4 = 256/4=64. Since 97 >64, this would imply that the subgraph must contain a directed 3-cycle. Therefore, if our initial assumption that the entire graph has no direcCD. For three numbers, is there a similar formula? I found in some resources that: For three numbers a, b, c: \[ a \times b \times c = \text{LCM}(a, b, c) \times \text{GCD}(a, b) \times \text{GCD}(a, c) \times \text{GCD}(b, c) \div \text{GCD}(a, b, c)^2 \] But I need to verify this. Let me take an example. Suppose a=2, b=3, c=5. Then LCM=30, GCD=1. Product: 235=30 Using the formula: LCM(a,b,c) GCD(a,b) GCD(a,c) GCD(b,c) GCD(a,b,c)^2 = 30 1 1 1 1 = 30. Correct. Ano+ 3 (I Sin[x])^10 Cos[x]^6 + 8 (I Sin[x])^8 Cos[x]^8) + Exp[-11 I y] (18 (I Sin[x])^5 Cos[x]^11 + 18 (I Sin[x])^11 Cos[x]^5 + 36 (I Sin[x])^7 Cos[x]^9 + 36 (I Sin[x])^9 Cos[x]^7 + 22 (I Sin[x])^6 Cos[x]^10 + 22 (I Sin[x])^10 Cos[x]^6 + 38 (I Sin[x])^8 Cos[x]^8 + 8 (I Sin[x])^4 Cos[x]^12 + 8 (I Sin[x])^12 Cos[x]^4 + 2 (I Sin[x])^3 Cos[x]^13 + 2 (I Sin[x])^13 Cos[x]^3) + Exp[-9 I y] (214 (I Sin[x])^8 Cos[x]^8 + 110 (I Sin[x])^10 Cos[x]^6 + 110 (I Sin[x])^6 Cos[x]^10 + 170 (I Sin[x])^7 Cos[x]^9 + 170 (I Sin[x])^9 Cos[x]^7 + 52 (I Sin[x])^11 Cos[x]^5 + 52 (I Sin[x])^5 Cos[x]^11 + 14 (I Sin[x])^12 Cos[x]^4 + 14 (I Sin[x])^4 Cos[x]^12 + 2 (I Sin[x])^3 Cos[x]^13 + 2 (I Sin[x])^13 Cos[x]^3) + Exp[-7 I y] (354 (I Sin[x])^6 Cos[x]^10 + 354 (I Sin[x])^10 Cos[x]^6 + 492 (I Sin[x])^8 Cos[x]^8 + 446 (I Sin[x])^7 Cos[x]^9 + 446 (I Sin[x])^9 Cos[x]^7 + 200 (I Sin[x])^5 Cos[x]^11 + 200 (I Sin[x])^11 Cos[x]^5 + 88 (I Sin[x])^4 Cos[x]^12 + 88 (I Sin[x])^12 Cos[x]^4 + 26 (I Sin[x])^3 Cos[x]^13 + 26 (I Sin[x])^13 Cos[x]^3 + 5 (I Sin[x])^2 Cos[x]^14 + 5 (I Sin[x])^14 Cos[x]^2) + Exp[-5 I y] (1124 (I Sin[x])^9 Cos[x]^7 + 1124 (I Sin[x])^7 Cos[x]^9 + 361 (I Sin[x])^11 Cos[x]^5 + 361 (I Sin[x])^5 Cos[x]^11 + 746 (I Sin[x])^6 Cos[x]^10 + 746 (I Sin[x])^10 Cos[x]^6 + 1232 (I Sin[x])^8 Cos[x]^8 + 126 (I Sin[x])^12 Cos[x]^4 + 126 (I Sin[x])^4 Cos[x]^12 +er than85 are 100. So r=100. Let's check k=5 and r=100: p=35, q=85, r=100. Check if q=85 < r=100: yes. Compute x=5*(7*100 +7*100 +17*100 +17)=5*(7,000,000 +70,000 +1,700 +17)=5*(7,071,717)=35,358,585 x=35,358,585. Now, con `nlHsApp` nlHsLit (mkHsString (showSDocOneLine (ppr tycon))) `nlHsApp` nlList constrs genAuxBind loc (MkDataCon dc) -- $cT1 etc = (mkHsVarBind loc rdr_name rhs, L loc (TypeSig [L loc rdr_name] sig_ty)) where rdr_name = mk_constr_name dc sig_ty = nlHsTyVar constr_RDR rhs = nlHsApps mkConstr_RDR constr_args constr_args = [ -- nlHsIntLit (toInteger (dataConTag dc)), -- Tag nlHsVar (mk_data_type_name (dataConTyCon dc)), -- DataType nlHsLit (mkHsString (occNameString dc_occ)), -- String name nlList labels, -- Field labels nlHsVar fixity] -- Fixity labels = map (nlHsLit . mkHsString . getOccString) s-plan-cul-albertville plan cul payant plan cul avec jeune femme vierge rouen videos de fessees couple baise sous la douche Webcam Amateur Porno Plan Cul Gay 34 Lieux De Sexe Sodomie La Premiere Logis De La Colle videos of ladyboys chat video amateur secretaire gros sein sites de pornos gratuit Video Sexe Vieille Femme Annon Plan Cul Utiac Transexuelle Asiatique Escort Girl Vip rencontre coquine black mature toulouse femme enceinte et sexe xxx pono sex plan cul sur ploeren ce soir Rencontre Plan Cul Chambery Whatsapp Plan Cul Clairmarais Porno Femme Mature Francaise Qu'il porn quality aixoise aimant plan cul.,(PN-MK8.4M100,HAMILTON [] H42211855 ;5012mm(50)..? ,Mens GUESS,, ,Mens GUESS,Mens GUESS.6000!Mens GUESS,.Mens GUESS,w/tourcase.AR4002-17Asmtb-tk! .
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Because the subsequence must be if parents and teachers are neglected by the youths, this has caused our social system, a great collapse, and no wonder there is so much decay in the system nowadays.Simonette, on the ground floor, is the hotels hopping, very charming French-style bistro. Its open all day and is equally good for a quick croissant and an espresso or a breakfast cocotte in the morning as it is for moules-frite and a glass of ros for lunch or dinner. Dishes are classic and pretty straight-forward, yet this is California, so you know the ingredients in that Nioise salad are going to amp things up. The Simonette Bar mixes some of the most technical craft cocktails in town, such as the Right Brigade (Fords Gin, Campari, white vermouth and amontillado sherry, filtered through coffee). , , (1908 .), (1933 .) (1964 .). , , . , .Agreed Meangreen - Calling pot smoking a history of "drug trouble" is very misleading when we have to states legalizing it and the United States government turning a blind eye. It's not a question or whether or not he has a drug problem. I'ts a question of whether or not he has a discipline issue which reflects on character. If someone tells you not to do something, no matter how trivial, then he has to be able not to do it. So maturity and discipline is his problem and not some goofy pot addiction.Zwarte Tieten Xvideos Saaie Seks Beste Jav Pornosites De Borkeld Latino Online Dating Clitoris Neuken Bikini Voor Kleine Borsten College Gf Creampie Latex Dildo Brouwhuis free chat canada size queen phone sex caught on security cam live africa lesbian numbers hausfrauensex privat fremdling 1080p full hd porG5NvpiWpcr4vkn0UdIOZuQ7y4j35Gc NsgXL9BHEx6F/BIr+hQpZzfyXSzyHfFXIn9VQt9N+lAf9Ev0QVLOCvAtBV8h OSRfRZ9OeBb6rLbIpySH9H+OfEfP3Yz1xqKvJ3w58ukm5FOSM8PQnpLfxtD+ Em+AfNoI+ZTk1DDcX6l/QfiDhfAHKeeeYR8q5VyGv3VGvpNy4J+euGcgOaWR T5shn5Kc1vD/l8h3JMca5+Ut7iUIt0c+tUU+JTkOOHc65EHiN0LebIa8SfzB X8vX5/V0ZLfcj6tY3W1d1GFBd3beBwGv0D0PVwkvOKjPk41pL5Ubkc5tqoTe UpZ4L3RxN3+lFPz2ucfKks+VUdM2356XHi/n+75fj52tWvC1ct7X3ynrwknJ t5xo9reGxzMloLDl6ZrsczHwlS3NHg0LTlVyuyX+dGefJKfUlraeldjznPwv NdzEnk/8+H0Hbs5mz9M99Azhz5e/3/vXZOnrCyfVFU39LvPnE/9C5am7d9Q/ pvruWRpe+9u//NXby9bz2ZeiDhpmHdhta4bkB5heirumu6eOL13/8oWwNMm3 MA2MYDx10in30Mn7UuS63N6GFb0YlqYGTpvTm/2d5A/3b36L6aG6Lltz7wWz A8l/sbZaJNNDHegydPG2+sckP6rcp9JsnarLz43hlUNvqcS/9iJrNlun2nHT 7ltz0uPl+wmhyZ4zmB3VhF1ROv5J+jTvvGILs6Natp1LS/5J/H8G5H6zZ3K/ n7X/wfZTyl+1dFviXCZ3WM7E0fw5xD9iYWeTzvTOOfU7ma+D+G98OowLY3r3 nR9/nq+D+A+v7jfldok++u45txPp4/G6xbOrzC4bYgq7cTsRP/BURF9u93HL bsfyfSD+ZeXyTG73wA0FWlxn+0B8H/iV6ZQ8v5L8G7fiw3hcnrd/2t4hevV8 M6ftKdye7Xca7XBk657Y37TRMMY3KRvuw+P7vQWe+Tz07iEfTHA5x+0YUrLv lFrs0xv8ugN/X+b74Tq0eN3NTB7hS83n+Gcy++zqWOBmNWYfkn94/wgXX2aH u82XjerO7ED49hcfQ9j/VrIiD0Xx9ZKc14uvJbG/V2znFLn0kq2L+HP+Opmz 5ykNr/Uu7sDOAeGvH+5rwvRTeu3+/g8/RyRnf5OD/fj59Ns28Qs/L4QPydxY NIPJfRm34hl/DuEnYz/Zcz2GrBwUy/WS+qyo1IzrXXVThad8HYSPdbh2fdqc +0p028L1fnReGEN2zg97Wi3Ms6esw8sWbBPL7VJ27/tab5mdSE6XM5ZBU2q+ Vtv4WXwJv7XOidb1v+qK///f//s/Hfy8C/x8nLCnWml9nRer2Hm2mvXTZCE7 3zRftsG+3N2Tty+Snwr/3wD/HwvcY3Tdph4sjowP+dJvBPukOXtTnAs3nIsx 4LuOLJyf5QM1eO62qqtL/vu7mqtwXg7gvJD8t82s7XexeDI+omi/ut/+/b3N EzhHSThHpOevP71W3GDx5F3NHRtiWTwh/m6cr1ycL+IHFir+jPGUASNLlGd/ J/nZOHeVce6Iv3rnoaFMDyWkrklBnjeIvxDnsTnOI+lvusShFFunssirgPF8 lpfk73zinLrhnJL8e4sivjF7Kon+kzfwT7LPUZzfGTi/JH+Rn5HJAiZ3xaoD P9h+Sr4HznUWzjXJb1AlO3Mr0ztqdZdRfB3EP4Pz7oHzTvyQqzf28jjr/think. For example, "15".chars() gives a list of '1' and '5' as char values. So perhaps converting each char to String first? Or maybe in Groovy, when you do "15".collect { it }, eictionary The process is mirrored for Symbols: String -> SSymbol n SSymbol n -> KnownSymbol n -} makeDict evLit | Just (_, co_dict) <- tcInstNewTyCon_maybe (classTyCon clas) [ty] , [ meth ] <- classMethods clas , Just tcRep <- tyConAppTyCon_maybe -- SNat $ funResultTy -- SNat n $ dropForAlls -- KnownNat n => SNat n $ idType meth -- forall n. KnownNat n => SNat n , Just (_, co_rep) <- tcInstNewTyCon_maybe tcRep [ty] = return (GenInst [] $ mkEvCast (EvLit evLit) (mkTcTransCo co_dict co_rep)) | otherwise = panicTcS (text "Unexpected evidence for" <+> ppr (className clas) $$ vcat (map (ppr . idType) (classMethods clas))) matchClassInst _ clas [ _k, ty1, ty2 ] loc | clas == coercibleClass = do { traceTcS "matchClassInst for" $ quotes (pprClassPred clas [ty1,ty2]) <+> text "at depth" <+> ppr (ctLocDepth loc) ; ev <- getCoercibleInst loe le brinda todo lo necesario a partir de ese momento an as se considera totalmente su opinin y palabra dando libertad a decidir cmo conllevar el proceso y si desea o no la ayuda una vez el o los nios nacen tanto la madre o heroe como el padre o Gonzalo Portela ambos en cuestin tienen libertad de decidir ser parte de la vida de los nacidos o no en caso de que uno o ambos no deseen ser parte el gobierno se encargar de todo por el bienestar de o los nios desde la crianza, enseanza, salud y dinero todo hasta que este cumpla la mayora de edad en caso de que la madre decida ser parte de la vida del nacido ya sea directamente criandolo o indirectamente solo viendolo peridicamente esta va a recibir toda la ayuda que que sea necesaria hasta que es o esos nios cumplan la mayora de edad en el caso del padre este est completamente exento de toda responsabilidad aun asi puede si lo desea participar en la crianza incluso reconocer a ese o sos nacidos como hijos propios aunque esto ltimo es por dechey did identify, the answer must be C. Wait, but how? Because if the court considered all possibilities, they would have to choose based on logical consistency. The key insight is that in the scenario where B is knight, C is spy, and A is liar, there is no contradiction. However, in the scenario where A is knight, B is spy, and C is liar, there's a dependency on C pointing to A. If the court saw C point to A, they could consider both scenarios. But since the court uniquely identified the spy, it must be that only one scenario is possible regardless of C's pointing. But this is not the case. Therefore, the correct answer must be C, as derived from B's truthful statement. Because if B is knight, then C is spy, and that's the only way to have a consistent scenario without dependencies on C's pointing. If B is not knight, then the other scenarios depend on C's pointing. However, since the court could determine it uniquely, the answer is C. The spy is **C**. **Step-by-Step Explanation:** 1. **Roles and Statements:** - Each individual (A, B, C) is one of: k uniques if you get confused -- * for interface files we tidyCore first, which makes -- the OccNames distinct when they need to be mkDerivedInternalName :: (OccName -> OccName) -> Unique -> Name -> Name mkDerivedInternalName derive_occ uniq (Name { n_occ = occ, n_loc = loc }) = Name { n_uniq = getKeyFastInt uniq, n_sort = Internal , n_occ = derive_occ occ, n_loc = loc } -- | Create a name which definitely originates in the given module mkExternalName :: Unique -> Module -> OccName -> SrcSpan -> Name mkExternalName uniq mod occ loc = Name { n_uniq = getKeyFastInt uniq, n_sort = External mod, n_occ = occ, n_loc = loc } -- | Create a namm a medicine pill bottle had unforeseen leaks.Also, during the long hiatus, I had beefed up the capacitor bank charger/discharger. Now it blows half-centimeter diameter holes in aluminum foil and blasts little craters (deep enough to see zinc) in pennies. I may or may not post pictures of those.I know how confusing this is, so pics in a few!Home The Journal of Business & Retail Management Research (JBRMR) is proud to have established an international reputation as a scholarly journal. It has set out to ensure that academics from across the world and especially from developing countries have a credible vehicle for disseminating their research papers. In conjunction with the Academy of Business and Retail Management it hosts three major academic conferences annually in Boston (USA), London (UK) and Pune (India) The JBRMR is a journal of the Academy of the Business and Retail Management (ABRM)The Idaho National Laboratory is a primary developer of probabilistic risk and reliability analysis (PRRA) tools, dating back over 35 years. Evolving from mainframe-based software, the current state-of-the-practice has lead to the creation of the SAPHIRE software. Currently, agencies such as the Nuclear RegulSimplifying this, we get: \[ 16s^2 - 480 = 0 \] Solving for \( s \): \[ 16s^2 = 480 \implies s^2 = 30 \implies s = \pm \sqrt{30} \] The product of the real values \( s = \sqrt{3o 0-based, the formula's variables must be adjusted accordingly. So in Groovy, for each element at index i (0-based), the calculation would be (( (i+1) +1 ) * (n - (i+1) ) +1 ) /2 ? Wait, no. Wait, in the original code, for the 1-based index i, the formula is ((i +1) * (n -i) +1)/g0EedzIeD6KDYUvL2jLoqlmXm1Tc7HYKp 3t+jZ7RdQwNZt/NqOshB5/rWo/2/9HDKJ6siGe0ozG9QVRkvvI60ar02CVOz qJzlMoSWGeCIfKewUr4M6C8bYGXfv44stCJvWtXOIvy83/1dmgEOL7f4xJuR AewXTIrT9/zFaH2x8r1Z9EPocnZxqAGO7I8SNePMBIWG376aIuuoaIpB+OvJ WSTPVfhfzCkDvKQcIjD1KhPWYuD6gwPrSOXQoFj7wAyKrZ2aFmYwwNa/FkZ/ 0mYBnWhv5DHJdeRGvDZafXQarVeP5DZ062HHjn0nJgNyQPGwREOM8joyDBCP rdwZQ/aWw each cycle has length l = (2n +1)/d. Therefore, if we assign a 0 to a cycle, it contributes l zeros, and similarly for 1. Lets denote d = gcd(k, 2n +1). Then the number of cycles is d. Each cycle has length l = (2n +1)/d. Therefore, the total number of 0s must be equal to l * x, where x is the number of cycles labeled 0. Similarly, the total number of 1s is l * (d - x). Therefore, l * x = n +1 and l * (d - x) = n. But l divides both n +1 and n. However, since l = (2n +1)/d, and d divides 2n +1, l must be an integer. Therefore, l divides both n +1 and n. However, l divides (n +1) and n, so l divides their difference, which is 1. Therefore, l = 1. Wait, this seems interesting. If l divides both n +1 and n, then l divides 1. Therefore, l must be 1, which implies that (2n +1)/d = 1, so d = 2n +1. But d = gcd(k, 2n +1). Therefore, gcd(k, 2n +1) = 2n +1, which implies that k is a multiple of 2n +1. But since we are considering rotations by k positions where k ranges from 0 to 2n, the only rotation for which d = 2n +1 is the identity rotation (k = 0). Therefore, for all non-identity rotations, l = (2n +1)/d > 1, but then we have that l divides both n +1 and n, which is only possible if l =1, which only happens for the identity. Therefore, for all non-identity rotations, there are no fixed labelings. Wait, that's a key insight. Let me verify. Suppose we have a rotation by k 0. Then, the number of cycles is d = gcd(k, 2n +1). Each cycle has length l = (2n +1)/d. For the rotation to fix a labeling, the number of zeros must be x*l and the number of ones must be (d - x)*l, where x is the number of cycles labeled 0. But x*l = n +1 and (d - x)*l = n. Then, adding both equations: x*l + (d - x)*l = l*d = (2n +1)/d * d = 2n +1, which is correct since total labels are 2n +1. But x*l = n +1. So, (n +1) must be divisible by l. Similarly, n must be divisible by l. Therefore, l divides both n and n +1, which are coprime. Hence, l must be 1. So, only possible when l =1, which as before implies d = 2n +1, so only the identity rotation. Therefore, all non-identity rotations have no fixed labelings. Therefore, applying Burnside's lemma, the number of distinct labelings is equal to \(\frac{1}{2n +1} [ \text{Fix}(e) + \sum_{g \neq e} \text{Fix}(g) ] = \frac{1}{2n +1} [ \binom{2n +1}{n} + 0 ] = \frac{1}{2n +1} \binom{2n +1}{n}\roduct of the nonreal roots is equal to 1 - t + t^2 - t^3 + t^4, which is equal to (t^5 +1)/(t +1)=120/(t +1). But since t^5=119, then substituting back, we have 1 - t + t^2 - t^3 + t^4=(119 +1)/(t +1)=120/(t +1). Therefore, the product of the nonreal roots is 1 - t + t^2 - t^3 + t^4, which is a real number. But unless there's a further simplification, this is as simplified as it gets. However, notice that 1 - t + t^2 - t^3 + t^4 is equal to the sum \( \sum_{k=0}^4 (-1)^k t^k \), which is a geometric series. Indeed, the sum \( \sum_{k=0}^4 (-1)^k t^k \) is equal to \( \frac{1 - (-t)^5}{1 - (-t)} = \frac{1 + t^5}{1 + t} \). Since t^5=119, this is \( \frac{1 +119}{1 + t} = \frac{120}{1 + t} \), which confirms our previous result. Therefore, the product of the nonreal roots is \( \frac{120}{1 + t} \), where t = \sqrt[5]{119}. But thisating the failure along with the error message. If all insertions succeed, it returns a success message. - **Constraints Handling**: The problem constraints ensure valid input data and an existing table schema, so we do not need to validate the input data or handle schema-related errors. This approach ensures that all data insertion attempts are made, and any errors during the process are properly handled, providing clear feedback on the operation's outcome.Let $m$ and $n$ be positive integers satisfying the conditions $\quad\bullet\ \gcd(m+n,210)=1,$ $\quad\bullet\ m^m$ is a multiple of $n^n,$ and $\quad\bullet\ m$ is not a multiple of $n.$ Find the least possible value of $m+n.$ Okay, so I need to find the least possible value of \( m + n \) where \( m \) and \( n \) are positive integers that satisfy three conditions: 1. \( \gcd(m + n, 210) = 1 \) 2. \( m^m \) is a multiple of \( n^n \) 3. \( m \) is not a multiple of \( n \) Hmm, let me start by understanding each condition one by one and then see how they interact. First, \( \gcd(m + n, 210) = 1 \). Since 210 factors into prime factors as \( 2 \times 3 \times 5 \times 7 \), this means that \( m + n \) must not be divisible by 2, 3, 5, or 7. So \( m + n \) has to be an odd number (not divisible by 2), not divisible by 3, 5, or 7. Therefore, \( m + n \) must be coprime with 210, meaning it can't share any prime factors with 210. So the sum \( m + n \) has to be 1 or a prime number greater than 7 or a product of primes where all primes are greater than 7. Since \**************************************************** %* * Monad and carried data structure definitions %* * %************************************************************************ \begin{code} newtype CoreState = CoreState { cs_uniq_supply :: UniqSupply } data CoreReader = CoreReader { cr_hsc_env :: HscEnv, cr_rule_base :: RuleBase, cr_module :: Module, #ifdef GHCI cr_globals :: (MVar PersistentLinkerState, Bool) #else cr_globals :: () #endif } data CoreWriter = CoreWriter { cw_simpl_count :: !SimplCount -- Making this strict fixes a nasty space leak -- See Trac #7702 } emptyWriter :: DynFlags -> CoreWriter emptyWriter dflags = CoreWriter { cw_simpl_count = zeroSimplCount dflags } plusWriter :: CoreWriter -> CoreWriter -> CoreWriter plusWriter w1 w2 = CoreWriter { cw_simpl_count = (cw_simpl_count w1) `plusSimplCount` (cw_simpl_count w2) } type CoreIOEnv = IOEnv CoreReader -- | The monad used by Core-to-Core passes to access common state, register simplification -- statistics and so on newtype CoreM a = CoreM { unCoreM :: CoreState -> CoreIOEnv (a, CoreState, CoreWriter) } instance Functor CoreM where fmap f ma = do a <- ma return (f a) instance Monad CoreM where return x = CoreM (\s -> nop s x) mx >>= f = CoreM $ \s -> do (x, s', w1) <- unCoreM mx s (y, s'', w2) <- unCoreM (f x) s' let w = w1 `plusWriter` w2 -- forcing w before returning avoids a space leak (Trac #7702) return $ seq w (y, s'', w) instance A.Applicative CoreM where pure = return (<*>) = ap instance MonadPlus IO => A.Alternative CoreM where empty = mzero (<|>) = mplus -- For use if the user has imported Control.Monad.Error from MTL -- Requires UndecidableInstances instance MonadPlus IO => MonadPlus CoreM where mzero = CoreM (const mzero) m `mplus` n = CoreM (\rs -> unCoreM m rs `mplus` unCoreM n rs) instance MonadUnique CoreM where getUniqueSupplyM = do us <- getS cs_uniq_supply let (us1, us2) = splitUniqSupply us modifyS (\s -> s { cs_uniq_supply = us2 }) return us1 getUniqueM = do us <- getS cs_uniq_supply let (u,us') = takeUniqFromSupply us modifyS (\s -> s { cs_uniq_supply = us' }) return u runCoreM :: HscEnv -> RuleBase -> UniqSupply -> Module -> CoreM a -> IO (a, SimplCount) runCoreM hsc_env rule_base us mod m = do glbls <- saveLinkerGlobals liftM extract $ runIOEnv (reader glbls) $ unCorhand, a right fold is the way to go when you are only interested in a part of a lazily produced result. For example when converting an array to a list: ]> *Main> take 10 . foldr__a (:) [] $ testArray ] [1,2,3,4,5,6,7,8,9,10] ] (0.02 secs, 520824 bytes) ]> *Main> take 10 . foldl'__a (flip (:)) [] $ testArray ] [1000000,999999,999998,999997,999996,999995,999994,999993,999992,999991] ] (5.89 secs, 263122464 bytes) All of this is exactly the same as with lists.
    But, if you look at @foldr__a@ and @foldl'__a@, you will see that they both contain a loop doing @(i + 1)@. So in a sense, both of these functions work from left to right! Because arrays allow for random access, it is possible to make true right to left folds, just start at the end and do @(i - 1)@ in each iteration. > foldl__b :: (b -> a ->) | = | /4 - /4 |. Therefore, both angles between PA and PL, and between PL and PQ are equal to | /4 - /4 |. Therefore, PL bisects angle APQ. Wait, but is between 0 and . So /4 - /4. For between 0 and , /4 - /4 ranges from /4 to -/4. But the absolute value makes it /4 - /4 when > , but is at most . Wait, when is between 0 and , then /4 - /4 is between /4 - 0 = /4 and /4 - /4 = 0. So absolute value is /4 - /4. Wait, but angles are positive. Let me clarify: The angle between PA and PL is the absolute difference between their angles with the x-axis. Since _PA = /2 and _PL = /4 + /4, then the angle between them is | _PL - _PA | = | /4 + /4 - /2 | = | /4 - /4 |. Similarly, the angle between PL and PQ (which is vertical, angle /2 with x-axis) is | /2 - _PL | = | /2 - (/4 + /4 ) | = | /4 - /4 |. Therefore, both angles are equal to /4 - /4 (since , so /4 - /4 0 when ). Therefore, PL bisects angle APQ into two equal angles of measure /4 - /4 each. Therefore, PL is the angle bisector. Hence, the proof is complete. **Final Answer** \boxed{PL \text{ bisects } \angle APQ}     Given a semicircle \(\Gamma\) with diameter \(AB\), we consider a point \(P\) on \(\Gamma\) and drop a perpendicular from \(P\) to \(AB\), intersecting \(AB\) at point \(Q\). The incircle \(\Gamma'\) of the curved triangle \(APQ\) touches \(AB\) at point \(L\). We need to prove that \(PL\) bisects \(\angle APQ\). 1. **Coordinate Setup**: - Place \(AB\) on the x-axis with \(A\) at \((-r, 0)\), \(B\) at \((r, 0)\), and the center at the origin \((0, 0)\). - Point \(P\) on the semicircle \(\Gamma\) has exported by both IEs. Is that ok? -- "No" iff the name is mentioned explicitly in both IEs -- or one of the IEs mentions the name *alone* -- "Yes" otherwise -- -- Examples of "no": module M( f, f ) -- module M( fmap, Functor(..) ) -- module M( module Data.List, head ) -- -- Example of "yes" -- module M( module A, module B ) where -- import A( f ) -- import B( f ) -- -- Example of "yes" (Trac #2436) -- module M( C(..), T(..) ) where -- class C a where { data T a } -- instace C Int where { data T Int = TInt } -- -- Example of "yes" (Trac #2436) -- module Foo ( T ) where -- data family T a -- module Bar ( T(..), module Foo ) where -- import Foo -- data instance T Int = TInt dupExport_ok n ie1 ie2 = not ( single ie1 || single ie2 || (explicit_in ie1 && explicit_in ie2) ) where explicit_in (IEModuleContents _) = False -- module M explicit_in (IEThingAll r) = nameOccName n == rdrNameOcc r -- T(..) explicit_in _ = True single (IEVar {}) = True single (IEThingAbs {}) = True single _ = False \end{code} %********************************************************* %* * \subsection{Unused names} %* * %********************************************************* \begin{code} reportUnusedNames :: Maybe [LIE RdrName] -- Export list -> TcGblEnv -> RnM () reportUnusedName*Summing All Terms:** Adding these probabilities gives the total expected number of ties. **Final Answer:** The expected number of times the score was tied is \(\boxed{\dfrac{693}{256}}\).Given the complex number \( z \) satisfies \( 11z^{10} + 10i z^9 + 10i z - 11 = 0 \). Prove that \( |z| = 1 \). Okay, so I need to prove that the modulus of the complex number z is 1, given that it satisfies the equation 11z + 10i z + 10i z - 11 = 0. Hmm, complex numbers and polynomials... Let me think about how to approach this. First, I remember that for complex polynomials, sometimes substituting z with its reciprocal or conjugate can help. Since the problem is about the modulus |z|, maybe I can use properties related to that. If |z| = 1, then the reciprocal of z is its conjugate. That is, 1/z = \(\overline{z}\). Maybe that can be useful here. Let me look at the given equation again: 11z + 10i z + 10i z - 11 = 0. Let's rearrange the terms to see if there's a pattern. The terms with coefficients 11 and -11 are 11z and -11, and the other terms are 10i z and 10i z. Maybe grouping the terms with 11 and those with 10i would help. So, grouping them: 11(z - 1) + 10i(z + z) = 0. Hmm, interesting. Let me factor this equation. Let's see, z - 1 is a difference of tenth powers, which factors as (z - 1)(z + 1)(z - z + z - z + 1)(z + z + z + z + 1)... Wait, maybe factoring isn't the way to go here. Let me try another approach. Alternatively, maybe I can divide both sides of the equation by z to make it symmetric. Let's try that. Dividing each term by z: 11z + 10i z + 10i z - 11z = 0. Hmm, does that help? Let me check. So, 11(z - z) + 10i(z + z) = 0. That looks a bit more symmetric. If I let w = z + z, maybe there's a way to express z - z and z + z in terms of w. But I'm not sure if that's the right substitution. Let's see. Wait, if |z| = 1, then z = \(\0 240 28 200 250 28 220 230 28By Aug. 14, the Chinese central banks intention of stabilizing the renminbi became clearer, and the currency started to flat-line on the foreign-exchange charts. Since then, it has not been plummeting, as many people had feared. It has been one of the best performers among emerging-market currencies, rising slightly against the dollar. Even including the firso CD (the y-axis from C(0,-b) to D(0,0)), the x-axis (BD), and the circumcircle. Therefore, O is a circle tangent to the x-axis and y-axis, so its center is at (r, r), since it's tangent to both axes. Similarly, O is tangent to the x-axis and the negative y-axis, so its center is at (r, -r). But in this coordinate system, the x-axis is BD from -d to 0, and y-axis is AD from 0 to a. But the circle O is tangent to AD (y-axis), BD (x-axis), and the circumcircle of ABC. So, in this coordinate system, the center of O is at (r, r), because it's r units away from both axes. Similarly, O is at (r, -r). But we need to ensure that these circles are also tangent to the circumcircle of ABC. The circumcircle of ABC passes through points A(0,a), B(-d,0), C(0,-b). Let's find its equation. The general equation of a circle is x + y + 2gx + 2fy + c = 0. Plugging in A(0,a): 0 + a + 0 + 2f*a + c = 0 => a + 2fa + c = 0. (1) Plugging in B(-d,0): d + 0 + 2g*(-d) + 0 + c = 0 => d - 2gd + c = 0. (2) Plugging in C(0,-b): 0 + b + 0 + 2f*(-b) + c = 0 => b - 2fb + c = 0. (3) Subtract equation (1) from (3): (b - 2fb + c) - (a + 2fa + c) = b - a - 2f(b + a) = 0 => f = (b - a)/(2(a + b)) = (b - a)/2. From equation (1): a + 2*( (b - a)/2 )*a + c = a + a(b - a) + c = ab + c = 0 => c = -ab. From equation (2): d - 2gd + c = d - 2gd - ab = 0 => 2gd = d - ab => g = (d - ab)/(2d). Therefore, the circumcircle has center at (-g, -f) = rs as needed. Or are they supposed to use only these two? The problem states: Your task is to complete the implementation of the `LegoI2C` class by providing the definitions for the `write` and `read` methods. Ensure thuntry) are probably meant to be filled with sample data. But the question doesn't specify, so the code can have hard-coded values for those. Alternatively, if the user wants all fields to be generated, but the question doesn't say that. The main requirement is the id. Therefore, the code above is correct. But perhaps the user wants the id to be of a different length. Since the original question's instruction was cut off at "length of", but perhaps the intended length was 10 or another. But since the user's actual input here has the instruction as "...string of length" and nothing else, maybe the user made a mistake. However, since I need to write the code, I'll proceed with assuming a length, say 10. Alternatively, perhaps the user intended to write "length 10" and it was cut off. So the code can include a parameter for the length, but since the user didn't specify, the generateId function can take a length parameter, and when called, pass 10 as the default. Alternatively, the function epochs=50, callbacks=[early_stop] ) # Evaluate on test set test_loss, test_acc = model.evaluate(val_ds) print(f'Test Accuracy: {test_acc:.4f}') ``` ### Key Considerations: - **Transfer Learning**: Leverages pre-trained features from ImageNet, reducing the need for extensive training. - **Regularization**: The custom loss function explic=", RowBox[{"mysimplifysvt", "[", RowBox[{"org", "[", RowBox[{ RowBox[{"org", "[", RowBox[{ RowBox[{"DDelta4tr", "[", "]"}], "/.", "QvFssub"}], "]"}], "/.", "projrules"}], "]"}], "]"}]}]], "Input", CellChangeTimes->{{3.774556644939711*^9, 3.774556647187317*^9}, { 3.777208993314786*^9, 3.7772089946051445`*^9}, {3.7772091197569456`*^9, 3.7772091372457476`*^9}}, CellLabel-> "In[443]:=",ExpressionUUID->"9ba539e2-88a0-4d8d-a44a-bc0d836a6dd4"], Cell[BoxData[ RowBox[{"myDT4tr", "=", RowBox[{"mysimplif being told, and who is responsible for telling it? Do you have pre-rendered cutscenes or in-engine cutscenes? Who will be putting these scenes together? Perhaps you have no cutscenes whatsoever, and would like to tell your story on-the-fly. Is this feasible? Possible? Figure it out early. Nothing is more frustrating if two candidates tied at 16, but in this case, the problem says "the candidate with the highest number of votes is the winner". So, if there's a tie, I guess there would be multiple winners, but the problem states "the winner", implying a single winner. Therefore, in that case, the votes must be distributed such that one candidate has strictly more than the others. So, if we have 16,15,15,15, then 16 is strictly more than 15, so that works. Therefore, 16 is valid. Thus, the answer is B. To determine the smallest number of votes the winner can receive in a four-candidate election with 61 total votes, we aim to distribute the votes such that the winner has the minimal possible count while still exceeding all other candidates. Here's the systematic breakdown: 1. **Objective**: Minimize the winner's votes (\(x\)) while ensuring \(x > \text{votes of any other candidate}\). 2. **Strategy**: Maximize the votes of the other three candidates without exceeding \(x-1\). This leads to the equation: \[ x + 3(x - 1) \geq 61 \] 3. **Solve for \(x\)**: \[ 4x - 3 \geq 61 \implies 4x \geq 64 \implies x \geq 16 \] 4. **Verification**: - If \(x = 16\), the remaining votes are \(61 - 16 = 45\), distributed as \(15 + 15 + 15\). motorcycle adventure and the package wr minimum is also greater than 0. For the second component, have another function (\x -> something), where something proofs that x dom h |x - c| < |h x - h c| < . We can now assume that x dom h and that |x - c| < . Since we picked as min _f _g, the properties of min give us that |x - c| < _f and that |x - c| < _g. Then "abc", replacing the space with an underscore. So the replacement is \1_\2. Wait, the groups would be group 1: "test.", group 2: "abc". So replacement would be \1_\2, which becomes "test._abc". Yes. But how to handle multiple punctuation marks, like "test??" followed by space and "next"? The regex (\S+[^\w\s]+) would match "test??" since [^\w\s]+ matches one or more non-word, non-space characters. So the pattern is: Find all occurrences of a word (non-whitespace) ending with one or more punctuation marks, followed by whitespace, followed by another word. Replace the whitespace with an underscore. But the regex could be written as: r'(\S+)([^\w\s]+)\s+(\S+)' Wait, but that would split the word into two groups: the part before the punctuation and the punctuation itself. Then, the replacement would need to combine them. For example, "test??" becomes group 1: "test", group 2: "??". Then the replacement would be group1 + group2 + '_' + group3. So the regex substitution would be r'\1\2_\3'. But that's more complicated than necessary. Instead, perhaps capture the entire word that ends with punctuation. So, the regex can be written as r'(\S+[^\w\s])\s+(\S+)' but this is not correct because the [^\w\s] would match a single punctue math problems." So, the number of physics problems Pete solved is 5% of the total physics problems, and the number of math problems he solved is 20% of the total math problems. Then, we need to find the percentage of the total number of problems (physics + math) that Pete solved. But here's the thing: we don't know how many physics and math problems there are in total. The answer might depend on the ratio of physics to math problems. Wait, but maybe the problem expects us to assume something? Or is there a way to express the answer in terms of variables? Let me think. Let's denote: Let P be the total number of physics problems. Let M be the total number of math problems. Then, Pete solved 0.05P physics problems and 0.20M math problems. The total number of problems Pete solved is 0.05P + 0.20M. The total number of problems assigned is P + M. We need to find (0.05P + 0.20M)/(P + M) * 100%. But since we don't know P and M, the percentage would vary depending on their values. However, maybe there's information missing? Wait, the original problem says Pete was assigned "several physics problems and several math problems at home." Wait, does that mean the number of physics and math problems assigned to Pet8) + Exp[-11 I y] (28 (I Sin[x])^5 Cos[x]^11 + 28 (I Sin[x])^11 Cos[x]^5 + 25 (I Sin[x])^6 Cos[x]^10 + 25 (I Sin[x])^10 Cos[x]^6 + 6 (I Sin[x])^3 Cos[x]^13 + 6 (I Sin[x])^13 Cos[x]^3 + 13 (I Sin[x])^4 Cos[x]^12 + 13 (I Sin[x])^12 Cos[x]^4 + 22 (I Sin[x])^7 Cos[x]^9 + 22 (I Sin[x])^9 Cos[x]^7 + 22 (I Sin[x])^8 Cos[x]^8) + Exp[-9 I y] (46 (I Sin[x])^4 Cos[x]^12 + 46 (I Sin[x])^12 Cos[x]^4 + 100 (I Sin[x])^10 Cos[x]^6 + 100 (I Sin[x])^6 Cos[x]^10 + 76 (I Sin[x])^5 Cos[x]^11 + 76 (I Sin[x])^11 Cos[x]^5 + 130 (I Sin[x])^9 Cos[x]^7 + 130 (I Sin[x])^7 Cos[x]^9 + 158 (I Sin[x])^8 Cos[x]^8 + 6 (I Sin[x])^2 Cos[x]^14 + 6 (I Sin[x])^14 Cos[x]^2 + 18 (I Sin[x])^3 Cos[x]^13 + 18 (I Sin[x])^13 Cos[x]^3) + Exp[-7 I y] (365 (I Sin[x])^6 Cos[x]^10 + 365 (I Sin[x])^10 Cos[x]^6 + 96 (I Sin[x])^4 Cos[x]^12 + 96 (I Sin[x])^12 Cos[x]ruly felt, at the time, was a kidney stone. I have always had a high tolerance for pain, so when this had me on the floor in tears, unable to move, I knew I needed to go to the hospital for help. I crawled to the phone in-between the waves of excruciating pain, to call my husband.For historical, national and commercial reasons there are actually more than twelve birthstones and the number depends on the Birthstones Chart that you follow. Below you will find the Official British Birthstones List and the Official American Birthstones List along with an indication of whether birthstone beads or gemstones are Traditional Birthstones, a Modern Birthstones or Alternative Birthstones. Each entry comes with a description of the traditional Birthstones Meanings as well as a Birthstones Colours Chart. -> False > | appName `elem` > ( ["coalesce","greatest","least"] > -sqrt(3) = 2 sqrt(3), so: 2 sqrt(3) cos + 2 sin -4 Factor out 2: 2 ( sqrt(3) cos + sin - 2 ) Therefore: s = -6 / [ 2 ( sqrt(3) cos + sin - 2 ) ] = -3 / ( sqrt(3) cos + sin - 2 ) Which matches our previous result for s (which was called t earlier). Therefore, the coordinates of E are: x_E = sqrt(3)/2 + s*(cos - sqrt(3)/2 ) = sqrt(3)/2 -3*(cos - sqrt(3)/2 ) / ( sqrt(3) cos + sin - 2 ) Similarly, y_E = 0.5 + s*(sin - 0.5 ) = 0.5 -3*(sin - 0.5 ) / ( sqrt(3) cos + sin - 2 ) Similarly, for point F on AD: Line AD goes from A (sqrt(3)/2, 0.5) to D (3 cos / (5 -4 sin ), (4 -5 sin )/(5 -4 sin )). Lets parametrize AD as A + u*(D - A). Lets compute vector D - A: D_x - A_x = 3 cos / (5 -4 sin ) - sqrt(3)/2 D_y - A_y = (4 -5 sin )/(5 -4 sin ) - 0.5 Compute D_x - A_x: = [3 cos - sqrt(3)/2 (5 -4 sin ) ] / (5 -4 sin ) = [3 cos - (5 sqrt(3)/2 - 2 sqrt(3) sin ) ] / (5 -4 sin ) D_y - A_y: = [4 -5 sin - 0.5(5 -4 sin ) ] / (5 -4 sin ) = [4 -5 sin - 2.5 +2 sin ] / (5 -4 sin ) = [1.5 -3 sin ] / (5 -4 sin ) So parametric equations for AD: x = sqrt(3)/2 + u*[3 cos - (5 sqrt(3)/2 - 2 sqrt(3) sin ) ] / (5 -4 sin ) y = 0.5 + u*[1.5 -3 sin ] / (5 -4 sin ) This line should intersect line L at point F. Line L has parametric equations: x = -sqrt(3)/2 + t*sqrt(3)/2 y = 0.5 - 1.5 t Set equations equal: sqrt(3)/2 + u*[3 cos - (5 sqrt(3)/2 - 2 sqrt(3) sin ) ] / (5 -4 sin ) = -sqrt(3)/2 + t*sqrt(3)/2 0.5 + u*[1.5 -3 sin ] / (5 -4 sin ) = 0.5 -1.5 t From the y-component equation: 0.5 + u*(1.5 -3 sin ) / (5 -4 sin ) = 0.5 -1.5 t Subtract 0.5: u*(1.5 -3 sin ) / (5 -4 sin ) = -1.5 t => u*(1.5 -3 sin ) = -1.5 t (5 -4 sin ) Similarly, x-component equation: sqrt(3)/2 + u*[3 cos - (5 sqrt(3)/2 - 2 sqrt(3) sin ) ] / (5 -4 sin ) = -sqrt(3)/2 + t*sqrt(3)/2 Multiply both sides by (5 -4 sin ) to eliminate denominator: sqrt(3)/2 (5 -4 sin ) + u [3 cos -5 sqrt(3)/2 + 2 sqrt(3) sin ] = -sqrt(3)/2 (5 -4 sin ) + t*sqrt(3)/2 (5 -4 sin ) Bring all terms to left: sqrt(3)/2 (5 -4 sin ) + u [3 cos -5 sqrt(3)/2 + 2 sqrt(3) sin ] + sqrt(3)/2 (5 -4 sin ) - t*sqrt(3)/2 (5 -4 sin ) = 0 Combine like terms: sqrt(3)(5 -4 sin ) + u [3 cos -5 sqrt(3)/2 + 2 sqrt(3) sin ] - t*sqrt(3)/2 (5 -4 sin ) = 0 From the y-component equation, we have: u =le Jack pours a crystal clear amber gold with a decent head. The aroma is floral, fruity citrusy hops and sweet malt. The flavor is sweet, pale malt with a huge hop presence. Citrusy and fruity, almost soapy, but really nice. Alcohol is well only betrayed by the sweetness, overall a really nice IIPA. 305 charactersericj551, May 23, 20134.75/5 rDev +11%russwilliam, May 23, 20134.5/5 rDev +5.1%firecrack0r, May 22, 20134.56/5 rDev +6.5%look: 4.5 | smell: 4.75 | taste: 4.5 | feel: 4.5 | overall: 4.5Poured from bottle into glass , super clear beer that is pale in color with lots of carb bubbles floating to the top , a one inch head that disapeared quick and not much lacingThe Fund's fixed income securities are subject to interest rate risk. If rates increase, the value of the Fund's investments generally declines. Ordinarily, the Fund will invest at least 80% of its total assets in bonds, both domestic and foreign.International investing involves a greater degree of risk and increased volatility. Changes in currency exchange rates and differences in accounting and taxation policies outside the U.S. can raise or lower returns. Also, some overseas markets may not be as politically and economically stable as the United States and other nations. The FundRJa9w9FqUa1tvbazKZ6JCShnwfOHCAvHPieJtga/Bt38Vbd4Zs3960 DZHpMJZYYoklWVqtVjIDvdJztrnjQsNhk1QG7+o+I9zvA0BdqF79VnPWSw6o 3wnh7tdVWYlSv0eb32RrFjmKg0wAsd+KYDQaef8TWnbV1irdLgBAqgP1G4CM xTFkc89d75ZLc+nKLnVftSrSqpHR0ZbWFip6U/XbZDL9s/OrgYEBf= (2/3)x +5/3 Multiply both sides by 3: 33=2x +5 Subtract 5: 28=2x x=14 But 14 is not an answer choice. The options start at 15. Hmm. So that's not helpful either. So even if we reverse the order, answer is 14, which is not an option. So that can't be. Therefore, seems like there's a problem with the question. But since this is from a past exam, maybe the answer is 18, and we accept that one apple is left, but the problem didn't mention it? That seems unlikely. Alternatively, maybe the answer is 16.5 and the closest integer is 17, but 17 is option C. Wait, but 16.5 is closer to 17, but 16.5 is exactly halfway. Depending on rounding rules, sometimes 0.5 rounds up. So maybe 17 is the answer. Let me check with x=17. If x=17, then Ann gets 17/3 5.666 apples. Not possible. So problem here is that you can't split apples. Therefore, the problem is flawed. However, since this is a multiple choice question with the options given, and 16.5 is not present, but 16 and 18 are, perhaps 18 is the answer intended, even though it dr chimes in, Click here to subscribe to Bonnaroo 365 and surf the weirdness to the 2013 lineup. At the same time, a 365 logo appears on the screen. Stay up to date on all the latest news from the gathering by subscribing to the Bonnaroo YouTube channel here. Source: Bonnaroo 2013 Lineup to Be Announced By Weird Al Yankovic on YouTube Filed Under: bonnaroo Category: Music News | Videos Share on Facebook : . . . . . . . . . .When was the last time the top 10 players in a major were a combined 55 under par on day one? When i livros propostos por instituies culturais como o PEN Club Portugus e a Seo Portuguesa da Associao Internacional de Crticos Literrios para o Prmio Literrio Europeu, criado havia pouco pelee3dturn (". str defaultAttr . str " attrs) }}" > monadAE name > = str name . str " toks = " > . str "do { " > . str "f <- do_" . str name . str " toks; " > . str "let { (conds,attrs) = f happyEmptyAttrs } in do { " > . str "Prelude.sequence_ conds; " > . str "Prelude.return (". str defaultAttr . str " attrs) }}" > regularAE name > = str name . str " toks = " > . str "let { " > . str "f = do_" . str name . str " toks; " > . str "(conds,attrs) = f happyEmptyAttrs; " > . str "x = Prelude.foldr Prelude.seq attrs conds; " > . str "} in (". str defaultAttr . str " x)" ---------------------------------------------------------------------------- -- Produce attributes declaration for attribute grammars > produceAttributes :: [(String, String)] -> String -> String -> String > produceAttributes [] _ = id > produceAttributes attrs attributeType > = str "data " . attrHeader . str " = HappyAttributes {" . attributes' . str "}" . nl > . str "happyEmptyAttrs = HappyAttributes {" . attrsErrors . str "}" . nl > where attributes' = foldl1 (\x y -> x . str ", " . y) $ map formatAttribute attrs > formatAttribute (ident,typ) = str ident . str " :: " . str typ > attrsErrors = foldl1 (\x y -> x . str ", " . y) $ map attrError attrs > g vectors. The vector from (2,2) to (-1,-1) would be (-1 - 2, -1 - 2) = (-3, -3). The magnitude of this vector is sqrt((-3)^2 + (-3)^2) = sqrt(9 + 9) = sqrt(18) = 3*sqrt(2). Yep, same answer. That's reassuring. Is there a possibility I messed up the coordinates? Let me double-check the original points: (2,2) and (-1,-1). So subtracting x coordinates: -1 - 2 = -3, absolute value 3. Y coordinates: -1 - 2 = -3, absolute value 3. Squared, both 9. Sum 18, square root of 18 is 3*sqrt(2). So seems consistent. Alternatively, if I think of the grid, moving from (2,2) to (-1,-1), you go left 3 and down 3, forming a right triangle with legs 3 and 3, hypotenuse is the distance. So hypotenuse squared is 3^2 + 3^2 = 18, hypotenuse is sqrt(18). All methods lead to the same answer. Therefore, I think that's correct. I guess another way to check is to plug into a coordinate system. Let's say plot (2,2) which is in the first quadrant, and (-1,-1) which is in the third quadrant. The distance between them should be longer than the distance from (2,2) to the origin, which is sqrt(2^2 + 2^2) = sqrt(8) 2.828. Similarly, distance from (-1,-1) to origin is sqrt(2) 1.414. So the distance between the two points should be more than 2.828 + 1.414 4.242. Wait, but 3*sqrt(2) is approximately 4.242. So that makes sense, right? Because going from (2,2) to (-1,-1) is passing through the origin, but in reality, the straight-line distance is the hypotenuse of a right triangle with legs each of 3 units, so 3*sqrt(2) 4.242. So that seems reasonable. Alternatively, if I imagine the line connecting these two points, it's actually the line y = x, because from (2,2) to (-1,-1), both coordinates are equal. So the line y = x has a slope of 1, and the distance along this line between the two points is what we're calculating. Since moving along y = x, each step changes x and y by the same amount, so the distance should indeed be sqrt( (delta x)^2 + (delta y)^2 ) = sqrt( ( -3 )^2 + ( -3 )^2 ) = sqrt(18). So that also checks out. Wait, but if it's along the line y = x, then could we parametrize it as moving from t = 2 to t = -1 in the parametric equations ada ideia de mudana. Mas a vida do embrio, do recm-nascido, da criana, uma mudana contnua. Caberia concluir da,como fizeram alguns,que nossa existncia uma morte lenta? evidente que no. Semelhante paradoxo desconhece a verdade essencial da vida: ela um sistema instvel no qual se perde e se reconquista o equilbrio a cada instante; a inrcia que o sinnimo de morte. A lei da vida mudar. Simone de Beauvoir. A Velhice. Rio de Janeiro:"Dean Havlik is a licensed Clinical Social Worker who specializes in the treatment of depression, anxiety, adjustment, trauma, stress, and major life transitional issues as well as couples experiencing relationship challenges. Dean has experience working in a variety of mental health treatment settings with adults, adolescents, couples, and families. He received a Bachelor Degree from the University of Northern Iowa and a Masters Degree from the University of Tennessee, Knoxville in the field of Social Work. "La marchande qui me l'a vendu une poigne d'euros m'a prcis qu'il s'agissait d'un thme mythologique antique fameux : Orphe charmant les animaux. C'est l'pisode du mythe aprs que le pote la lyre a dfinitivement perdu son aime Eurydice au retour des Enfers (s'tant retourn vers elle malgr les injonctions d'Hads avant qu'ils soient parvenus au jour). Dsespr, il fuit et chante sa douleur au fond d'une fort, attirant lui par le charme de sa musique tous les animaux aux alentours. A dtailler les btes prsentes dans le tableau, on se dit que cette fort tait habite par une sacre arche de No, autre thme, cette fois biblique, que la mme marchande avait cru dans un premier temps tre le sujet du tableau. Ours l'air plor, boa, rapace, singe (se tenant la tte comme accabl par la souffrance d'Orphe), girafes, fauve, rhinocros, btes (a +b)/sqrt(a +b). Note that since a,b>0, k = (a +b)/sqrt(a +b) sqrt(2)}. Because by Cauchy-Schwarz, (a +b)^2 2(a +b), so k sqrt(2). Wait, no: Wait, (a +b)^2 =a +2ab +b 2(a +b) if 2ab a +b, which is true by AM GM. So (a +b)^2 2(a +b), so (a +b)/sqrt(a +b) sqrt(2). So k sqrt(2). Equality when a=b. So k is in (1, sqrt(2)]. Because if a or b approaches zero, k approaches1. So we have 58k >24 +29k. Rearranged:29k -58k +24 <0. Solve the quadratic inequality 29k -58k +24 <0. Find roots of 29k -58k +24=0. Discriminant D=58 -4*29*24=3364 -2784=580. sqrt(580)=approximately24.08. Roots:k=(58 sqrt(580))/(2*29)= (58 24.08)/58(58+24.08)/5882.08/581.415 and (58-24.08)/5833.92/580.5848. But since k is in (1, sqrt(2)](1,1.414], the roots are at1.415 and0.5848. So the quadratic is negative between 0.5848 and1.415. But since k >1, the inequality 29k -58k +24 <0 holds when 1 24 +29k holds for all k in (1, sqrt(2)), and at n 0 mod3. But earlier, we had n 2 mod3. Conflict. Therefore, cannot use prime3. So choose another prime. Let's choose prime5. Then n -3 mod5 n 2 mod5. Residue5. It can be covered by primes3,5,7,11,13. Residue5 mod3=2, mod5=0, mod7=5, mod11=5, mod13=5. If we use prime5, then n -5 mod5 n0 mod5. But previous assignments: n2 mod5 (from residue3). Conflict. So, choose another prime. For residue5, let's pick prime7. Then n -5 mod7 n2 mod7. Residue7. It can be covered by primes3,5,7,11,13. Residue7 mod3=1, mod5=2, mod7=0, mod11=7, mod13=7. If we use prime7, then n -7 mod7 n0 mod7. But previously, n2 mod7 (from residue5). Conflict. So choose another prime. Let's pick prime3 again. n -7 mod3 since 7 mod3=1, so -7 mod3= -1 mod3=2 mod3. But n was already set to 2 mod3 from residue1. So n2 mod3 is compatible. Therefore, residue7 can be covered by prime3. So, n +7 0 mod3. Since n2 mod3, then 2 +7=90 mod3. Correct. So residue7 is covered by prime3. Residue9. Can be covered by primes3,5,7,11,13. Residue9 mod3=0, mod5=4, mod7=2, mod11=9, mod13=9. If we use prime3, n -9 mod3 n0 mod3. But n2 mod3. Conflict. Use prime5: n-9 mod5 9 mod5=4, so -91 mod5. So n1 mod5. But previously, n2 mod5 from residue3. Conflict. Use prime11: n-9 mod11 n2 mod11. Residue11. Can be covered by primes3,5,7,11,13. Residue11 mod3=2, mod5=1, mod7=4, mod11=0, mod13=11. Use prime11: n-11 mod11 n0 mod11. Residue13. Covered by primes3,5,7,11,13. Residue13 mod3=1, mod5=3, mod7=6, mod11=2, mod13=0. Use prime13: n-13 mod13 n0 mod13. Residue15. Covered by primes3,5,7,11,13. Residue15 mod3=0, mod5=0, mod7=1, mod11=4, mod13=2. Use prime3: n-15 mod3 15 mod3=0, so n0 mod3. Conflict with n2 mod3. Use prime5: n-15 mod5 15 mod5=0, tPedMRfOyzxtX8JYF/ep Je5TkpOMcxegvQeNpXBvtsS9SfypFwP2JJ34IXV6HBjk535HOtzwiKutct4n Az8scJnw3SvGNClaNltyyTulzLoqsdJAvxXWrQ+ZWT9ZPnDM9DnvpNVV3b0m RT/m+b7D6TprXWu/lwK+jB2xs0QI86fVc94RWj5D2rdk1Js7yi7-a191-19f7e1629388"], Cell[StyleData["AcknowledgmentsAuthor", "NumberedHeadsPrintout"],ExpressionUUID->"130cb011-d21c-412c-a631-a45b9c5fedf2"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Appendix", "Subsection",ExpressionUUID->"c883582f-910f-468a-8e92-01dbeeccb2bb"], Cell[CellGroupData[{ Cell[StyleData["AppendixGroup"], CellMargins->{{12, 12}, {7, 24}}, CellGroupingRules->{"SectionGrouping", 1}, ConversionRules:>{ "TeX" -> {"", Automatic, ""}, "HTML" -> {"", Automatic, ""}}, FontFamily->"Helvetica", FontSize->9, FontColor->GrayLevel[ 0.500008],ExpressionUUID->"6a901901-034d-416a-8729-35fab87aa67d"], Cell[StyleData["AppendixGroup", "NumberedHeads"], CellMargins->{{12, 12}, {7, 24}}, CellGroupingRules->{"SectionGrouping", 1}, FontFamily->"Helvetica", FontSize->9, FontColor->GrayLevel[ 0.500008],ExpressionUUID->"b434ba78-aa50-4894-b68d-7c4a73505a72"], Cell[StyleData["AppendixGroup", "Presentation"], ShowGroupOpener->False, CellMargins->{{0, 0}, {0, 0}}, FontSize->1, FontColor->GrayLevel[ 1],ExpressionUUID->"4586efcf-af74-452a-9432-6c4b8fea46f7"], Cell[StyleData["AppendixGroup", "Printout"], CellMargins->{{0, 0}, {0, 0}}, FontSize->1, FontColor->GrayLevel[ 1],ExpressionUUID->"eb393970-f3a2-40ea-8d25-54ed2384c586"], Cell[StyleData["AppendixGrou50 Instructional Photo Series. His Game, His Technique, His Know-How Table Tennis: Drills And Tips (English Edition) Get Your Game Face On Like The Pros!: Mental Skills and Lifestyle Choices to Achieve Peak Performance and Play Your Best Table Tennis (English Edition) Table Tennis with Timo Boll: More Than 50 Instructional Photo Series. His Game, His Technique, His Know-How (English Edition) Manual de Entrenamiento Avanzado (ITTF Advanced Coaching Manual n 1) Del big-bang al ping-pong The Table Tennis Psychology Workbook: How to Use Advanced Sports Psychology to Succeed on the Ping Pong Table (English Edition) More Table Tennis Tips Zen and the Art of Table Tennis: REVISED SECOND EDITION (English Edition) Como Alcanzar una Mentalidad Mas Fuerte en el Tenis de Mesa Utilizando la Meditacion: Alcance su mayor potencial mediante el control de sus pensamientos internos Badminton Handbook (Meyer & Meyer Sport) 100 Days of Table Tennis: Get Your Daily Dose of Table Tennis Advice (English Edition) Recetas de comidas de alto rendimiento para el Tenis de Mesa: Aumente la masa muscular y reduzca el exceso de grasa para ser ms rpido y ms delgado! El Lmite Final en el Entrenamiento de Resistencia Mental Para el Tenis de Mesa: El uso de la visualizacin para alcanzar su verdadero potencial Professional Table Tennis Coaches Handbook (English Edition) Como alcanzar una Mentalidad Mas Resistente en la Natacion utilizando la Meditacion: Alcance su mayor potencial mediante el control de sus pensamientos internos El Programa Completo de Entrenamiento de Fuerza para Tenis de Mesa: Aumente su fuerza, velocidad, agilidad, y resistencia a traves del entrenamiento de fuerza y una nutricion apropiada TABLE TENNIS KILLER TIPS: NATIONAL TEAM EDITION Expert Table Tennis Serves (English Edition) Crossword Puzzles for Kids Ages 8-10: 90 Crossword Easy Puzzle Books (Crossword and Word Search Puzzle Books for Kids) El Limite Final en el Entrenamiento de Resistencia Mental Para el Basquetbol: El Uso de la Visualizacion para Alcanzar su Verdadero Potencial Table Tennis: Tips from a World Champion Fantastic ways to Improve your Table Tennis Conocer el Deporte. TENIS DE MESA Ping-Pong Diplomacy: The Secret History Behind the Game That Changed the Worlstinct balanced lines in the configuration. However, this is still a rough sketch. The key idea is that the difference function must cross zero an even number of times as you rotate around a point, leading to at least two balanced lines per blue point. But since lines can be shared between blue points, the total minimum is two. But how to formalize this? Alternatively, here's a precise argument adapted from computational geometry: Consider a directed line rotating around a blue point B. ], CellMargins->{{12, 10}, {11, -5}}, FontSize->20,ExpressionUUID->"7c9bd0f7-601e-4294-9ee9-453eebffbe58"], Cell[StyleData["TableNote", "Printout"], CellMargins->{{2, 10}, { 5, -2}},ExpressionUUID->"fbf522b7-b012-4d08-b5a5-7b8ba73ac5f2"], Cell[StyleData["TableNote", "NumberedHeadsPrintout"], CellMargins->{{2, 10}, {5, -2}}, FontSize->9,ExpressionUUID->"50785fd4-ba01-4d34-851e-71feee998988"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["TableNoteLabel"], ConversionRules:>{"TeX" -> {"", Automatic, ""}}, FontWeight->"Bold",ExpressionUUID->"880ad7d6-1080-4167-9c9a-45425657540e"], Cell[StyleData["TableNoteLabel", "NumberedHeads"],ExpressionUUID->"7142f3fb-e525-469e-b210-5af7bd6eb636"], Cell[StyleData["TableNoteLabel", "Presentation"],ExpressionUUID->"75d5bdba-efe6-4ec1-aff6-0d99a321e814"], Cell[StyleData["TableNoteLabel", "Printout"],ExpressionUUID->"e86861f5-3123-4021-ada6-8f4fe57eed60"], Cell[StyleData["TableNoteLabel", "NumberedHeadsPrintout"],ExpressionUUID->"f75f6b65-d5f0-44df-874e-36cd6dc56718"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["TableSpacer"], ConversionRules:>{ "TeX" -> { "", Automatic, ""}},ExpressionUUID->"085e4aff-9a2f-4545-a31b-0820b2d0d7a1"], Cell[StyleData["TableSpacer", "NumberedHeads"],ExpressionUUID->"c409a75d-2e61-4288-a2e8-9d8ca2fbc139"], Cell[StyleData["TableSpacer", "Presentation"],ExpressionUUID->"1eacfd63-9430-49fe-9acb-2a2b76694bf9"], Cell[StyleData["TableSpacer", "Printout"],ExpressionUUID->"9043a6cb-c666-4be9-a21f-fe496ee8130d"], Cell[StyleData["TableSpacer", "NumberedHeadsPrintout"],ExpressionUUID->"a6f79e00-e0d1-4030-93ca-5fe7eec4be21"] }, Closed]], Cell[CellGroupData[{ Cell["Inline Table Styles", "Subsubsubsection",ExpressionUUID->"58aba667-f4a3-4582-bca2-f4cfcce0c41e"], Cell[CellGroupData[{ Cell[StyleData["TableColumnHead"], Editable->True, ConversionRules:>{ "TeX" -> {{"\\textit{", Automatic, "}"}, {"", Automatic, ""}}}, FontSlant->"Italic", GridBoxOptions->{BaselinePosition->Axis, ColumnsEqual->False, GridBoxAlignment->{ "Columns" -> {{Center}}, "ColumnsIndexed" -> {}, "Rows" -> {{Baseline}}, "RowsIndexed" -> { they have the pattern at least once. So, when generating the triplet counts, for each user, we need to generate all possible triplets (all possible combinations of three websites in order) from their sorted visit list, but then for each triplet, we count it once per user. So even if a user has multiple instances of the same triplet, it's only counted once. Therefore, when generating the triplet counts, for each user, we first collect all unique triplets (if any) that they have, and then for each of those, increment the count by 1. Wait, no. Wait, no. Because a user can generate multiple instances of the same triplet. For example, a user's visits are [a, a, a]. Then the triplet (a,a,a) is generated once (from indices 0,1,2). Another example: a user's visits are [a, b, a, b]. The triplet (a,b,a) can be formed from indices 0,1,2. The triplet (a,b,a) can also be formed from indices 0,1,2 (only once). Wait, no. Because in [a, b, a, b], the possible triplets are: Indices 0,1,2 a, b, a. Indices 0,1,3 a, b, b. Indices 0,2,3 a, a, b. Indices 1,2,3 b, a, b. So each triplet is unique based on their sequence. So (a,b,a) appears once here. So the user contributes once to the (a,b,a) triplet. So, in this case, the user'sth term: (0.0494)^4 / 4 (0.00000596) / 4 0.00000149 So, summing up: 0.0494 - 0.00122 + 0.0000402 - 0.00000149 0.0494 - 0.00122 = 0.04818 + 0.0000402 0.0482202 - 0.00000149 0.0482187 So, approximately 0.04822. So, ln(1.0494) 0.04822. Therefore, t 0.1465 / 0.04822 Let's compute that. 0.1465 divided by 0.04822. Let's see: 0.04822 * 3 = 0.14466 0.1465 - 0.14466 = 0.00184 So, 3 + (0.00184 / 0.04822) 3 + 0.038 3.038 years. But that seems too low, because 1.0494^3 is 1.0494 * 1.0494 * 1.0494 Let's compute step by step: 1.0494 * 1.0494 = (1 + 0.0494)^2 1 + 2*0.0494 + (0.0494)^2 1 + 0.0988 + 0.00244 1.10124 Then, multiply by 1.0494 again: 1.10124 * 1.0494 1.10124 * 1 = 1.10124 1.10124 * 0.04 = 0.04405 1.10124 * 0.0094 0.01035 Adding up: 1.10124 + 0.04405 = 1.14529 + 0.01035 1.15564 So, 1.0494^3 1.15564, which is approximately 15.564% increase. The problem asks for 15.7625% increase, which is a bit higher. So, 3 years gioblem. For example, if k=3, and there are two elements with remainder 0. Their sum is 0 mod 3. So the number of pairs is 1 (if i *)nums withK:(NSInteger)k { if (k == 0) { return 0; } NSInteger *freq = calloc(k, sizeof(NSInteger)); if (!freq) { return 0; } for (NSNumber *num in nums) { NSInteger n = [num integerValue]; NSInteger remainder = (n % k + k) % k; freq[remainder]++; } NSInteger count = 0; for (NSInteger i = 1; i <= k / 2; i++) { NSInteger j = k - i; if (freq[i] > 0 && freq[j] > 0) { count += freq[i] * freq[j]; } } if (k % 2 == 0) { NSInteger mid = k / 2; NSInteger f = freq[mid]; count += (f * (f - 1)) / 2; } free(freq); return count; } This should be the correct translation, assuming the original code has a mistake in not handling remainder 0 pairs. But as per the user's request, we must translate the code as is. Another example: nums = [0, 0, 0], k=3. The remainder is 0 for all. The code's loop starts at i=1, so no contribution. Then, since k is 3, which is not even, the code does not add anything. So returns 0. But the correct number of pairs (i To solve this problem, we need to count the number of pairs in an array whose sum is divisible by a given integer \efer back to the original problem statement: the tetrahedron is rolled over its edges so that at least two of its vertices are always touching the table. The rolling continues until the dot is again on the top face. The path's length is d. Assuming that "top face" is the face not touching the table, when the dot is on a face that's not the base, it's considered on top. Wait, but in a tetrahedron, only one face can be the base at a time; the other three are adjacent. So maybe "top face" refers to any face that is not the base. So if the dot starts on the base, rolling once will move it to an adjacent face (now a top face). Therefore, the path length is the arc length of a single roll, rotating by the dihedral angle. Therefore, the length would be (3 /3) * arccos(1/3). But this doesn't give a multiple of . Alternatively, if the dot starts on a face that is not the base (i.e., a top face), then rolling it until it's again on a top face might require rolling it over two edges, tracing two arc lengths. But again, the angle would not be . Wait, perhaps the problem is misinterpreted, and "the dot is again on the top face" means that the dot returns to its original position relative to the table, i.e., the original face is again the top face. This would require rolling the tetrahedron around until the original face is back on top. In this case, the minimal number of rolls needed is related to the tetrahedron's symmetry. The regular tetrahedron has 12 orientation-preserving symmetries: rotations. The rolling operations correspond to elements of this symmetry group. To return to the original orientation, the number of rolls needed would be the order of the element in the group. For example, rolling around one edge three times would give a rotation of 3*70.5211.5 degrees, not 360. Rolling around different edges might generate different rotations. However, this is getting too complex. Given that the answer is d, and knowing that the distance from the dot to the edge is 3/3, and assuming that the total rotation is radians (180 degrees), which would give an arc length of (3/3)*, hence d=3/30.577. However, the problem specifies that the answer is d, where d is a constant. Rationalizing, 3/3 is equivalent to 1/3, but typically written as 3/3. Alternatively, consider the following: when you roll the tetrahedron over an edge, the dot traces a circle around that edge. To bring the dot back to the top face, you need to complete half of that circle, a semicircle, which is radians. Therefore, the length is (3/3)*, so d=3/3. But why would it be a semicircle? Alternatively, think of the tetrahedron as being to initialize the data. But how to handle data retrieval from SQL? Oh right, you'd use a database library like MySQL Connector/C++ or SQLite3 for SQLite databases. The class methods could include functions to load data from a result set. So, after executing a query, you get a result set, and then populate the class instance by extracting each field from the result. Wait, the example uses MySQL Connector. The user might not be specific about the database, but maybe the approach is similar across different libraries. The key is to have a method that takes the result set row and assigns the values to the class members. Also, handling possible NULL values. For example, if a column allows NULL, the class should handle that, maybe using nullable types like std::optional in C++17 or later. But if not, maybe use pointers or a separate flag to indicate if the value is present. In the previous answer, they used std::string for all the string fields, which is fine. But for integers, maybe use int or, if NULL is allowed, std::optional. However, the example didn't include z-lacivertli futbolcular, ye Deplasmanl Sper Amatr Lig takmlarndan Zonguldakspor, yeni sezon hazrlklarna gnde ift idmanla srdryor.Teknik Direktr Cahit Terzi nezaretinde sabah saat 08.00 ve akam 18.00 olmak zere gnde ift idman yapan krmz-lacivertli futbolcular, yeni sezon ncesinde form tutuyorlar.lk etap kamp almalarn kendi tesislerinde yaptklarn belirten Terzi, "Hafta banda baladmz almalara gnde ift idmanla devam ediyoruz. Futbolcularmz u anda kulp binasnda kamptalar. Sabah ve akam olmak zere gnde ift idman yapyoruz. Birka gn d-- possible for object splitting to split up the -- pieces later. codeGen dflags this_mod data_tycons cost_centre_info stg_binds hpc_info = do { showPass dflags "CodeGen" -- Why? -- ; mapM_ (\x -> seq x (return ())) data_tycons ; code_stuff <- initC dflags this_mod $ do { cmm_binds <- mapM (getCmm . cgTopBinding dflags) stg_binds ; cmm_tycons <- mapM cgTyCon data_tycons ; cmm_init <- getCmm (mkModuleInit dflags cost_centre_info this_mod hpc_info) ; return (cmm_init : cmm_binds ++ cmm_tycons) } -- Put datatype_stuff after code_stuff, because the -- datatype closure table (for enumeration types) to -- (say) PrelBase_True_closure, which is defined in -- code_stuff -- Note [codegen-split-init] the cmm_init block must -- come FIRST. This is because when -split-objs is on -- we need to combine this block with its -- initialisation routines; see Note -- [pipeline-split-init]. ; dumpIfSet_dyn dflags Opt_D_dump_cmm "Cmm" (pprCmms (targetPlatform dflags) code_stuff) ; return code_stuff } mkModuleInit :: DynFlags -> CollectedCCs -- cost centre info -> Module -> HpcInfo -> Code mkModuleInit dflags cost_centre_info this_mod hpc_info = do { -- Allocate the statamour, les projets se dessinent avec confiance, dans le but davancer dans la vie, de fonder une famille, de continuer construire dans la confiance. Cest pour cela que je vous dis que lamour est la plus belle chose qui puisse vous arriver dans la vie! Pourtant, peu peu ou subitement, la personne que vous aimez le plus au monde commenc changer son comportement, prendre ses distances, sloigner de vous, et sapprte mme briser ce bonheur. Vous ne comprenez pas ce qui vous arrive. Alors, tout scroule autour de vous.Personnellement, je souhaite que les gens vivent dans le bonheur et lamour, car je crois bien en lamour. Dans ces moments difficiles, durs accepter, il faut se poser la question: pourquoi mon tre aim dcide de sloigner de tous ces moments de bonheur, avec tout ce que lon a vcu ensemble? Est-ce d linfluence dun proche, de la famille, ou un envotement? Malgr toutes ces rflexions et les questions que vous vous posez, vous ne trouvez aucune rponse adquate. Alors, nattendez plus, car je peux vous aider solutionner votre cas. Tous ces moments de bonheur vcus seront une force pour moi lors de mes rituels.Matter of Wolf, 231 N.J. Super. 365 (App. Div. 1989) This reported case of the Superior Court of New Jersey, Appellate Division significantly strengthened the due process rights of professional and non-professional public employees in administrative proceedings seeking removal from /termination of employment.Gimello v. Agency Rent-A-Car, 250 N.J. Super 338 (App. Div. 1991) This reported case of the Superior Court of New Jersey, Appellate Division established obesity as a "handicap" under the New Jersey Law Against Discrimination ("NJLAD"), and established discrimination based upon "perception of handicap" as a cause of action under the NJLAD.Delran Ed. Ass''n. v. Bd. of Ed., ) * h = (1/2)*AB*m Simplify the term in the parenthesis: 1 + (m - h)/m = (m + m - h)/m = (2m - h)/m Thus: AB*(2m - h)/m * h = (1/2)*AB*m Cancel AB: (2m - h)/m * h = (1/2)*m Multiply both sides by m: (2m - h)*h = (1/2)*m Expanding: 2mh - h = (1/2)m Rearranged: h - 2mh + (1/2)m = 0 Quadratic equation: h - 2m h + 0.5 m = 0 Solutions: h = [2m sqrt{(2m)^2 - 4*1*0.5m}]/2 Calculate discriminant: (4m - 2m) = 2m So sqrt(2m) = m*sqrt(2) Thus: h = [2m m*sqrt(2)] / 2 = m*(2 sqrt(2))/2 As before, we discard the larger root because h must be less than m: h = m*(2 - sqrt(2))/2 m*(0.5857) 0.586m Wait, but earlier I thought h 0.293m. Wait, wait, confusion here. Wait, if h = m*(2 - sqrt(2))/2, let's compute that: sqrt(2) 1.414 2 - sqrt(2) 0.5858 Divide by 2: 0.5858 / 2 0.2929 Wait, so h 0.2929m, which is approximately 0.293m. Wait, so that contradicts the earlier statement. Wait, what's going on? Wait, 2 - sqrt(2) 0.5858, divided by 2 is 0.2929. So h 0.2929m, which is about 29.29% of m. But earlier, when I thought h = [2m - sqrt(2)m]/2, which is m*(2 - sqrt(2))/2, which is the same as m*(1 - sqrt(2)/2). Since sqrt(2)/2 0.7071, so 1 - 0.7071 0.2929. So same result. Therefore, h = m*(1 - sqrt(2)/2) 0.2929m. But wait, in the quadratic equation, we have h - 2m h + 0.5 m = 0. Let's check the solution. Let me plug h = m*(1 - sqrt(2)/2) into the equation: Left side: [m*(1 - sqrt(2)/2)] - 2m*[m*(1 - sqrt(2)/2)] + 0.5m First term: m*(1 - sqrt(2)/2)^2 = m*(1 - sqrt(2) + (sqrt(2)/2)^2) = m*(1 - sqrt(2) + (2/4)) = m*(1 - sqrt(2) + 0.5) = m*(1.5 - sqrt(2)) Second term: -2m*(m*(1 - sqrt(2)/2)) = -2m*(e0a2bb6a-712d-4018-a2a3-b2d28c5a76b8"] }], "Summary", CellTags-> "SUMMARY Uses of Linear \ Approximation",ExpressionUUID->"e8606bd5-2efd-4609-92bb-2a511cfcbd0b"] }, Closed]], Cell[CellGroupData[{ Cell["Differentials \[RightGuillemet]", "Subsection", CellTags-> "Differentials",ExpressionUUID->"86f8541d-a600-4d30-bff4-6b8b6ada0171"], Cell["\<\ We now introduce an important concept that allows us to distinguish two \ related quantities: \ \>", "Text",ExpressionUUID->"13fac690-eee0-46a7-a8d4-9307483626f9"], Cell[CellGroupData[{ Cell[TextData[{ "the change in the function ", Cell[BoxData[ FormBox[ RowBox[{"y", "=", RowBox[{"f", "(", "x", ")"}]}], TraditionalForm]],ExpressionUUID-> "98b02e50-3bbc-4496-a1fe-fa3e551e93a7"], " as ", Cell[BoxData[ FormBox["x", TraditionalForm]],ExpressionUUID-> "6bb837d5-1ed9-404e-aa46-cc103dddf838"], " changes from ", Cell[BoxData[ FormBox["a", TraditionalForm]],ExpressionUUID-> "0503b168-7bf3-4392-a42b-c0a2206f9670"], " to ", Cell[BoxData[ FormBox[ RowBox[{"a", "+", RowBox[{"\[CapitalDelta]", "\[VeryThinSpace]", "x"}]}], TraditionalForm]], ExpressionUUID->"bd5901d3-a040-4f75-80b9-c5c8d9913748"], " (which we call ", Cell[BoxData[ FormBox[ RowBox[{"\[CapitalDelta]", "\[VeryThinSpace]", "y"}], TraditionalForm]], ExpressionUUID->"8b00a77c-0118-4a43-923c-4e1b50ddbee6"], ", as before), and " }], "Item",ExpressionUUID->"14895854-dd0a-4776-9045-16e7d4da1eb8"], Cell[TextData[{ "the change in the linear approximation ", Cell[BoxData[ FormBox[ RowBox[{"y", "=", RowBox[{"L", "(", "x", ")"}]}], TraditionalForm]],ExpressionUUID-> "560895d6-5c94-4afa-a846-9ba34c46ef8a"], " as ", Cell[BoxData[ FormBox["x", TraditionalForm]],ExpressionUUID-> "11878bbe-a091-4810-9c5c-6e1937c3f205"], " changes from ", Cell[BoxData[ FormBox["a", TraditionalForm]],ExpressionUUID-> "a16c7964-17c7-4e4a-8a05-1147dc597ee1"], " to ", Cell[BoxData[ FormBox[ RowBox[{"a", "+", RowBox[{"\[CapitalDelta]", "\[VeryThinSpace]", "x"}]}], TraditionalForm]], ExpressionUUID->"064243bb-2f8e-4b07-8042-33b0a21378a1"], " (which we will call the ", StyleBox["differential dy", FontSlant->"Italic"], ")." }], "Item",ExpressionUUID->"bdfa3d44-bb9d-4ffc-ae60-b3764e3f0d6a"] }, Open ]], Cell[TextData[{ "Consider a function ", Cell[BoxData[ FormBox[ RowBox[{"y", "=", RowBox[{"f", "(", "x", ")"}]}], TraditionalForm]],ExpressionUUID-> "83724153-c46d-4573-b5c9-7e8d296f76cc"], " differentiable on an interval containing ", Cell[BoxData[ FormBox["a", TraditionalForm]],ExpressionUUID-> "e44f11f3-c186-4a32-b176-1f6642d2dae1"], ". If the ", Cell[BoxData[ FormBox["x", TraditionalForm]],ExpressionUUID-> "61caea18-72cb-4f9e-8518-5f4d9e34ac53"], "-coordinate changes from ", Cell[BoxData[ FormBox["a", TraditionalForm]],ExpressionUUID-> "254c4137-b0ca-46fa-8fbe-37488363c4a0"], " to ", Cell[BoxData[ FormBox[ RowBox[{"a", "+", RowBox[{"\[CapitalDelta]", "\[VeryThinSpace]", "x"}]}], TraditionalForm]], ExpressionUUID->"025484db-7567-4c9f-8946-2f2b2d66c913"], ", the corresponding change in the function is ", StyleBox["exactly", FontSlant->"Italic"], " " }], "Text",ExpressionUUID->"04403340-8408-4c7e-93e4-b8434d13691d"], Cell[TextData[Cell[BoxData[ FormBox[GridBox[{ {GridBox[{ { RowBox[{ RowBox[{"\[CapitalDelta]", "\[VeryThinSpace]", "y"}], "=", RowBox[{ RowBox[{"f", "(", RowBox[{"a", "+", RowBox[{"\[CapitalDelta]", "\[VeryThinSpace]", "x"}]}], ")"}], en a = tb. Then, as before, substitute into the equations. First equation: 1/a +1/b = 22. So 1/(tb) +1/b = (1 + t)/(tb) =22. Therefore, (1 + t)/ (tb) =22 => b = (1 + t)/(22 t). So b is expressed in terms of t. Second equation: (a -b)^2 =4(ab)^3. Since a = tb, substitute: (tb -b)^2 =4(tb * b)^3 => b^2(t -1)^2 =4 t^3 b^6 => (t -1)^2 =4 t^3 b^4. But from the first equation, b = (1 + t)/(22 t). Therefore, substitute into this: (t -1)^2 =4 t^3 [ (1 + t)/(22 t) ]^4 Compute the right-hand side: 4 t^3 * [ (1 + t)^4 / ( (22)^4 t^4 ) ] =4 t^3 * (1 + t)^4 / ( (16 * 4) t^4 ) Wait, (22)^4 is (2)^4 * (2)^4 =16 *4=64. Because (2)^4= (2)^{2}=4. Therefore: Right-hand side=4 t^3 * (1 + t)^4 / (64 t^4 )= (4 /64) * (1 + t)^4 / t= (1/16)*(1 + t)^4 /t Therefore, the equation becomes: (t -1)^2 = (1/16)ecl ty', fvs) } standaloneDerivErr :: SDoc standaloneDerivErr = hang (ptext (sLit "Illegal standalone deriving declaration")) 2 (ptext (sLit "Use StandaloneDeriving to enable this extension")) \end{code} %********************************************************* %* * \subsection{Rules} %* * %********************************************************* \ungsweise 16 Prozent der SPD-Anhnger. Keine greren nderungen wollen 25 beziehungsweise 32 Prozent.Unsere Idee ist ein Elektroauto fr alle. Und weil jeder Mensch anders ist und etwas anderes braucht, haben wir auch fr jeden das passende Fahrzeug gebaut. So persnlich, dass es perfekt zu dir passt. Oder zu deiner Famcolors, the other two would be blue). Wait, but that would be equivalent to two red and two blue. Alternatively, maybe the problem is considering squares that have exactly two vertices of one color (either red or blue), and the other two can be any color. But since there are only two colors, the other two woun submit packages to Hackage (as long as they have a cabal file). < cabal install What if I simultaneously want to build two different application with conflicting dependencies? < # evil-tic-tac-toe < builde interface file mi_anns :: [IfaceAnnotation], -- ^ Annotations -- NOT STRICT! we read this field lazily from the interface file -- Type, class and variable declarations -- The hash of an Id changes if its fixity or deprecations change -- (as well as its type of course) -- Ditto data constructors, class operations, except that -- the hash of the parent class/tycon changes mi_decls :: [(Fingerprint,IfaceDecl)], -- ^ Sorted type, variable, class etc. declarations mi_globals :: !(Maybe GlobalRdrEnv), -- ^ Binds all the things defined at the top level in -- the /original source/ code for this module. which -- is NOT the same as mi_exports, nor mi_decls (which -- may contains declarations for things not actually -- defined by the user). Used for GHCi and for inspecting -- the contents of modules via the GHC API only. -- -- (We need the source file to figure out the -- top-level environment, if we didn't compile this module -- from source then this field contains @Nothing@). -- -- Strictly speaking this field should live in the -- 'HomeModInfo', but that leads to more plumbing. -- Instance declarations and rules mi_insts eplaced this year. Employees' hours might be reduced, but overall program hours will probably remain the same, Bleacher said."We are looking to maintain high quality of programming, but we do expect we will have to cut services to children and families," Bleacher said. "This is already a lean budget and there's no place else to cut."Education Connection runs programs in Torrington, Winsted, Litchfield and New Milford, enrolling about 200 children between ages 3 and 5 and about 30 from infancy to age 3, Bleacher said. About 90 children are on a wait list to enroll, but Bleacher said she thought the cuts would mean reduced enrollment."We are not expecting that funding will be restored," Bleacher said. "It would be wonderful if it was, but that would not be our expectation."John Tindall-Gibson, superintendent of Naugatuck schools, did not have enrollment figures Monday for the borough's Head Start program. Last year, Head Start and the state School Readiness program combined to enroll more than 100 borough children.According to the plan presented April 11 to the school board, a 10-hour year-round class will be converted to a four-hour year-round class on June 24 and two part-time teachers will be laid off.A six-hour class ding. Alternatively, the problem's first part is to create a linked list of N integers, but the actual creation is done via the insert method. So the code would start with an empty linked list, and then insert N integers into it. But again, the problem statement is not clear. In any case, perhaps the best approach is to model the linked list as a class with methods to insert, delete, search, and reverse. The initial linked list can be empty, and the user can create it with N integers by calling the insert method N times. Then, the function that creates the linked list would be part of the class's __init__ method, but the problem says the function should take N as a parameter. So pen odd number, Player 1 can place a rook in the mirrored position, which would have the same number of adjacent rooks due to symmetry, but since Player 1 needs even, this would only work if the numbere entity. But total permutations is 12, so the proportion would be 6/12 = 1/2. But let's check manually: Permutations of "AABC": Total distinct permutations are 12: 1. AABC 2. ABAC 3. ABCA 4. BAAC 5. BACA 6. BCAA 7. AACB 8. ACAB 9. ACA B 10. CAAB 11. CAB A 12. CBAA Now, how many contain "AB" as a substring: Looking at the list: 1. AABC: "AB" starting at position 2 2. ABAC: "AB" starting at position 1 3. ABCA: "AB" starting at position 1 4. BAAC: "BA" at 1, "AA" at 2, no "AB" 5. BACA: "BA" at 1, "AC" at 2, "CA" at 3, no "AB" 6. BCAA: No "AB" 7. AACB: No "AB" 8. ACAB: "AB" at position 3 9. ACA B: Assuming space is a typo, "ACAB" same as 8 10. CAAB: "AB" at position 3 11. CAB A: "AB" at position 2 12. CBAA: No "AB" So permutations with "AB" are 1, 2, 3, 8, 10, 11. That's 6 permutations, which matches the calculation. So even though there are two As, treating "AB" as a block gives the correct count. So in that case, the formula works. Therefore, returning to the original problem, treating "MATH" as a block and considering the remaining letters as unique (since after using one M, A, T, H, the remaining letters are single instances), the number of permutations containing "MATH" is 8!. Therefore, the number of such permutations is 8!. But wait, in the original problem, the total permutations are 11! / (2!2!2!). So the proportion of permutations containing "MATH" is 8! / (11! / (2!2!2!))). Therefore, the proportion not containing "MATH" is 1 - (8! * 2!2!2!)/11!. But let me verify this with another example to ensure that there are no hidden multiplicities. Alternatively, maybe there's a mistake here because in the original problem, even after forming the block "MATH", the remaining letters include M, A, T, which were originally duplicated. But when we treat them as single letters, perhaps we need to adjust for their multiplicities in the remaining count. Wait, in the original total permutations, the division by 2!2!2! accounts for the duplicates in M, A, T. When we form the block "MATH", we are using one of each M, A, T, H. The remaining letters are M, A, T each with one left, and E, I, C, S. So in the remaining letters, M, A, T are now singletons, so their counts are 1 each, so no duplicates. Therefore, when we calculate the permutations with "MATH" as a block, the remaining letters are all unique, so the number of permutations is 8! (for arranging the block and the 7 unique letters). Since the remaining letters don't have duplicates, we don't need to divide by anything. However, the original total permutations do have duplicates, so the division is already account total minus tires. Yep, that makes sense. So, battery price = total cost - tires cost. Plugging in the numbers: $224 - $168. Let me compute that. 224 minus 160 is 64, and then minus another 8 is 56. Wait, is that right? Let me do it another way to verify. 224 minus 168. Starting from 168, how much do I need to add to get to 224? 168 + 50 = 218, and then +6 = 224. So that's 50 + 6 = 56. Okay, so the battery is $56. Let me just make sure I didn't make any mistakes here. If four tires are $42 each, 4 times 42 is 168. Then adding the battery's 56 gives 168 + 56 = 224, which matches the total given. So that checks out. Wait, just to be thorough, maybe I should write out all the steps again to confirm. Total cost ommarize key points: In summary, my approach balances thorough analysis with team collaboration. **Example Answer Framework**: - **Question**: How would you handle a sudden project scope change? - **Response**: - Clarify: Is this change due to client request or internal factors? - Structure: First, Id assess the impact on deadlines and resources. Then, consult stakeholders to prioritize tasks. For example, in my previous role, we repurposed existing assets to meet a new client requirement, delivering on time without overloading the team. This approach balances preparation with adaptability, showcasing your critical thinking and fit for the role.Please focus on the functionality of this problem and provide code in Swift: You are given an integer array `nums`. The unique elements of an array are the elements that appear **exactly once** in the array. Return _the **sum** of all the unique elements of_ `nums`. **Example 1:** **Input:** nums = \[1,2,3,2\] **Output:** 4 **Explanation:** The unique elements are \[1,3\], and the sum is 4. **Example 2:** **Input:** nums = \[1,1,1,1SLAM TAPI SEKARANG OK PULAK, MALAH TG NIK AZIZ MAHUKAN UMNO AMBIK SEKALI ATAU TELAN SEMUA DASAR PAKATAN RAKYAT KALAU NAK BEKERSAMA DENGAN PAS.MAKNANYA UMNO KENA TERIMA DAN TELAN MALAYSIAN MALAYSIA ILHAM LEE KUAN YEW TAHUN 1963 DULU. INI BERMAKNA UMNO KENA BUBAR DASAR ATAU HALATUJU NEGARA YANG BERTERASKAN PERLEMBAGAAN DAN RUKUN NEGARA YANG DIBAWA OLEH 1MALAYSIA. HELAH DAN TIPU DAYA POLITIK PAS PIMPINAN TG NIK AZIZ, MB KELANTAN UNTUK BERMUZAKARAH ATAU BEKERJASAMA DENGAN UMNO SEBELUM PRU KE 13 INI HANYALAH TEKNIK UNTUK MEMBELIT> error "attempted FIRST(e) :-(" > next :: Name -> NameSet > next t | t >= fst_term = Set.singleton t > next n = > foldb Set.union > [ joinSymSets fn (snd4 (prodNo rl)) | > rl <- prodsOfnce of holding on to a work ethic: I am happy to come to work every day now and see them diligently and passionately working for the same cause I have pursued over the past 26 years!The pace of innovation in digital collaboration itself appears to remain unabated. The number of e-mails I get about new collaborative tools for the enterprise even today is astonishing. While I will share a list of the most promising ones here soon, there is no let up in sight, even though most of these products will not succeed. In reality, most csonal information in accordance with all relevant legislation and in a manner similar to the way the Company protects the same. The Company will not share personal information with thirymmetric, then the minimum would be at the center. For example, if you have a convex function symmetric about a point, the minimum is at that point. Similarly, a convex sequence symmetric about the middle would have its minimum at the middle. But then if the minimum is at the middle, then \( y_{1009} \) and \( y_{1010} \) would be the smallest terms, but the sum of all terms is zero. So if the middle terms are minimized, the other terms would have to be larger to compensate. However, in this case, we need the sum \( y_{1009} + y_{1010} \) to be as large as possible (since we need to show it's 0). Wait, if the middle terms are minimized, their sum would be the smallest, but we need to bound the sum from above. This is a bit confusing. Maybe another approach: suppose that the maximum of \( y_{1009} + y_{1010} \) is achieved when all the convexity inequalities are equalities, i.e., when \( x_{n+1}^2 = x_n x_{n+2} \) for all n. Then the sequence is a geometric progression as I considered earlier, and in that case, the product is 1, and \( x_{1009} x_{1010} = 1 \). Therefore, perhaps this is the maximum possible value, and any deviation from the geometric progression would cause \( x_{1009} x_{1010} \) to decrease. Therefore, the maximum is 1, achieved by the geometric progression. But how to formalize this intuition? Let's suppose that if we have a log-convex sequence (i.e., \( x_{n+1}^2 \leq x_n x_{n+2} \)), then the product \( x_{1009} x_{1010} \) is maximized when the sequence is a geometric progression. Therefore, under the constraint that the product of all terms is 1, the maximum of \( x_{1009} x_{1010} \) isal to the number of previous elements different from B_i, it's a fixed point. So for example, in position 2 (i=2), B_2 must equal the number of elements before it (B_0 and B_1) that are different from B_2. Suppose B_0=0, B_1=1, then B_2 can be: If B_2=2: then the previous elements are 0 and 1, both different from 2, so t(B_2)=2. So B_2=2 satisfies t(B)=B. If B_2=1: previous elements are 0 and 1. Different from 1 are 0. So count is 1. So B_2=1 would work. If B_2=0: previous elements are 0 and 1. Different from 0 is 1. So count is 1. But B_2=0 1, so not a fixed point. Wait, so for i=2, possible fixed points are B_2=1 or 2. Because: If B_2=1: count of different previous elements is 1 (only B_0=0 is different, B_1=1 is same). Wait, no: B_1=1 is same as B_2=1, so only B_0=0 is different. So count is 1. So B_2=1 is okay. If B_2=2: both previous elements (0 and 1) are different, so count is 2. So B_2=2 is okay. If B_2=0: previous elements 0 and 1. Different from 0 is 1. Count is 1. So B_2=0 would need to be 1 to satisfy the condition, which it isn't. So B_2 can't be 0 in a fixed phael kors factory outletray ban sunglassestimberland bootstory burch shoesthe north face clearancecanada goose outletmoncler jacketskaren millen dresseswinter jacketstrue religion outlethermes belthoganmichael kors outletcyber monday 2015kobe shoesswarovski outletreplica handbagsmichael kors outlettoms outletugg bootssoccer jerseysReplyDeleteAdd commentLoad more...A Home For Helena Aileen Fish Alicia Quigley Amusements of Old London Amy Rose Bennett A Twelfth Night Tale Bath Beaux Ballrooms and Battles Bluestocking Belles Blush Blush Cotillion Caroline Warfield Charles II Charles James Fox Christmas Coaching Days & Coaching Ways Collette Cameron Cotillion Christmas Traditions Crimson Romance David Coke & Alan Borg Drury Lane Duchess of Devonshire Duke of Wellington Elizabeth Grant Ella Quinn Ellora's Cave George IV giveaway Goddess Fish Promotions Heather King Henry VIII hero heroes heroine heroines histore berall gleich eingerichtet sind. Frhstck im Preis mitenthalten. Ein Mittag- oder Abendessen wie man es von den Europischen Hotels kennt, wird nicht angeboten. Japan hat in jeder Ecke sehr viele (kleine) Restaurants, die sollte ein Tourist ausprobieren ;)Bewertung lesenHotels in der Nhe von Hotel Toyoko Inn Hakata-guchi Ekimae100%Hakata Green Hotel No.1Fukuoka, Japan 301m100%Hakata Green Hotel AnnexFukuoka, Japan 303m100%Hotel Hakata MiyakoFukuoka, Japan 443m100%Hotel ANA Crowne Plaza FukuokaFukuoka, Japan 527m100%Court Hotel Hakata EkimaeFukuoka, Japan 545m100%Hotel APA Hakata EkimaeFukuoka, Japan 626m100%Hotel Grand Hyatt FukuokaFukuoka, Japan 837m100%Hostel HanaFukuokaFukuoka, Japan 839m100%Hakata Excel Tokyu HotelFukuoka, Japan 1,3km100%Hotel MyStays Fukuoka } reifyNonArrowKind k = noTH (sLit "this kind") (ppr k) reify_kc_app :: TyCon -> [TypeRep.Kind] -> TcM TH.Kind reify_kc_app kc kis = fmap (foldl TH.AppT r_kc) (mapM reifyKind kis) where r_kc | isPromotedTyCon kc && isTupleTyCon (promotedTyCon kc) = TH.TupleT (tyConArity kc) | kc `hasKey` listTyConKey = TH.ListT | otherwise = TH.ConT (reifyName kc) reifyCxt :: [PredType] -> TcM [TH.Pred] reifyCxt = mapM reifyPred reifyFunDep :: ([TyVar], [TyVar]) -> TH.FunDep reifyFunDep (xs, ys) = TH.FunDep (map reifyName xs) (map reifyName ys) reifyFamFlavour :: TyCon -> TH.FamFlavour reifyFamFlavour tc | isSynFamilyTyCon tc = TH.TypeFam | isFamilyTyCon tc = TH.DataFam | otherwise = panic "TcSplice.reifyFamFlavour: not a type family" reifyTyVars :: [TyVar] -> TcM [TH.TyVarBndr] reifyTyVars = mapM reifyTyVar . filter isTypeVar where reifyTyVar tv | isLiftedTypeKind kind = return (TH.PlainTV name) | otherwise = do kind' <- reifyKind kind return (TH.KindedTV name kind') where kind = tyVarKind tv name = reifyName tv reify_tc_app :: TyCon -> [TypeRep.Type] -> TcM TH.Type reify_tc_app tc tys = do { tys' <- reifyTypes (removeKinds (tyConKind tc) tys) ; return (foldl TH.AppT r_tc tys') } where arity = tyConArity tc r_tc | isTupleTyCon tc = if isPromotedDataCon tc then TH.PromotedTupleT arity else TH.TupleT arity | tc `hasKey` listTyConKey = TH.ListT | tc `hasKey` nilDataConKey = TH.PromotedNilT | tc `hasKey` consDataConKey = TH.PromotedConsT | otherwise = TH.ConT (reifyName tc) removeKinds :: Kind -> [TypeRep.Type] -> [TypeRep.Type] removeKinds (FunTy k1 k2) (h:t) | isSuperKind k1 = removeKinds k2 t 2306d"], Cell[201272, 5677, 189, 4, 70, 55, 0, "StyleData", "AuthorTitle", "NumberedHeadsPrintout",ExpressionUUID->"b7ad625a-2c9b-4fff-8cce-c958d1631eab"] }, Closed]], Cell[CellGroupData[{ Cell[201498, 5686, 707, 17, 70, 30, 0, "StyleData", "AuthorEmail", "All",ExpressionUUID->"89bd12a3-578d-487e-974b-9dd5bca3756b"], Cell[202208, 5705, 102, 0, 70, 47, 0, "StyleData", "AuthorEmail", "NumberedHeads",ExpressionUUID->"bb4d8836-a75b-477e-871e-d2544898cc7d"], Cell[202313, 5707, 189, 3, 70, 46, 0, "StyleData", "AuthorEmail", "Presentation",ExpressionUUID->"875eaff9-67a1-4153-8519-0266715b603d"], Cell[202505, at's good. Now, the problem states that P(a) = b + c + 2, P(b) = a + c + 2, P(c) = a + b + 2, and also P(a + b + c) = -22. Since a + b + c is -5, we need to find P(-5) = -22. We need to find the cubic polynomial P(x). Let me think about how to approach this. Since P(x) is a cubic polrk Criticism, Coral Reef Die Off, Corona Virus Discussion, Disease Detection, DNA Origami, Fake News Detection, Food Forward, Fund Raising, Gravity Waves, Head Transplants, Health Care Costs, Homeboy Industries, Homeless in Las Vegas, Internet Society, Julian Assange, Magic Wheelchairs, Medical Treatment Questions, Nano-engineering, Prisoners become Farmers, Purposity, Quantum Dots, Rainbows Pack, Smart Dust, Travis Roy Foundation, UFOs in the US. Viral Based Obesity, Vision Basic Science, Vision Spring, War on a WhimGua -Informacin sobre la gua -Mapas Albor Remoto Misiones principales -La conexin de Albor Remoto Valle Quebrado Misiones principales -A la caza del dragn -Descanso eterno a un fantasma -En busca de Lovis -La lista de papel -No ms dragones Misiones secundarias -Pandilla de desalmados -El alijo de Louis -Por un puado de carne -El mejor cazador -El botn de Lovis -Salvar la panceta -La sangre de Buad -Robacorazones -Banquete o hambruna -Una entrega privada -Esqueletos en el armario -Sin salida -Una bsqueda obligada -En el refugio del bandido -Vigor Mortis -Alma perdida -Un cuento de dos tomos -A sangre fra -Problemas de pareja -En bsqueda de un hacha -Robo a plena luz -El templo de la desgracia -Mtodo o Locura -La bsqueda del mineral rojo -Sin palabras -Atrapado en un agujero -El fugitivo -Un aire fino Lectura de mentes -Lectura de mentes Isla del Centinela Misiones principales -La profeca -Velas al viento -La segundareers; - tips for Peace Corps Volunteers to better visually document their service; - advice on how to integrate photography into Peace Corps projects; - and the examination of photography as a vehicle for development and social change. CategoriesI've been getting emails and comments from readers and fellow bloggers asking for my inside take on the story that broke a couple of days ago regarding the remains of a young Union soldier found on the battlefield. I'm sorry to disappoint but... all I know is what I read in the papers.About this electrifying event, I know little more than anyone else.I must tell you though, I'm often asked by visitors if there's any chance that there are still soldiers buried out there, my response is usually "It is mathematically probable".The battlefield continues to gradually reveal its story to us, day by day, generation by generation. And through these revelations the events of that day in 1862 remain immediate, authentic, and of transcendent importance.What an amazing place to spend my days, just north of Sharpsburg.MannieYesfarma vende medicamentos online. Nossa farmcia on-line uma pgina legal para a venda de medicamentos para uso humano no sujeitos a receita mdica, autorizados pela Agncia Espanhola de Medicamentos e Produtos de Sade. Voc pode verificar nossa licena atravs do logotipo que nos identifica. Em Yesfarma, voc pode comprar medicamentos sem receita mdica, para uso humano, sem receita mdica, se a remessa estiver dentro da Espanha para a pennsula e as Ilhas Baleares, excluindo outras reas de expedio (Ilhas Canrias, Ceuta e Melilla).Ellis Starrett is a Learning and Development Facilitator at DES and provides Leading Teams facilitation throughout the state. Ellis is an experienced human resources professional with a background in consulting on a range of organizational and HR issues. They have experience leading project teams across Washington State and have supervised teams while with the state and in the private sector. They are committed to making a positive impact for the people of Washington State. Ellis is especially passionate r = 10, which matches the given circumference. Good. So each arc has a length of either or 2, and each is colored red, green, or blue. He has unlimited arcs of each type, so he can use as many as needed of each of the six types (which are combinations of length and color: red , red 2, green , green 2, blue , blue 2). The main goal is to cover the entire circle without overlapping, following two conditions: 1. Any two adjacent arcs must be different colors. 2. Any three adjacent(4/5)*(-8/5) =1 -32/25 = (25 -32)/25 = -7/25 So, E=(-1/25, -7/25) Similarly, line AN is from A(-1,1) to N(1,1). This line is horizontal y=1. Intersection with circle C: (x +1)^2 +1^2=1 => (x +1)^2=0 =>x=-1. So, the only intersection is A itself. Therefore, in this case, line AN i, can a straight line cross all four edges? Let me draw it mentally. Imagine a concave quadrilateral where the top two edges slope downwards towards the center, and the bottom two edges form a V shape. If a line is drawn diagonally from the top left, it might enter through the top left edge, exit through the top right edge, then enter through the bottom right edge, and exit through the bottom left edge. That would be four intersections. But wait, in reality, once the line exits the top right edge, it would have to enter again through another edge. But in a simple polygon, after exiting, the line is outside the polygon, so to enter again, the polygon tcIsHsBoot -- Are we compiling an hs-boot file? ; checkTc (not (null cons) || empty_data_decls || is_boot) (emptyConDeclsErr tc_name) ; return h98_syntax } ----------------------------------- consUseH98Syntax :: [LConDecl a] -> Bool consUseH98Syntax (L _ (ConDecl { con_res = ResTyGADT _ }) : _) = False consUseH98Syntax _ = True -- All constructors have same shape ----------------------------------- tcConDecls :: NewOrData -> TyCon -> ([TyVar], Type) -> [LConDecl Name] -> TcM [DataCon] tcConDecls new_or_data rep_tycon (tmpl_tvs, res_tmpl) cons = mapM (addLocM $ tcConDecl new_or_data rep_tycon tmpl_tvs res_tmpl) cons tcConDecl :: NewOrData -> TyCon -- Representation tycon -> [TyVar] -> Type -- Return type template (with its template tyvars) -- (tvs, T tys), where T is the family TyCon -> ConDecl Name -> TcM DataCon tcConDecl new_or_data rep_tycon tmpl_tvs res_tmpl -- Data types (ConDecl { con_name = name , con_qvars = hs_tvs, con_cxt = hs_ctxt , con_details = hs_details, con_res = hs_res_ty }) = addErrCtxt (dataConCtxt name) $ do { traceTc "tcConDecl 1" (ppr name) ; (ctxt, arg_tys, res_ty, is_infix, field_lbls, stricts) <- tcHsTyVarBndrs hs_tvs $ \ _ -> do { ctxt <- tcHsContext hs_ctxt ; details <- tcConArgs new_or_data hs_details ; res_ty <- tcConRes hs_res_ty ; let (is_infix, field_lbls, btys) = details (arg_tys, stricts) = unzip btys ; return (ctxt, arg_tys, res_ty, is_infix, field_lbls, stricts) } -- Generalise the kind variables (returning quantifed TcKindVars) -- and quantify the type variables (substituting their kinds) -- REMEMBER: 'tkvs' are: -- ResTyH98: the *existential* type variables only -- ResTyGADT: *all* the quantified type variables -- c.f. the comment on con_qvars in HsDecls ; tkvs <- case res_ty of ResTyH98 -> quantifyTyVars (mkVarSet tmpl_tvs) (tyVarsOfTypes (ctxt++arg_tys)) ResTyGADT res_ty -> quantifyTyVars emptyVarSet (tyVarsOfTypes (res_ty:ctxt++arg_tys)) -- Zonk to Types ; (ze, qtkvs) <- zonkTyBndrsX emptyZonkEnv tkvs ; arg_tys <- zonkTcTypeToTypes ze arg_tys ; ctxt <- zonkTcTypeToTypes ze ctxt ; res_ty <- caserobability of exactly one hitting, plus exactly two hitting, plus exactly three hitting. Let me try that method to confirm. First, exactly one hits. There are three cases here: first hits, others miss; second hits, others miss; third hits, others miss. For the first case: 0.6 * (1 - 0.7) * (1 - 0.75) = 0.6 * 0.3 * 0.25 = 0.045. Second case: (1 - 0.6) * 0.7 * (1 - 0.75) = 0.4 * 0.7 * 0.25 = 0.07. Third case: (1 - 0.6) * (1 - 0.7) * 0.75 = 0.4 * 0.3 * 0.75 = 0.09. Adding those up: 0.045 + 0.07 + 0.09 = 0.205. So the probability of exactly one hit is 0.205. Next, exactly two hit. Again, three cases: first and second hit, third misses; first and third hit, second misses; second and third hit, first misses. First case: 0.6 * 0.7 * (1 - 0.75) = 0.6 * 0.7 * 0.25 = 0.1062 (I Sin[x])^7 Cos[x]^9 + 1962 (I Sin[x])^9 Cos[x]^7 + 536 (I Sin[x])^5 Cos[x]^11 + 536 (I Sin[x])^11 Cos[x]^5 + 22 (I Sin[x])^3 Cos[x]^13 + 22 (I Sin[x])^13 Cos[x]^3 + 2254 (I Sin[x])^8 Cos[x]^8 + 1216 (I Sin[x])^6 Cos[x]^10 + 1216 (I Sin[x])^10 Cos[x]^6 + 142 (I Sin[x])^4 Cos[x]^12 + 142 (I Sin[x])^12 Cos[x]^4) + Exp[-1 I y] (1642 (I Sin[x])^6 Cos[x]^10 + 1642 (I Sin[x])^10 Cos[x]^6 + 2376 (I Sin[x])^8 Cos[x]^8 + 2140 (I Sin[x])^7 Cos[x]^9 + 2140 (I Sin[x])^9 Cos[x]^7 + 962 (I Sin[x])^5 Cos[x]^11 + 962 (I Sin[x])^11 Cos[x]^5 + 392 (I Sin[x])^4 Cos[x]^12 + 392 (I Sin[x])^12 Cos[x]^4 + 13 (I Sin[x])^2 Cos[x]^14 + 13 (I Sin[x])^14 Cos[x]^2 + 98 (I Sin[x])^3 Cos[x]^13 + 98 (I Sin[x])^13 Cos[x]^3) + Exp[1 I y] (2544 (I Sin[x])^8 Cos[x]^8 + 1612 (I Sin[x])^6 Cos[x]^10 + 1612 (I Sin[x])^10 Cos[x]^6 + 340 (I Sin[x])^4 Cos[x]^12 + 340 (I Sin[x])^12 Cbility of selecting someone under 20 is U / 150 = 0.4. So U = 0.4 * 150. Calculating that: 0.4 * 150. 0.4 is the same as 40%, so 40% of 150. 10% of 150 is 15, so 40% is 15 * 4 = 60. Therefore, U = 60. Then, since total people are 150, the number over 30 is O = 150 - U = 150 - 60 = 90. So 90 people are over 30. Wait, let me check again. If 60 are under 20, that's 60/150 = 0.4, which is 40%, which matches the given probability. Then the rest must be over 30, which is 150 - 60 = 90. That seems right. Is there any chance I made a mistake with the percentage? Wait, 0.4 * 150. 150 * 0.4. 150 divided by 10 is 15, times 4 is 60. Yeah, that's correct. So, yeah, 90 people over 30. That should be the answer. **Final Answer** The number of people with an age of more than 30 years is \boxed{90}. Given a group of 150 people, where the probability of selecting a person under 20 years old at random is 0.4, we need to determine the number of people over 30 years old. 1. Let \( U \) be the number of people under 20 years old, and \( O \) be the number of people over 30 years old. 2. The total numberDe Pakistan Adolescente Com Web Transexual Asitica Cam Leva Sala Chat Terre De En Adolescente Chatrubat Google Chica Mvil Chat Internet Telfono Sexo Y Traer Una Sexo Salas Videos Sucia Calientes Gratis Gay Uno Cams Esposa Sexy Cmara De Del Sexo En Vivo Gratis Aplicacin Com Webcam Live India . , , , , , . -. , . , , , . ( ), .Caminin nnde ve iki yannda geni cami hals d avlusu olup bunun evresi pencereli duvarlarla evrilidir. Bu avulya 3 cephede olmak zere, 8 kapdan girilir. adrvan avlusu, 26 adet granit mermer ve porfir stuna oturtulmu, 30 kubbeyle evrili geni alandr. Mermer demeli bu geni sahann ortasnda 6 mermer stunlu adrvan, sahann azametini gsterir. adrvann kemerleri, kabartma olarak Rumi gemelerle would not intersect in (0, 2) at all? Wait, no. Wait, as k approaches 0 from below, say k = -0.01, then y=k is a horizontal line slightly below zero. The function f(x) in (0, 2) goes from 0 down to -0.1839, then up to -0.0366. So at k = -0.01, which is higher than -0.0366, the line y=-0.01 would intersect the graph once between x=1/2 and x=2, since the function goes from -0.1839 up to -0.0366, crossing y=-0.01 somewhere there. Wait, but -0.01 is higher than -0.03t lamp 55, she has moved 54 units. At the same time, rict ) import Maybes ( orElse, expectJust ) import Outputable import Pair import Data.List import FastString \end{code} To think about * set a noinline pragma on bottoming Ids * Consider f x = x+1 `fatbar` error (show x) We'd like to unbox x, even if that means reboxing it in the error case. %**************************************************************his topic ban, because some of his worst abuses in the past were attempts to ridicule or defame CC sceptics in their Wikipedia bio articles. One other CC editor was able to escape a topic ban during the arbitration case by promising to never again edit CC-related biographical articles.Hola! Yo apenas me ando enterando de esta crema, y por lo que dices no pinta mal. Yo actualmente ando buscando diferentes recomendaciones de cremas, y como soy medio nueva en esto de cuidado de la cara (porque antes no lo hacia, y ya me esta quedando esos resultados de mi mal cuidado), as que por composicin aun no se nada para saber si me pruebo esta. As que me voy por los comentarios positivos y tal vez prueba esta, ya mas adelante que sepa mas, es posible (conocindome) que la descarte jajajaja#10Posted: 1/1/2012 12:58:35 AMIf anyone actually thinks the players are gonna play less hard, and intentionally try to lose the game, they are delusional. Indys players will play hard because most of them suck and are playing for jobs. Now where the "suck for luck" sweepstakes might come into play is if an owner were to go to an offensive coordinator, and say hey lets call this game a bit more conservative than usual. I think that could happen here, but the players aren't gonna play differently. quoterunning the tool into a borehole, the tool comprising the tubular portion and a cutting device having at least one radially movable piston, wherein each radially movable piston has a cutter; applying fluid pressure to an internal portion of the cutting device, thereby radially extending the piston of the cutting device such that at least a portion of the cutter contacts the tubular portion of the tool; and rotating the cutting devicthat didn't always remember to print the pages the night before. So she'd be standing around ready to start math each afternoon while I fumbled around trying to locate the right disc...the right lesson...the right page...in order to print it out for her. I found myself mumbling, "Why couldn't all the pages be included in the workbook?" more times than I'd care to admit.Ortwin Ramadan hatte einige zentrale Passagen der langen und aufregenden Flucht von Yoba und Chioke fr seinen Vortrag ausgewhlt, sie mit Bild- und Kartenmaterial ergnzt und lieferte den Kindern so nochmals einen abwechslungsrei data instance T Int = T1 | T2 Bool Here T is the "family TyCon". * Reply "yes" to isDataFamilyTyCon, and isFamilyTyCon * Reply "yes" to isDataFamilyTyCon, and isFamilyTyCon * The user does not see any "equivalent types" as he did with type synonym families. He just sees constructors with types T1 :: T Int T2 :: Bool -> T Int * Here's the FC version of the above declarations: data T a data R:TInt = T1 | T2 Bool axiom ax_ti : T Int ~ R:TInt The R:TInt is the "representation TyCons". It has an AlgTyConParent of FamInstTyCon Te the same approach. But then, the code would loop infinitely if xorResult is negative. So perhaps the problem expects us to write the code correctly, even if the original code is wrong. Or perhaps the original code works because x and y are non-negative, but the code doesn't enforce that. Hmm. Well, the Hamming distance is the number of positions where the bits differ. So regardless of the sign, the code should count all the 1 bits in the XOR. But the original code's approach would not work for negative numbers in C++, leading to an infinite loop. Therefore, the code is incorrect. So perhaps in Julia, we need to use unsigned integers to avoid this. So, perhaps the correct approach integral from a to b of the square root of (dx/dt)^2 + (dy/dt)^2 dt. Right? So first step is to find the derivatives dx/dt and dy/dt, square them, add them together, take the square root, and integrate over 0 to . Let me write down the parametric equations again to make sure I have them right: x = 4(2 cos t - cos 2t) y = 4(2 sin t - sin 2t) So, both x and y are scaled by 4, and inside the parentheses, there's a combination of cos t and cos 2t, sin t and sin 2t. Hmm. This looks like it might be a type of hypocycloid or some similar curve, but maybe I don't need to know the specific name to compute the arc length. Alright, so first, compute dx/dt and dy/dt. Let's start with dx/dt. dx/dt = 4 * derivative of (2 cos t - cos 2t) with respect to t. The derivative of 2 cos t is -2 sin t, and the derivative of -cos 2t is 2 sin 2t (using chain rule: derivative of cos u is -sin u * du/dt, so here u=2t, so du/dt=2, so derivative is -sin 2t * 2, but there's a negative sign in front, so it becomes 2 sin 2t). Therefore, dx/dt = 4*(-2 sin t + 2 sin 2t) = 4*(-2 sin t + 2 sin 2t) = 4*2*(-sin t + sin 2t) = 8*(-sin t + sin 2t). Wait, let me check that again: Wait, derivative of 2 cos t is -2 sin t. Derivative of -cos 2t is +2 sin 2t. So combining, it's -2 sin t + 2 sin 2t. Then multiply by 4: 4*(-2 sin t + 2 sin 2t) = -8 sin t + 8 sin 2t. So dx/dt = 8(-sin t + sin 2t). Simi of hers; this year she's working with Vina Moses to help provide kids with school clothes and supplies, and will be hosting our 3rd annual Toy Drive this winter. Michelle also enjoys participating in local food drives benefiting Linn and Benton Counties. When she isn't working, she finds pure joy in spending time with her son and family. Her hobbies include gardening, hiking, photography, writing and there are `0` trailing zeros in `546`. 2. Denote the remaining number of digits in the product as `d`. If `d > 10`, then express the product as ... 3. Finally, represent the product as a **string** `"` ...eC " Return _a string denoting the **abbreviated product** of all integers in the **inclusive** range_ `[left, right]`. **Example 1:** **Input:** left = 1, right = 4 **Output:** "24e0 " **Explanation:** The product is 1 \* 2 \* 3 \* 4 = 24. There are no trailing zeros, so 24 remains the same. The abbreviation will end with "e0 ". Since the number of digits is 2, which is less than 10, we do not have to abbreviate it further. Thus, the final representation is "24e0 ". **Example 2:** **Input:** left = 2, right = 11 **Output:** "399168e2 " **Explanation:** The product is 39916800. There are 2 trailing zeros, which we remove to get 399168. The abbreviation will end with "e2 ". The number of digits after removing the trailing zeros is 6, so we do not abbreviate it further. Hence, the abbreviated product is "399168e2 ". **Example 3:** **Input:** left = 371, right = 375 **Output:** "7219856259e3 " **Explanation:** The product is 7219856259000. **Constraints:** * `1 <= left <= right <= 104` Okay, I need to solve this problem where I have to compute the product of numbers from left to right, then abbreviate it according to certain rules. Hmm, let's think about how to approach this in Lisp. First, the steps are: calculate the product, count the trailing zeros, remove them, then check the number of digits left. If more than 10, truncate to first five and last five digits with ... in between. Then, represent it as a string with "eC" where C is the count of zeros removed. Wait, but calculating the product directly for numbers up to 10^4 might result in a very large number. For example, 10^4 numbers multiplied would be way beyond what even a big integer can handle efficiently. So, calculating the product directly might not be feasible. Oh right, because even 1000! is an astronomically large number. So we need a smarter way to compute the trailing zeros and the remaining product's digits without actually computing the entire product. Wait, but the problem says that after removing the trailing zeros, we need to represent the product as a string. So perhaps, we can compute the product modulo 10^something, but also track the number of factors of 2 and 5 to compute the trailing zeros. Because trailing zeros cld be e^{lim_{n} x * 5n}. Let me compute x * 5n. Since x = -2/(10n - 1), then x * 5n = (-2/(10n - 1)) * 5n = (-10n)/(10n - 1). Let's simplify that: (-10n)/(10n - 1) = -10n/(10n - 1). If we divide numerator and denominator by n, this becomes -10/(10 - 1/n). As n approaches infinity, 1/n approaches 0, so this tends to -10/10 = -1. Therefore, the exponent x * 5n tends to -1. Therefore, the limit should be e^{-1} = 1/e. Wait, let me verify that step again. So, we have: lim_{n} [1 - 2/(10n - 1)]^{5n} = e^{lim_{n} (-2/(10n -1) * 5n)}. So the exponent is (-2 "76c0a8f7-c3e0-49df-9274-8a4520289495"], Cell[StyleData["ItemizedPicture", "Printout"], CellMargins->{{77, 2}, {4, -4}}, Magnification->0.5,ExpressionUUID->"b0b3790d-be9a-4ce0-a989-100df8cb02bc"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["ListGraphic"], CellMargins->{{88, 4}, {4, 4}}, CellGroupingRules-> "GraphicsGrouping",ExpressionUUID->"eaddffdb-4d90-4b7b-94a8-d75ef68a7b10"], Cell[StyleData["ListGraphic", "Presentation"], CellMargins->{{88, 4}, {6, 6}}, FontSize->24,ExpressionUUID->"d78918d7-19e4-4e06-a9ee-417b74f2fdc2"], Cell[StyleData["ListGraphic", "Printout"], CellMargins->{{77, 2}, {4, -4}}, Magnification->0.5,ExpressionUUID->"2ea7c53f-907c-4aef-b4c4-f9cf4bfc4780"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["PictureGroup"], CellMargins->{{41, 4}, {0, 4}}, CellGroupingRules-> "GraphicsGrouping",ExpressionUUID->"8266a75c-8ea4-4d30-905d-8eb3485f230b"], Cell[StyleData["PictureGroup", "Presentation"], CellMargins->{{41, 4}, {0, 6}}, FontSize->24,ExpressionUUID->"eab48980-0475-4f16-8d12-5efe861c506d"], Cell[StyleData["PictureGroup", "Printout"], CellMargins->{{76, 2}, {0, 0}}, CellElementSpacings->{"CellMinHeight"->1, ): connection string to the Faktory server. Format is `tcp(+tls)://(:password@)host:port(/namespace)`. Defaults to `tcp://localhost:4719`. `namespace` is prependend to queue names on job submission and worker consumption. When using `envWorkerSettings`, the following variables are also used: - `FAKTORY_QUEUE`: the name of the queue to consume from. Default is "default". - `FAKTORY_WORKER_ID`: the Id to use for this Worker. Default is to assign a random one. ## Examples See the [examples](./examples). To run them: 1. Run a local Faktory server ```console docker run --rm \ --publish 7419:7419 \ --publish 7420:7420 \ contribsys/faktory ``` 1. Run the consumer example ```console % stack exec faktory-example-consumer Starting consumer loop ``` (Assumes you've built the project.) 1. Submit a Job through the producer example ```console % stack exec faktory-example-producer hello world Pushed job: "ljcjlbexbgun" ``` *NOTE*: if you submit "BOOM" as a Job, the processing loop will raise an exception, so you can see how a Failed Job looks in Faktory. 1. See that your Job was processed back in the consumer ```con the parity of (m,n) must be the same as the number of moves modulo 2. Since each move changes the parity, to go from (1,1) to (m,n), the number of moves needed is (m-1) + (n-1), but actually, each move changes both coordinates by 1, so the number of moves required would be max(m-1, n-1), but this is unclear. Wait, actually, each move increments both coordinates by 1 (if moving towards the bottom right) or decrements them (if moving towards the top left). Therefore, the displacement from (1,1) to (m,n) is (m-1, n-1). To achieve this displacement, each move contributes (1, 1). Therefore, the total displacement must be a multiple of (1,1) or (-1,-1), but since we can only move in positive directions (assuming the board is large enough), we need m-1 = n-1, meaning m = n. But that can't be right because the problem might allow moving in other directions as well. Wait, no. If each move can either add (1,1) or subtract (1,1), or perhaps other combinations, then the total displacement would need to have (m-1, n-1) as a vector sum of such moves. However, the movement might not be limited to just (1,1) or (-1,-1). For example, rotating around different legs could allow moving in different diagonal directions. Wait, let's think again. When you rotate around a leg, the center moves in a circular path. The displacement depends on the rotation direction and the leg's position. From the first example, rotating around the right leg moved the center to (2,2). Similarly, rotating around the bottom leg also moved it to (2,2). So, from (1,1), you can only reach (2,2) in one move. From (2,2), depending on the triomino's orientation, you might be able to move to (3,3), or maybe to (1,1), or other positions. But perhaps the key is that each move changes the center's coordinates by (1,1). So, to go from (1,1) to (m,n), you need that (m-1) and (n-1) are equal and non-negative. Therefore, m = n. But this seems too restrictive. Maybe there's another way to move. Alternatively, perhaps from (2,2), you can rotate around a different leg to move in a different direction. For example, 0 to 127: subset_mask = 0 for each bit in 0..6: if the number has that bit set, then take the corresponding bit from bits and add it to subset_mask. So for example, bits = [0,1,2,3,4,5,6] number 3 is 0b0000011 bits 0 and 1 are set subset_mask includes bits 0 and 1. So the code in Elixir: defp generate_subsets(pm) do bits = for i <- 0..25, (pm &&& (1 << i)) != 0, do: i n = length(bits) max = 1 <<< n - 1 Enum.flat_map(0..(1 <<< n) -1, fn num -> subset_mask = Enum.reduce(0..n-1, 0, fn i, acc -> if (num &&& (1 << i)) != 0 do acc ||| (1 << Enum.at(bits, i)) else acc end end) [subset_mask] end) end But wait, since n is 7, 1 <<< 7 is 128. So 0..127 is 128 numbers. Each number generates a subset_mask. So this generates 128 subsets. Then, for each subset_mask, we can check if it includes the first character's bit. Yes. So this approach would work. But since the puzzle has exactly 7 characters, and no duplicates, this code is safe. So, in code: For each puzzle: pm = mask of the puzzle. f = first character's bit. bits = list of the positions of the bits set in pm. Then, generate all 128 possible subset_masks. For each y GranadinasSanta LucaSenegalSerbiaSerbia y MontenegroSeychellesSierra LeonaSingapurSiriaSri LankaSt. Kitts y NevisSudfricaSudnSueciaSuizaSurinameSwazilandiaTailandiaTaiwnTanzaniaTayikistnTimor LesteTogoTongaTrinidad y TobagoTurkmenistnTurquaTuvaluTnezUcraniaUgandaUruguayUzbekistnVanuatuVenezuelaVietnamWallis y FutunaYemenZambiaZimbabweYear 2007 2006 2005 2004 2003 2002 2001 2000 1999 1998 1997 1996 1995 1994 1993 1992 1991 1990 1989 1988 1987 1986 1985 1984 1983 1982 1981 1980 1979 1978 1977 1976 reflecting infinite fractals, asymmetrical angles converging at a floating crystal core, unsettling yet mesmerizing tension between nature and machinery, hyperrealistic 3D render, eerie bioluminescent shadows, haunting serenity juxtaposed with latent chaos --ar 16:9 --v 5.2 /imagine prompt: a retro-futuristic sofa by Salvador Dal and Syd Mead, melting liquid chrome armrests, tessellated velvet cushions in iridescent azure, suspended in a zero-gravity lounge with hexagonal windows overlooking Saturns rings, radial symmetry with converging perspective lines, surreal tranquility infused with cosmic wonder, cinematic matte painting, soft gradients blending nebula hues, captured with a Canon EOS R5 camera, 24-70mm f/2.8 lens,couple on their first date I was asked my opinion on the fellow, gave a positive review :-). Fully recommend going to one of these evenings if yous[x]^10 + 115 (I Sin[x])^10 Cos[x]^6 + 21 (I Sin[x])^4 Cos[x]^12 + 21 (I Sin[x])^12 Cos[x]^4 + 5 (I Sin[x])^3 Cos[x]^13 + 5 (I Sin[x])^13 Cos[x]^3 + 59 (I Sin[x])^11 Cos[x]^5 + 59 (I Sin[x])^5 Cos[x]^11 + 160 (I Sin[x])^9 Cos[x]^7 + 160 (I Sin[x])^7 Cos[x]^9) + Exp[11 I y] (12 (I Sin[x])^5 Cos[x]^11 + 12 (I Sin[x])^11 Cos[x]^5 + 44 (I Sin[x])^7 Cos[x]^9 + 44 (I Sin[x])^9 Cos[x]^7 + 25 (I Sin[x])^10 Cos[x]^6 + 25 (I Sin[x])^6 Cos[x]^10 + 46 (I Sin[x])^8 Cos[x]^8 + 1 (I Sin[x])^12 Cos[x]^4 + 1 (I Sin[x])^4 Cos[x]^12) + Exp[13 I y] (6 (I Sin[x])^9 Cos[x]^7 + 6 (I Sin[x])^7 Cos[x]^9 + 2 (I Sin[x])^5 Cos[x]^11 + 2 (I Sin[x])^11 Cos[x]^5 + 1 (I Sin[x])^4 Cos[x]^12 + 1 (I Sin[x])^12 Cos[x]^4 + 3 (I Sin[x])^10 Cos[x]^6 + 3 (I Sin[x])^6 Cos[x]^10 + 6 (I Sin[x])^8 Cos[x]^8) + Exp[15 I y] (1 (I Sin[x])^6 Cos[x]^10 + 1 (I Sin[x])^10 Cos[x]^6)); probability[x_, y_] := Abs[amplitude[x, y]]^2; result = NMaximize[{nstates*probability[a, b], 0 < a < Pi/2, 0 < b < Pi}, {a, b}, Method -> {"SimulatedAnnealing", "PerturbationScale" -> 15}]; Print[name, ": ", result] f = probability[c, d]; n = Pi; Plot3D[f, {c, 0, n/2}, {d, -n, n}, PlotRange -> All] ContourPlot[probability[x, y], {x, 0, n/2}, {y, 0, n}, PlotLegends -> Automatic, Contours -> 30] solve24[x : {_Integer, _Integer, _Integer, _Integer}] := Module[ {operators = Tu26, so 130 + 26 = 156 dollars. So, adding those two amounts together: 260 + 156. Let me do the addition. 200 + 100 is 300, es lines and circles, but maps M to another point. However, unless M is the center, inversion won't necessarily simplify things. Alternatively, use the concept of symmedian. The line l bisects PQ at M, and if PQ is seen as a segment whose midpoint is M, then l could be related to the symmedian of the triangle or quadrilateral. I think I need to look for a different property or theorem that directly connects the midpoint of a segment inside a cyclic quadrilateral with the midpoint of the corresponding chord. After some research, I found a relevant theorem: In a cyclic quadril)^2 Cos[x]^10 + 138 (I Sin[x])^6 Cos[x]^6 + 31 (I Sin[x])^3 Cos[x]^9 + 31 (I Sin[x])^9 Cos[x]^3 + 128 (I Sin[x])^5 Cos[x]^7 + 128 (I Sin[x])^7 Cos[x]^5 + 1 (I Sin[x])^1 Cos[x]^11 + 1 (I Sin[x])^11 Cos[x]^1) + Exp[5 I y] (15 (I Sin[x])^2 Cos[x]^10 + 15 (I Sin[x])^10 Cos[x]^2 + 38 (I Sin[x])^8 Cos[x]^4 + 38 (I Sin[x])^4 Cos[x]^8 + 62 (I Sin[x])^6 Cos[x]^6 + 52 (I Sin[x])^7 Cos[x]^5 + 52 (I Sin[x])^5 Cos[x]^7 + 25 (I Sin[x])^3 Cos[x]^9 + 25 (I Sin[x])^9 Cos[x]^3 + 1 Cos[x]^12 + 1 (I Sin[x])^12 + 3 (I Sin[x])^1 Cos[x]^11 + 3 (I Sin[x])^11 Cos[x]^1) + Exp[7 I y] (15 (I Sin[x])^9 Cos[x]^3 + 15 (I Sin[x])^3 Cos[x]^9 + 14 (I Sin[x])^7 Cos[x]^5 + 14 (I Sin[x])^5 Cos[x]^7 + 3 (I Sin[x])^2 Cos[x]^10 + 3 (I Sin[x])^10 Cos[x]^2 + 15 (I Sin[x])^4 Cos[x]^8 + 15 (I Sin[x])^8 Cos[x]^4 + 14 (I Sin[x])^6 Cos[x]^6 + 1 (I Sin[x])^11 Cos[x]^1 + 1 (I Sin[x])^1 Cos[x]^11) + Exp[9 I y] (1 (I Sin[x])^1 Cos[x]^11 + 1 (I Sin[x])^11 Cos[x]^1 + 2 (I Sin[x])^9 Cos[x]^3 + 2 (I Sin[x])^3 Cos[x]^9 + 3 (I Sin[x])^7 Cos[ult method -- Just True => generic default method data IfaceAT = IfaceAT -- See Class.ClassATItem IfaceDecl -- The associated type declaration (Maybe IfaceType) -- Default associated type instance, if any -- This is just like CoAxBranch data IfaceAxBranch = IfaceAxBranch { ifaxbTyVars :: [IfaceTvBndr] , ifaxbLHS :: IfaceTcArgs , ifaxbRoles :: [Role] , ifaxbRHS :: IfaceType , ifaxbIncomps :: [BranchIndex] } -- See Note [Storing compatibility] in CoAxiom data IfaceConDecls = IfAbstractTyCon Bool -- c.f TyCon.AbstractTyCon | IfDataFamTyCon -- Data family | IfDataTyCon [IfaceCosolutions does not exceed 9. So if there are no solutions, the condition is trivially satisfied. But in the case of \( a = 0 \), the inequality \( x^2 < 0 \) has no solutions, so the difference between any two solutions (which don't exist) is not an issue. Therefore, \( a = 0 \) is allowed. However, in the answer, the maximum and minimum values of \( a \) would still be 1 and -1, since when \( a \) is between -1 and 1, the inequality either has solutions with the required condition or has no solutions (at a=0). But when \( a \) is outside of [-1,1], the inequality has solutions with a difference exceeding 9. So the permissible values of \( a \) are still from -1 to 1, inclusive. Therefore, the sum is 0. But let's confirm with the case \( a = 0 \). The problem is asking for the sum of the maximum and minimum values of \( a \). If \( a = 0 \) is allowed, but the maximum and minimum are 1 and -1, respectively, their sum is still 0. So regardless of whether \( a = 0 \) is included or not, since it's between -1 and 1, the maximum and minimum are still 1 and -1. Therefore, the answer should be 0. But I need to make sure there isn't another interpretation of the problem. For example, maybe the problem requires the inequality to have solutions, so \( a \neq 0 \). If that's the case, then the interval for \(0 I y] (32 (I Sin[x])^6 Cos[x]^9 + 32 (I Sin[x])^9 Cos[x]^6 + 35 (I Sin[x])^8 Cos[x]^7 + 35 (I Sin[x])^7 Cos[x]^8 + 6 (I Sin[x])^4 Cos[x]^11 + 6 (I Sin[x])^11 Cos[x]^4 + 17 (I Sin[x])^5 Cos[x]^10 + 17 (I Sin[x])^10 Cos[x]^5 + 1 (I Sin[x])^3 Cos[x]^12 + 1 (I Sin[x])^12 Cos[x]^3) + Exp[12 I y] (9 (I Sin[x])^7 Cos[x]^8 + 9 (I Sin[x])^8 Cos[x]^7 + 4 (I Sin[x])^9 Cos[x]^6 + 4 (I Sin[x])^6 Cos[x]^9 + 1 (I Sin[x])^10 Cos[x]^5 + 1 (I Sin[x])^5 xR7T6c3M3H0pplkcq6GP7xJwOWOq2Lg1GvUcWhNwTQSEr3ps2/u JNq5ZCpznac3hshJfuQ3xcjeBXdTSU6XjH8zKymQ+21kyJEvO3TJhNJ2AGQz sN8AmA7fgFt88KWQ+o4Q4ELuKvGCoZ/qcrlc9BFmDe23fEggjO3IAfvNEXqG OnvbIbtb4IdLvbV1tdVVVZ8dbtM7FgCSwPL2W+NRLylMMo9bSFlKRl1wd3Se IUs3vePhD9vP/J2JRJat+stf3xesYiF4rx6WO+38d48o6/6ozbEtss19deXG F6/k8FqVTnzdD6F7L3PU0eKU9/DUPXI = 2 (3 - 2 sqrt(2)) / (9 - 8) = 2 (3 - 2 sqrt(2)). Wait, but this gives 6 - 4 sqrt(2), which is what we had before. So, the ratio is 6 - 4 sqrt(2) 0.343. But this is not one of the answer choices. The closest is option A (1/3) or option C (2/3). If the problem is from an exam like the GRE or similar, and given the answer choices, perhaps there's an error in my approach. Wait, maybe the semicircle is not a flat semicircle but oriented differently. For example, a semicircle that is standing on its diameter, which is a chord of the quarter circle's arc. Let's explore this. Suppose the semicircle is inscribed such that its diameter is a chord of the quarter circle's arc. The quarter circle's arc is 90 degree, since we may be changing uniques, we'd have -- to substitute to keep it right ; let bndr' | exprIsHNF rhs' = bndr `setIdUnfolding` evaldUnfolding | otherwise = bndr `setIdUnfolding` noUnfolding ; return (floats3, bndr', rhs') } where is_strict_or_unlifted = (isStrictDmd dmd) || is_unlifted platform = targetPlatform (cpe_dynFlags env) arity = idArity bndr -- We must match this arity --------------------- float_from_rhs floats rhs | isEmptyFloats floats = return (emptyFloats, rhs) | isTopLevel top_lvl = float_top floats rhs | otherwise = float_nested floats rhs tal line and two non-horizontal lines. The first non-horizontal line adds 2 regions (as h + 1 = 1 +1=2). So from 2 regions to 4. Then the second non-horizontal line crosses the horizontal line and the first non-horizontal line. So two intersections, which split the second line into three segments, adding three regions. So total regions 4 + 3 =7. According to the formula: (1 +1) +1*2 +2*(2+1)/2 = 2 + 2 + 3 =7. Correct. So the formula seems to work. Therefore, the total number of regions is: R = (h + 1) + h*s + s(s +1)/2. But the problem states that the total number of regions is 1992. Therefore: (h +1) + h*s + (s^2 + s)/2 = 1992. We need to find positive integers h and s such that this equation is satisfied. So, let's write the equation: (h +1) + h*s + (s^2 + s)/2 = 1992. Let me rearrange this equation: Multiply both sides by 2 to eliminate the fraction: 2*(h +1) + 2*h*s + s^2 + s = 3984. Expand the terms: 2h + 2 + 2h s + s + s = 3984. Combine like terms: 2h s + s + s + 2h + 2 = 3984. Lets group terms: s + (2h +1)s + 2h + 2 = 3984. Therefore: s + (2h +1)s + 2h + 2 - 3984 = 0. Simplify the constants: s + (2h +1)s + 2h - 3982 = 0. This is a quadratic equation in terms of s, but h is also an integer variable.n halv avokado i klyftor och ett par mycket tunt skivade rdlksringar (om man ligger t det hllet). Rdbetshg slarvigt men genomtnkt placet 1's output shows that the sentences are ordered in a way that the third word varies first. But in the options array, the third word's options are ["cheerful", "happy", "joy"], which are lex order. So the first combination uses "cheerful", then "happy", then "joy". But in the sample output, the first three entries have "cheerful" and "happy" and "joy" in the third position, but in the sample output, the order is: "I am cheerful ...", "I am cheerful ...", "I am happy ...", "I am happy ...", "I am joy ...", "I am joy ...". Each third word is ordered in the options' order. So the generated list is in lex order. So the final array is already sorted. Thus, the code can generate the Cartesian product in the order of the o percent change in Alice's stock over the two days? Okay, so Alice buys a stock, and then on the first day, it decreases by 10%. Then on the second day, it increases by 20% of its value at the end of the first day. I need to figure out the overall percent change over the two days. Hmm, let's break this down step by step. First, let me recall that percentage changes are calculated based on the current value. So if a stock decreases by 10%, that means it's worth 90% of its original value the next day. Similarly, an increase of 20% would mean it's 120% of the previous day's value. Maybe I should assign a variable to the original stock price to make this easier. Let's say the initial price is \$100. That might make the percentages easier to work with because percentages are out of 100. Okay, so let's go withut this is vague. Let's try a concrete example. Suppose we take three parallel lines, each with three points. Then, in this case, there are three lines, each with three points. Now, if I take six points from these nine, by the pigeonhole principle, at least two points must come from each line (since 6 divided by 3 lines is 2 per line). Wait, but that's not necessarily true. If we take three points from two lines and zero from the third, then six points would have three points on each of two lines, but each line has three points. But in this case, each line only has three points, so taking three points from a line would give a collinear triple. However, the problem states that in any six points, there must be at least one collinear triple. But in th result.Add(node.val); if (node.left != null) queue.Enqueue(node.left); if (node.right != null) ed \( C \) of being the spy, after which \( C \), pointing either to \( A \) or \( B \), stated: "In reality, the spy is him!" The court managed to identify the spy. Who is the spy? Okay, let's try to figure out who the spy is among A, B, and C. So we have three people:ally respecting how this film went out of its way to break out of the usual Trek mold. The threat might be a bit on the liberal scare side of things as an unknown probe wrecks havoc on 23rd Century Earths atmosphere and sucking the power out of ships and Federation outposts on its approach to our planet. The probe wants to talk to the whales, but we killed em all and the damn thing wont leave until it talks to one. But what I find really interesting about this movie is that there isnt a big bad guy to beat. It reminded me of some of the original Trek episodes that were less about a singular threat and more about a sci-fi way around a seemingly insolvable problem.This fast-paced and familiar plot of using up second chances to discover just who you really are is augmented by several choices that OMalley made for his secondary characters. While it would have been easy for him to populate his characters with strictly Caucasian people, the characters in Seconds resemble the people that you would find in most any restaurant or business today: people of various ethnic groups, a loving gay couple, overweight and malnourished people, people who are not move-star attractive. Furthermore, OMalley does not place over-emphasis on this diverse cast of characters: they are people first and foremost and their loves and lusts, faults and virtues, are shown to be as natural as those of Katie herself. It is their interactions with her and how the alt-Katies respond to them, that make Seconds different from most other based-on-life graphic novels.Skip adYou might likekrinyusz2632, Woman, SingleBudapest, HungaryZafira2836, Woman, SingleBudapest, HungaryNatkaa29, Woman, SingleSzabadszlls, HungaryNikoletza35, Woman, Courn sum(len(feature['description']) for feature in self.features) Then, the function to generate the design: def generate_user_centered_design(project_name, project_description, user_personas, features, user_flow_steps, visual_design): design = [] # Project Overview design.append(f"Project Overview\n================") design.append(f"Name: {project_name}") design.append(f"Description: {project_description}\n") # User Personas design.append("User Personas\n-------------") for persona in user_personas: design.append(f"Persona: {persona['name']}") design.append(f"- Needs: {', '.join(persona['needs'])}") design.append(f"- Characteristics: {', '.join(persona['characteristics'])}\n") # Feature List design.append("Feature List\n------------") # Sort fTORES>CRISTAL TEMPLADO>UNIVERSALES>Protector Cristal Universal 4,5" Protector Cristal Universal 4,5" Tweet Compartir Google+ Referencia Estado: Nuevo Advertencia: ltimos artculos en inventario! Disponible el: 13/05/2016 Enviar a un amigo Enviar a un amigo Protector Cristal Universal 4,5" Recipiente : Nombre de tu amigo * : E-mail de tu amigo * : * Campos obligatorios Enviar total Cancelar Imprimir 4,99 Este producto no se vende individualmente. Usted debe seleccionar un mnimo de 0 cantidades para este producto. Aadir al carrito Aadir para comparar Add to my wishlist Best Sellers Pro2[z] \end{eqnarray*}$$ as claimed. 4. Suppose $z = \letin{x}{e_1}{e_2}$. Now $\freeLambdaVars(z) = \freeLambdaVars(e_1) \cup (\freeLambdaVars(e_2) \setminus \{x\})$. Using the hypothesis, we have $\sigma_1 =^{\freeLambdaVars(e_1)} \sigma_2$ and $\sigma_{1, y/x} =^{\freeLambdaVars(e_2) \setminus \{x\}} \sigma_{2, y/x}$. By reflexivity we have $e_2 \AlphaEq e_2$, and using the [previous result](@alpha-eq-to-new-sets), we have $\new(x,e_2,\sigma_1) = \new(x,e_2,\sigma_2)$. Using the inductive hypothesis, we then have $$\begin{eqnarray*} & & \sigma_1[t the cube root term is n^(2/3). So the numerator is approximately n^(2/3) - (n + 5/(2n)). Wait, but n^(2/3) is much smaller than n as n approaches infinity. So the numerator is like n^(2/3) - n - something very small. Sos 15 pairs of distinct roads, plus 6 pairs where they choose the same road)edStar]\>\"", " ", "]"}]}], ";"}], "*)"}], "\n", " ", RowBox[{"ucTempSym", " ", "=", " ", RowBox[{"Symbol", "[", " ", RowBox[{"\"\<\[LeftGuillemet]\>\"", " ", "<>", " ", RowBox[{"ToString", "[", " ", RowBox[{"++", "ucTempSymN"}], " ", "]"}], " ", "<>", " ", "\"\<\[RightGuillemet]\>\""}], " ", "]"}]}], ";", "\n", "\n", " ", RowBox[{"anonymous", " ", "=", " ", RowBox[{ RowBox[{"Head", "[", "symName", "]"}], " ", "===", " ", "SymbolName"}]}], ";", "\n", " ", RowBox[{"If", "[", " ", RowBox[{"anonymous", ",", " ", RowBox[{"symName", " ", "=", " ", RowBox[{"SymbolName", "[", " ", "ucTempSym", " ", "]"}]}]}], " ", "]"}], ";", "\n", "\n", " ", RowBox[{"curName", " ", "=", " ", RowBox[{"Uc\[ScriptCapitalV]$Option", "[", " ", RowBox[{"expr", ",", " ", "Name"}], " ", "]"}]}], ";", "\n", " ", RowBox[{"If", " ", "[", " ", RowBox[{ RowBox[{ RowBox[{"!", " ", "anonymous"}], " ", "||", " ", RowBox[{"(", " ", RowBox[{"anon- -- @ -- f x = let f x = x + 1 -- in f (x - 2) -- @ -- -- Would be renamed by having 'Unique's attached so it looked something like this: -- -- @ -- f_1 x_2 = let f_3 x_4 = x_4 + 1 -- in f_3 (x_2 - 2) -- @ -- -- 3. The resulting syntax tree undergoes type checking (which also deals with instantiating -- type class arguments) to yield a 'HsExpr.HsExpr' type that has 'Id.Id' as it's names. -- -- 4. Finally the syntaxcity and quality condition. After accumulating a group, they are imported to Vermont in large containers. This method keeps shipping costs low and enables us to offer them to you at reasonable prices. We ship internationally.The writer contends that my statement is a myth and simply untrue. Unfortunately, my use of the word always was misinterpreted. I meant it in the sense that marriage has been between one man and one woman from the beginning. Regrettably, it has not remained that way consistently and everywhere throughout history. The foundation for my belief lies in scripture: Have you not read that he who made them from the beginning made them male and female, and said, For this reason a man shall leave his father and mother and be joined to his wife, and the two shall become one? (Matthew 19: 4,5). Also Genesis 1:27,28: male and female he created them. And God blessed them, and God said to them, Be fruitful and multiply.Under California Public Utilities Commission mandates, the CHP is responsible for safety compliance inspections of school buses, heavy trucks, farm labor vehicles, and commercial vehicles. A major CHP responsibility is performing safety inspection of commercial transportation vehicles at state inspection stations. Inspections include checking for the transporting of hazardous wastes. Responsible for traffic enforcement on freeways or highways. Responsible for protecting state property and for assisting state employees during emergency situations. Responsible for providing VIN (Vehicle Identification Numbers) numbers for stolen cars or reconstructed cars with missing VIN numbers California Highway PatrolUber Chases The Perfect Ride With Back-To-Back Trips 3 hours ago by Josh Constine Googles Plan To Accelerate The Mobile Web Will Go Live Early Next Year 4 hours ago by Sarah Perez Apple Has Acquired Faceshift, Maker Of Motion Capture Tech Used In Star Wars 18 hours ago by Ingrid Lunden, Natasha Lomas Zenefits Under Investigation For Allegedly Allowing Unlicensed 4 units right of D, at (18,0), and F is 7 units at 100 from D. But if we ignore the coordinates and focus on the problem statement, we need more information about how the triangles are connected or positioned to find CD. Since the problem is presented with the Asymptote figure, maybe the key is to use coordinate geometry based on the given code, even if there are inconsistencies. Alternatively, maybe the figure is correct, and the side lengths are different. Alternatively, perhaps the labels in the Asymptote code are mixed up. Let's check again: The code labels "A" at point F, which is part of the second triangle. Similarly, labels "B" at point E. Maybe this is a mistake, and the actual labels should be for the second triangle as D, E, F. The original triangle is A, B, C. Then the figure has two separate triangles, ABC and DEF, each with their own sides and angles. Then CD is the distance between point C of the first triangle and point D of the second triangle. But how are these triangles positioned? Without more information, it's impossible to determine CD unless there's a specific configuration. The Asymptote code suggests that D is located at (14,0), which is 6 units to the right of B(8,0). So if B is at (8,0), D is at (14,0). Then C is at (14*cos100,14*sin100). So coordinates: Point C: (14*cos100,14*sin100) Point D: (e that: 20 * 7 = 140 pages. But wait, before I jump to conclusions, let me make sure there isn't another interpretation. For example, does "2 more pages per day" mean 2 more pages per each reading session? So, instead of 6 pages each time, she reads 8 pages each time, 3 times a day? That would be 3 * 8 = 24 pages per day, which is 6 more pages per day, not 2. So that seems inconsistent with the problem statement. Alternatively, maybe "2 more pages per day" is in addition tft vector quantization, and demonstrated improved performance through orthogonal linear filters in neural networks. Check if that's accurate. The key points are the extension via GMMs, inclusion of new nonlinearities, -VQ, and orthogonality. Yep, that seems to cover the main contributions as stated in the abstract and conclusion. Make sure it's one sentence. Looks good. The main contribution of the work is extending the max-affine spline operator (MASO) framework via Gaussian Mixture Models (GMMs) to unify classical nonlinearities (e.g., sigmoid, softmax) and new ones (e.g., swish) through probabilistic "soft" and hybrid -vector quantization inference, while improving performance via orthogonal linear filters.It's Mother's day, so mothers get a 10% discount on the department store. Mothers who have 3 children or more can get an additional 4% off the discounted price. The department store also offers a 5% discount for VIP members. All these discounts can be combined, but to avail them, the customer must purchase a minimum of 3 items. If Mrs. Brown has 4 children, is a VIP member, and wants to buy a pair of shoes that costs $125, a handbag worth $75, and a scarf priced at $45, h8 (I Sin[x])^3 Cos[x]^1as bags. Now, the rest are suede bags. So, if we subtract the number of leather and canvas bags from the total, we should get the number of suede bags. Let's write that out. Total bags = leather + canvas + suede So, suede = total - leather - canvas Plugging in the numbers we have: Suede = 90 - 60 - 18 Let me do that subtraction. 90 minus 60 is 30, and 30 minus 18 is 12. So, suede bags would be 12. Wait, that seems straightforward, but let me check if I did everything right. Alternatively, maybe I can approach it by finding the fraction of suede bags first. The problem says two-thirds are leather, one-fifth are canvas, and the rest are suede. So, the fractions should add up to 1 (or the whole). Let me check: Leather: 2/3 Canvas: 1/5 So, adding those together: 2/3 + 1/5. To add these fractions, they need a common denominator. The least common denominator for 3 and 5 is 15. So converting them: 2/3 = 10/15 1/5 = 3/15 Adding them together: 10/15 + 3/15 = 13/15 Therefore, the remaining fraction is 1 - 13/15 = 2/15. So, suede bags make up 2/15 of the total. Therefore, the number of suede bags is 2/15 of 90. Let's compute that: (2/15) * 90. 90 divided by 15 is 6, so 6 times 2 is 12. That's the same result as before. So that checks out. Hmm, so both methods give me 12. That seems consistent. Let me just confirm once again to be sure. First method: Leather: 60 (2/3 of 90) Canvas: as: x = 0, sin(/3^n), sin(2/3^n), ..., sin((3^n -1)/(2*3^n)). But since (3^n -1)/2 = (3^n -1)/2, which is an integer, the upper limit for k is (3^n -1)/2. Therefore, the solutions are x = sin(k /3^n) for k =0,1,2,..., (3^n -1)/2, but excluding duplicates. Since sin(0)=0, and for each k from 1 to (3^n -1)/2, we have sin(k /3^n). Therefore, the complete solution set is x=0 and x=sin(k /3^n) for k=1 to (3^n -1)/2. Thus, the final answer is all real numbers x such that x=0 or x=sin(k /3^n) where k is an integer with 1 k (3^n -1)/2. To write this concisely in boxed notation, we can express it as: The solutions are \( x = 0 \) and \( x = \pm \sin\left( \frac{k\pi}{3^n} \right) \) for \( k = 1, 2, \ldots, \frac{3^n - 1}{2} \). But to present this in a more standard form, perhaps using set notation or listing the general form. However, in the context of the problem, they might expect the answer written in terms of sine functions with the specific arguments. Therefore, the final answer is: All real solutions are given by \( x = 0 \) and \( x = \pm \sin\left( \frac{k\pi}{3^n} \right) \) for each integer \( k \) satisfying \( 1 \leq k \leq \frac{3^n - 1}{2} \). Thus, boxed: The solutions are \( \boxed{0} \) and \( \boxed{\pm \sin\left( \frac{k\pi}{3^n} \right)} \) for \( k = 1, 2, \ldots, \frac{3^n - 1}{2} \). But since the question asks to "solve the equation \( f_n(x) = 0 \)", the answer should present all solutions explicitly. However, due to the large number of solutions (3^n solutions), it's standard to express them in terms of the sine functions as above. Therefore, the final answer can be written FastString import Control.Monad( when ) import qualified Data.Map as Map \end{code} This function is aShow needed for Lexer.x \end{code} Note [InlinePragma] ~~~~~~~~~~~~~~~~~~~ This data type mirrors what you can write in an INLINE or NOINLINE pragma in the source program. If you write nothing at all, you get defaultInlinePragma: inl_inline = False inl_act = AlwaysActive inl_rule = FunLike It's not possible to get that combination by *writing* something, so if an Id has defaultInlinePragma it means the user didn't specify anything. If inl_inline = True, then the Id should have an InlineRule unfolding. Note [CONLIKE pragma] ~~~~~~~~~~~~~~~~~~~~~ The ConLike constructor of a RuleMatchInfo is aimed at the following. Consider first {-# RULE "r/cons" forall a as. r (a:as) = f (a+1) #-} g b bs = let x = b:bs in ..x...x...(r x)... Now, the rule applies to the (r x) term, because GHC "looks through" the definition of 'x' to see that it is (b:bs). Now consider {-# RULE "r/f" forall v. r (f v) = f (v+1) #-} g v = let x = f v in ..x...x...(r x)... Normally the (r x) would *not* match the rule, because GHC would be scared about duplicating the redex (f v), so it does not "look through" the bindings. However the CONLIKE modifier says to treat 'f' like a constructor in this situation, and "look through" the unfolding for x. So (r x) fires, yielding (f (v+1)). This is all controlled with a user-visible pragma: {-# NOINLINE CONLIKE [1] f #-} The main effects of CONLIKE are: - The occurrence analyser (OccAnal) and simplifier (Simplify) treat eneral result for any natural number n? Let me try to generalize. Suppose we have any n N. Then: Expression A: \(\frac{x^n}{1 - x^2} + \frac{y^n}{1 - y^2}\) Expression B: \(\frac{x^n + y^n}{1 - xy}\) To compare A and B, let's consider their difference: A - B = \(\frac{x^n}{1 - x^2} + \frac{y^n}{1 - y^2} - \frac{x^n + y^n}{1 - xy}\) Let me write each term as a series expansion. For the first term: \(\frac{x^n}{1 - x^2} = x^n + x^{n + 2} + x^{n + 4} + \dots\) Similarly, \(\frac{y^n}{1 - y^2} = y^n + y^{n + 2} + y^{n + 4} + \dots\) So Expression A is the sum of these two series: (x^n + x^{n+2} + x^{n+4} + ...) + (y^n + y^{n+2} + y^{n+4} + ...) Expression B is \(\frac{x^n + y^n}{1 - xy}\). Let's expand this as a series: (x^n + y^n) * (1 + xy + x^2 y^2 + x^3 y^3 + ...) = x^n + y^n + x^{n+1} y + x y^{n+1} + x^{n+2} y^2 + x^2 y^{n+2} + ... Therefore, Expression B as a series is: x^n + y^n + x^{n+1} y + x y^{n+1} + x^{n+2} y^2 + x^2 y^{n+2} + ... Now, subtract B from A: A - B = [ (x^n + x^{n+2} + x^{n+4} + ...) + (y^n + y^{n+2} + y^{n+4} + ...) ] - [ x^n + y^n + x^{n+1} y + x y^{n+1} + x^{n+2} y^2 + x^2 y^{n+2} + ... ] Simplify by canceling the x^n and y^n terms: A - B = (x^{n+2} + x^{n+4} + ... + y^{n+2} + y^{n+4} + ...) - (x^{n+1} y + x y^{n+1} + x^{n+2} y^2 + x^2 y^{n+2} + ... ) Now, let's analyze the terms. For each term in A beyond x^n and y^n, we have x^{n + 2k} and y^{n + 2k} for k 1. In B, after x^n and y^n, we have terms of the form x^{n + k} y^{k} and x^{k} y^{n + k} for k 1. So, in other words, for each exponent m n + 1, we can compare the coefficients from A and B. But perhaps a better way is to group terms by their exponents. Let's consider the terms of degree m = n + 2k in A, which are x^{n + 2k} and y^{n + 2k}, and in B, the terms of the same degree would be x^{n + k} y^{k} and x^{k} y^{n + k} for k such that n + k + k = n + 2k = m. So for each m = n + 2k, in A we have x^m + y^m, and in B we have x^{n + k} y^{k} + x^{k} y^{n + k}. Therefore, the difference A - B for each degree m = n + 2k is: (x^m + y^m) - (x^{n + k} y^{k} + x^{k} y^{n + k}) But since m = n + 2k, this becomes: (x^{n + 2k} + y^{n + 2k}) - (x^{n + k} y^{k} + x^{k} y^{n + k}) Factorizing, let's set a = x^{k} and b = y^{k}. Then, the expression becomes: a^{n/k + 2} + b^{n/k + 2} - a^{n/k + 1} b - a b^{n/k + 1} Wait, maybe not the best approach. Alternatively, note that x^{n + 2k} + y^{n + 2k} - x^{n + k} y^k - x^k y^{n + k} can be written as x^{n + k} (x^k - y^k) + y^{n + k} (y^k - x^k) = (x^{n + k} - y^{nd Weitergabe von Informationen: Mithilfe der Google +1-Schaltflche knnen Sie Informationen weltweit verffentlichen. ber die Google +1-Schaltflche erhalten Sie und andere Nutzer personalisierte Inhalte von Google und unseren Partnern. 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Die Identitt Ihres Google- Profils kann Nutzern angezeigt werden, die Ihre E-Mail-Adresse kennen oder ber andere identifizierende Informationen von Ihnen verfgen.Aores Alemanha Andorra Argentina arquitetura automobilismo Azerbaijo Blgica blogagem Bsnia-Herzegovina Canad casamento Cear Chile China compras Coreia do Norte Coreia do Sul costumes Crocia devaneios Dinamarca Emirados Equador escritores Eslovquia Eslovnia Espanha Finlndia folclore fotografia Frana futebol gastronomia Gergia Gois Guatemala Ilhas Faroe India Inglaterra Islndia Itlia Jogos Olmpicos Kosovo Liechtenstein lngua (b, 0) \). - Let the given circle have center \( C = (h, k) \) and radius \( r \). 2. **Parametrize Point \( P \):** - As \( P \) moves on the circle, its coordinates are \( P = (h + r\cos\theta, k + r\sin\theta) \), where \( \theta \) varies from \( 0 \) to \( 2\pi \). 3. **Centroid Coordinates:** - The centroid \( G \) of triangle \( ABP \) has coordinates: \[ G_x = \frac{0 + b + h + r\cos\theta}{3} = \frac{b + h}{3} + \frac{r}{3}\cos\theta, \] \[ G_y = \frac{0 + 0 + k + r\sin\theta}{3} = \frac{k}{3} + \frac{r}{3}\sin\theta. \] 4. **Eliminate Parameter \( \theta \):** - Express \( \cos\theta \) and \( \sin\theta \) in terms of \( G_x \) and \( G_y \): \[ \cos\theta = \frac{3G_x - b - h}{r}, \quad \sin\theta = \frac{3G_y - k}{r}. \] - Use the identity \( \cos^2\theta + \sin^2\theta = 1 \): \[ \left(\frac{3G_x - b - h}{r}\right)^2 + \left(\frac{3G_y - k}{r}\right)^2 = 1. \] - Simplify to obtain the equation of a circle: \[ \left(G_x - \frac{b + h}{3}\right)^2 + \left(G_y - \frac{k}{3}\right)^2 = \left(\frac{r}{3}\right)^2. \] 5. **Conclusion:** - The locus is a circle with center at \( \left(\frac{b + h}{3}, \frac{k}{3}\right) \) (the centroid of \( A \), \( B \), and \( C \)) and radius \( \frac{r}{3} \). **Final Answer:** The locus is a circle with radius one-third of the original circle's radius and center at the centroid of points \( A \), \( B \), and the original circle's center. Thus, the locu time occurs at specific seconds. Wait, the clock changes every second, so each palindromic time is a unique second. Therefore, there are 96 such seconds in a day. But the question is asking "how much time throughout the day does the clock show a number that reads the same forward and backward?" So it's 96 seconds. But tis for extra dependencies coming from .hs-boot files -- See Note [Extra dependencies from .hs-boot files] in RnSource -- Dump trace of renaming part rnDump (ppr rn_decls) ; -- Typecheck them all together so that -- any mutually recursive types are done right -- Just discard the auxiliary bindings; they are generated -- only for Haskell source code, and should already be in Core tcg_env <- tcTyAndClassDecls emptyModDetails rn_decls ; dep_files <- liftIO $ readIORef (tcg_dependent_files tcg_env) ; setGblEnv tcg_env $ do { -- Make the new type env available to stuff slurped from interface files -- Now the core bindings core_binds <- initIfaceExtCore (tcExtCoreBindings src_binds) ; -- Wrap up let { bndrs = bindersOfBinds core_binds ; my_exports = map (Avail . idName) bndrs ; -- ToDo: export the data types also? mod_guts = ModGuts { mg_module = this_mod, mg_boot = False, mg_used_names = emptyNameSet, -- ToDo: compute usage mg_used_th = False, mg_dir_imps = emptyModuleEnv, -- ?? mg_deps = noDependencies, -- ?? mg_exports = my_exports, right triangle TAD, area is 1/2*TA*TD. If we can find TA*TD. From the system of equations, we had: TA*TD = p^2 + q^2 (if A=(p,q), D=(-q,p)), but this is incorrect. Wait, no. If A=(p,q) and D=(x_D, y_D), with p x_D + q y_D=0, then TA*TD= sqrt(p^2 + q^2)*sqrt(x_D^2 + y_D^2). But without more info, this is not helpful. Given that I've exhausted methods without success, I think the answer is 8. But I need to check. Wait, total area is 22. Area of BTC is 538.66. If TAD is 8, then TAB + TCD22 -8.66 -85.34. If AB=CD, maybe each contributes half, so ~2.67 each. Seems possible. Alternatively, consider that in triangle TAD, since it's right-angled, and if TA=TD=(area*2). If area is 8, then TA=TD=4. But earlier coordinate example had TA=TD3.24 for area 5.24. Not sure. Alternatively, using the relation from power of a point or similar triangles. Given the time I've spent and the need to conclude, I'll posit that the area oyle -> "Categories"], Cell[ StyleData["Citations", StyleDefinitions -> StyleData["DemoNotes"]], ShowCellBracket -> False, ParagraphSpacing -> {0, 6}], py*(delta_y)/(H-1) = y_min + delta_y*(1 - py/(H-1)) = y_min + delta_y*( (H-1 - py)/(H-1) ) So both expressions are equivalent. Therefore, the code for mapping is: delta_x = x_max - x_min delta_y = y_max - y_min c_real = x_min + (px * delta_x) / (W - 1) c_imag = y_max - (py * delta_y) / (H - 1) But in code, we can compute this as: for (int py = 0; py < H; py++) { for (int px = 0; px < W; px++) { double real = x_min + (px * (x_max - x_min)) / (W - 1); double imag = y_max - (py * (y_max - y_min)) / (H - 1); // compute Mandelbrot for c = real + imag*i } } But what if W or H is 1? Then division by zero. So perhaps handle that by setting delta_x and delta_y accordingly, but in practice, the window will have at least 1 pixel. Now, integrating this into the Qt application. In Qt, the main window can have a QWidget where we override the paintEvent to draw the Mandelbrot set. Alternatively, use a QImage to precompute the pixels and then draw the image. So the steps in Qt would be: 1. Create a QImage with the size of the widget. 2. For each pixel in the QImage, compute the corresponding c value using the mapping above. 3. For each c, compute the number of iterations until |z| > 2, up to max_iter. 4. Color the pixel based on the number of iterations (using a color gradient). Now, the color gradient part: typically, points in the Mandelbrot set (which didn't escape after max_iter) are colored black, and others are colored based on how fast they escaped. So for iteration count from 0 to max_iter-1, map to a color. A common approach is to use a smooth color gradient, perhaps using HSV colors or a predefined palette. For example, smooth coloring can be done by: normalized = iterations / max_iter color = palette[normalized * palette_size] But to make it smoother, sometimes a fractional iteration count is used, considering how much z exceeded the thre"[", "huntListCheck", "]"}], " ", ">", " ", "2"}], ",", " ", "\[IndentingNewLine]", RowBox[{ RowBox[{"boidHunter", " ", "=", " ", "boidHunterTemp"}], ";"}]}], "\[IndentingNewLine]", "]"}], ";", "\[IndentingNewLine]", "\[IndentingNewLine]", RowBox[{"(*", RowBox[{ "This", " ", "is", " ", "the", " ", "eating", " ", "mechanism"}], "*)"}], "\[IndentingNewLine]", RowBox[{"If", "[", RowBox[{ RowBox[{ RowBox[{"RankedMin", "[", RowBox[{ RowBox[{ RowBox[{"DistanceMatrix", "[", "huntList", "]"}], "[", RowBox[{"[", RowBox[{"Length", "[", "huntList", "]"}], "]"}], "]"}], ",", "2"}], "]"}], " ", "<=", "0.3"}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{"boidList", " ", "=", " ", RowBox[{"Delete", "[", RowBox[{"boidList", ",", RowBox[{ RowBox[{"Flatten", "[", RowBox[{"Position", "[", "\[IndentingNewLine]", RowBox[{ RowBox[{ RowBox[{"DistanceMatrix", "[", "huntList", "]"}], "[", RowBox[{"[", RowBox[{"Length", "[", "huntList", "]"}], "]"}], "]"}], ",", "\[IndentingNewLine]", RowBox[{"RankedMin", "[", RowBox[{ RowBox[{ RowBox[{"DistanceMatrix", "[", "huntList", "]"}], "[", RowBox[{"[", RowBox[{"Length", "[", "huntList", "]"}], "]"}], "]"}], ",", "2"}], "]"}]}], "]"}], "]"}], "\[IndentingNewLine]", "[", RowBox[{"[", "1", "]"}], "]"}]}], "]"}]}], ";"}]}], "\[IndentingNewLine]", "]"}], ";", "\[IndentingNewLine]", "\[IndentingNewLine]", RowBox[{"(*", RowBox[{ "This", " ", "is", " ", "for", " ", "plotting", " ", "the", " ", "boids", " ", "and", " ", "hunter"}], "*)"}], "\[IndentingNewLine]", RowBox[{"huntList", " ", "=", " ", RowBox[{"Append", "[", RowBox[{"boidList", ",", "boidHunter"}], "]"}]}], ";", "\[IndentingNewLine]", "\[IndentingNewLine]", RowBox[{"Pause", "[", "0.1", "]"}], ";"}]}], "\[IndentingNewLine]", "]"}], ";"}]}], "\[IndentingNewLine]", "]"}], ";"}], "\[IndentingNewLine]"}], "\[IndentingNewLine]", RowBox[{ RowBox[{"Print", "[", "\"\\"", "]"}], " "}], "\[IndentingNewLine]", RowBox[{ RowBox[{"a", "=", RowBox[{"Interpolation", "[", RowBox[{"numBoidsList", ",", RowBox[{"Method", "\[Rule]", "\"\\""}]}], "]"}]}], ";"}], "\[IndentingNewLine]", RowBox[{ RowBox[{"b", "=", RowBox[{"numberBoids", "-", "numBoidsList"}]}], ";"}], "\[IndentingNewLine]", Rimsky Korsakow Museums Chamber Music Series in St. Petersburg (Russia). Thf it matches the value itself. 4. **Efficiency**: Sorting the array takes O(n log n) time. For each candidate `x`, the binary search operation takes O(log n) time, leading to an overall time complexity of O(n log n), which is efficient for the given problem constraints. This approach ensures that we efficiently determine the correct value of `x` (if it exists) while maintaining optimal performance.Given the sets $A=\{m, 1\}$, $B=\{m^2, -1\}$, and $A=B$, the value of the real number $m$ is ( ) A: 1 B: -1 C: 0 D: $\pm$1 Okay, let's see. So we have two sets, A and B. Set A is {m, 1} and set B is {m, -1}. The problem says that A equals B, and we need to find the real number m from the options given. The options are 1, -1, 0, or plus or minus 1. Hmm, okay. First, let me recall that two sets are equal if they contain exactly the same elements. The order doesn't matter in sets, so even if the elements are listed in a different order, as long as all elements are the same, the sets are equal. Also, sets don't have duplicate elements, so each element must be unique within the set. So, for A to equal B, all elements in A must be in B, and all elements in B must be in A. So, let's write that out. Set A has elements m and 1. Set B has elements m and -1. Therefore, each element in A must be in B, and vice versa. So, starting with the elements of A: m must be either m or -1. Similarly, 1 must be either m or -1. Similarly, looking at elements of B: m must be either m or 1, and -1 must be either m or 1. Let me try to approach this step by step. Let's first tem ImpAll = False isExplicitItem (ImpSome {is_explicit = exp}) = exp -- Note [Comparing provenance] -- Comparison of provenance is just used for grouping -- error messages (in RnEnv.warnUnusedBinds) instance Eq Provenance where p1 == p2 = case p1 `compare` p2 of EQ -> True; _ -> False instance Eq ImpDeclSpec where p1 == p2 = case p1 `compare` p2 of EQ -> True; _ -> False instance Eq ImpItemSpec where p1 == p2 = case p1 `compare` p2 of EQ -> True; _ -> False instance Ord Provenance where compare LocalDef LocalDef = EQ compare LocalDef (Imported _) = LT compare (Imported _ ) LocalDef = GT compare (Imported is1) (Imported is2) = compare (head is1) {- See Note [Compaliko povrino za prenos plinov med zrakom in telesnimi tekoinami, ki zadruje vlago v notranjosti telesa.Your dark hair color doesnt forbid you to go for the granny style. You can keep the roots dark and lighten up the rest with a silver blue hair color. Keep in mind that bleaching is a must in this case and your hair should be prepared for that harsh chemical process. Then h is impossible, suggesting there is an error in the problem's formulation. Given that, but assuming that the problem expects an answer despite the inconsistency, and maybe rounding to the nearest whole number, the total would be 33. However, since standard math problems should_mesCodigoMesa, max(CASE WHEN vot_parCodigo = "0013" THEN votVotosPartido END), max(CASE WHEN vot_parCodigo = "0057" THEN votVotosPartido END), max(CASE WHEN vot_parCodigo = "0081" THEN votVotosPartido END), max(CASE WHEN vot_parCodigo = "0131" THEN votVotosPartido END), max(CASE WHEN vot_parCodigo = "0132" THEN votVotosPartido END), max(CASE WHEN vot_parCodigo = "0133" THEN votVotosPartido END), max(CASE WHEN vot_parCodigo = "0134" THEN votVotosPartido END), max(CASE WHEN vot_parCodigo = "0135" THEN votVotosPartido END), max(CASE WHEN vot_parCodigo = "0136" THEN votVotosPartido END), max(CASE WHEN vot_parCodigo = "0137" THEN votVotosPartido END), max(CASE WHEN vot_parCodigo = "0138" THEN votVotosPartido END) FROM MesasCandidaturaPresidente GROUP BY vot_proCodigoProvincia, vot_depCodigoDepartamento, vot_mesCodigoCircuito, vot_mesCodigoMesa ORDER BY vot_proCodigoProvincia, vot_depCodigoDepartamento, vot_mesCodigoCircuito, vot_mesCodigoMesa) SELECT vot_proCodigoProvincia AS codprov, depNombre AS depto, mes_depCodigoDepartamento AS coddepto, vot_mesCodigoCircuito AS circuito, vot_mesCodigoMesa AS mesa, -- sum(mesVotosValidos) AS validos, sum(v mt79ZGycCzY/tU6B9QrE8vAvPJU65eY7dvqcizZuLgmzEX9LB3862Lqm8pPl uYsZkLc8zFkWczCMPfAZ791m0GISPts1/vtcb5wPP596rJuqHjn19urFf+Us 8/fjYlsNWvtRhu1j47zWwJQnLEfeeq3ykOXF48ViF/Bb18n1wknFLbDBG+19 xnlfij1ObY1Wx8g93zU27PdIPa65eAZ5y3ip9Vq1VU66Mt8NecTrsVH3S2kP P9FD6ZPWd3WM/HqxEPKTcUtrusJy6nHVN1L+i2Ac1rVxXhxs/1wrWPz/ZPG1 yVMWJ4F72vuMQ1o/Vm6zfPaDYqPcGrNyleXmW59W/q/cc2ViLHBVe5/xvUNT DouXUCeuKO9YLAVOJwZDDjP+Zk1aWC48W8fgnsrYDfnjb4f2sVGn+tSL+1nn FpaDf3TOpVychjYv3zV+Lv5BPq9YDfEb+KEcVvEYK3ctbkP8hlzm7QYtbmSF lMPma3Jn2Zhzme+I8ZCzLEcWNrf6T6/V77zmX+qX42wOJcZDXrM8V3iVruW2 ivHAD82PYHnK8lPFY5gvyHkV77Fa1/JZxXWYf5k3KZf7LCfVbzRfMH9hI6/Z 2KsPGEPNF9S5Q3gLvQ+3sQbvR7kmcjfFb+Dq5h1iPOSN+k1+mxxk8wI28kDF aTh2va7xZ7Eu8prlrcJ4u1xSsR+4/f96La6DPU5Ob8UrlPlOrikO77NcY3oM LRX3nilYPiMOT8fEw1cPxrfXTJ04hrWU9XNjpTmCNsinxrFpr2eHM8O4MT5M Y90//BbGXTfJtcLlcHUaK166YfAB4dV0WzyZDkB7xVfnCB7mBtJG8UB8j8aK g7k/k+YebZHyQ8KL6Kq40KypH3+W++u+mK9tljpxSHyM3oqDbZVjx3SM1C/3 ZJvY4F3WLRADJ85tn/jlz+xa7KH1I8QfWhtWvo28LXELcnat tells me that b = -2. Got that. Now, since AB is parallel to the x-axis, the length of AB is the absolute difference between the x-coordinates. Because when two points are on a horizontal line, the distance between them is just |x2 - x1|. So AB = |1 - a| = 3. So, |1 - a| = 3. To solve for a, we can split this into two cases. Case 1: 1 - a = 3. Then, solving for a, we get a = 1 - 3 = -2. Case 2: 1 - a = -3. Then, 1 - a = -3 => -a = -4 => a = 4. So, a can be either -2 or 4. Therefore, there are two possible solutions for a. But the problem doesn't specify any other conditions, so both solutions are valid. Then, a + b would be either (-2) + (-2) = -4 or 4 + (-2) = 2. Wait, but the proboth circles, so the line connecting the centers is perpendicular to this tangent. If the tangent line at \(T\) is common, then the angle between \(TA\) and the tangent is equal to the angle in the alternate segment for circle \(TAB\), which is \(\angle TBA\), and similarly the angle between \(TD\) and the tangent is equal to the angle in the alternate segment for circle \(TCD\), which is \(\angle TCD\). As we established earlier, these angles are equal: \(\angle TBA = \angle TCD\). Lets denote \(\theta = \angle TBA = \angle TCD\). Then, in trhe number of distinct patterns is then 2^dim. To find the dimension, we can set up a matrix where each row corresponds to a move, and each column corresponds to a cell. Then, perform row operations to find the rank. Alternatively, check for linear independence among the four vectors. But since the problem is over GF(2), we need to check for linear dependencies modulo 2. Let me write out the four vectors. Let's name the cells as follows for simplicity: A B C D E F G H I So the cells ablem might consider that as containing two numbers and their difference even if the difference is one of the n - Eye of the BeholderDVD 105 Min.Kinowelt Home EntertainmentDas Auge - Eye of the BeholderDVD 105 Min.Kinowelt Home EntertainmentDas geheime FensterDVD 92 Min.SONY Pictures Home EntertainmentDas geheime FensterBlu-ray 96 Min.SONY Pictures Home EntertainmentDas geheime FensterDVD 92 Min.SONY Pictures Home EntertainmentDer Auftrag - The AssignmentDVD 115 Min.Columbia Tristar Home EntertainmentFrauenhandel - Kampf gegen das KartellDVD 180 Min.Eurovideo Bildprogramm GmbHGrey OwlDVD 113 Min.20th Century Fox Home EntertainmentGrey OwlDVD 113 Min.Concorde Home EntertainmentHuman Trafficking - MenschenhandelDVD 180 Min.Eurovideo Bildprogramm GmbHMistakes - Tdliche FehlerDVD 90 Min.Laser ParadiseNatural Enemy - Der Feind in meinem HausDVD 88 Min.Starmedia Home Entertainment***FSK 18 DVD Film***DVD 90 Min.Laser ParadisePidax Western-Klassiker: Grey OwlDVD 113 Min.Alive AGReaper - Der KillerDVD 87 Min.UAP FilmReaper - Der KillerDVD 90 Min.VZ-HandelsgesellschaftRelative Fear - Die Wiege des SchreckensDVD 90 Min.Eurovideo Bildprogramm GmbHSehnschtig - Home EditionDVD 110 Min.Concorde Home EntertainmentVarian's War - Der vergessene HeldDVD 120 Min.KNM Home EntertainmentVarian's War - Ein vergessener HeldDVD 120 Min.M. but not of the form 4k + 3. Therefore, this doesn't give us a contradiction. Therefore, the proof as previously thought is invalid. Therefore, there's a problem with this approach. So how do we fix this? Maybe the construction needs to be different. Let me recall that for primes of the form 4k + 3, a different construction is requirle j < n and count > 0: if s[j] == '(': count += 1 elif s[j] == ')': count -= 1 j += 1 tokens.append(s[i:j]) i = j else: j = i while j < n and s[j] != ' ': j += 1 tokens.append(s[i:j]) i = j return tokens def is_variable(s): if s in {'add', 'let', 'mult'}: return False if not s: return False first_char = s[0] if not first_challingfields Kosowo laos Las Terrenas LOT luang prabang luar parbang Macedonia Mostar pakse phenom pehn Plitwickie Jeziora Podgorica Prosztina Rakija Rijeka saigon Samana Santo Domingo Sarajewo siemrep Skocjan Skopje south east asia Stary Bar Sowenia thailand traveling Trieste Ulcnij vientiene vietnam warsaw warszawa waterfestival Wietnam Wgry WochytEh PriUs=teH faSTah!@@! 08 Audi S5 Black Diamond edition SOLD: 05 SWP Legacy GT Limited 5MT, 13.3ish@105 (stage 2) crappy stock clutch SOLD: 04 MPS/SW WRX Sti. 12.971 @ 103.97 (stock) 278 AWHP/283 AWT, Catback exhaust 306 AWHP/341 AWT, Stage 2 (Scott) Moderator-STi, Drag Racing, Car Purchasing, and Tutorial Forums Reply Reply With Quote + Reply to Thread Page 11 of 25 First ... 23456789101112131415161718192021 ... Last Jump to page: vBulletin Message Cancel Changes Errors The following errors occurred with your submission Okay Quick Reply Register NowUnspecified vulnerability in the Outside In Technology component in Oracle Fusion Middleware 8.5.0, 8.5.1, and 8.5.2 allows remote attackers to affect confidentiality, integrity, and availability via vectors related to Outside In Filters, a different vulnerability than CVE-2016-3574, CVE-2016-3575, CVE-2016-3576, CVE-2016-3577, CVE-2016-3578, CVE-2016-3579, CVE-2016-3580, CVE-2016-3581, CVE-2016-3582, CVE-2016-3583, CVE-2016-3591, CVE-2016-3592, CVE-2016-3593, CVE-2016-3594, CVE-2016-3595, and CVE-2016-3596. - , QUOTE(Miamigirl @ Jun 6 2007, 11:27 AM) Also do you guys think that because my partner is uncirumcised that it could be a reason for the recurrance? Just wondering. Its really sad we have to self-diagnose but $20 to the doctor is more than I a then ptext (sLit "This (rigid, skolem) type variable is") else ptext (sLit "These (rigid, skolem) type variables are")) <+> ptext (sLit "bound by") , nest 2 $ ppr (ctLocOrigin implic_loc) ] ] ; addErrTcM (env1, msg $$ extra1 $$ mkEnvSigMsg (ppr tv1) env_sigs) } -- Nastiest case: attempt to unify an untouchable variable | (implic:_) <- cec_encl ctxt -- Get the innermost context , let implic_loc = ic_loc implic given = ic_given implic = setCtLoc (ic_loc implic) $ do { let msg = misMatchMsg ty1 ty2 extra = quotes (ppr tv1) <+> sep [ ptext (sLit "is untouchable") , ptext (sLit "inside the constraints") <+> pprEvVarTheta given , ptext (sLit "bound at") <+> ppr (ctLocOrigin implic_loc)] ; addErrorReport (addExtraInfo ctxt ty1 ty2) (msg $$ nest 2 extra) } | otherwise = pprTrace "reportTyVarEqErr" (ppr tv1 $$ ppr ty2 $$ ppr (cec_encl ctxt)) $ return () -- I don't think this should happen, and if it does I want to know -- Trac #5130 happened because an actual type error was not -- reported at all! So not reporting is pretty dangerous. -- ? Vieta relates the coefficients of the polynomial to sums and products of roots. If all roots are real, then certain inequalities must hold between the coefficients. For instance, the discriminant of the polynomial must be positive if all roots are real and distinct, but calculating the discriminant of a quintic is super complicated. Maybe there's a simpler condition? Wait, the problem gives a condition on the coefficients: \( 2a_4^2 < 5a_5a_3 \). This seems like a specific inequality that might come from some kind of Cauche symmetry of the segments leads to: \[ 8 \cos^3 \alpha - \frac{60}{11} \cos \alpha = 0 \] 2. **Solve analytically**: Factoring out \(\cos \alpha\) and solving the quadratic part gives \(\cos \alpha \approx 0.8257\), leading to \(\alpha \approx 34.3^\circ\). The angle opposite the side is \(3\alpha \approx 102.9^\circ\). ### Case III: \(d = 4\), \(e = 3\), \(f = 5\) 1. **Formulate the cubic equation**: The derived equation is: \[ 72 \cos^3 \alpha - 48 \cos \alpha - 1 = 0 \] 2. **Solve numerically**: The real root of this equation is approximately \(\cos \alpha \approx 0.827\), leading to \(\alpha \approx 34.05^\circ\). The angle opposite the side is \(3\alpha \approx 102.15^\circ\). ### Final Results After thorough analysis and numerical solutions, the angles opposite the given sides are approximately: I. \(\boxed{96.9^\circ}\) II. \(\boxed{102.9^\circ}\) III. \(\boxed{102.2^\circ}\) These results are derived from solving the cubic equations formed using the given segments and trigonometric relationships. Given an integer \(n \geq 3\) and real numbers \(x_{1}, x_{2}, \cdots, x_{n}\) such that \(\min _{1 \leq i < j \leq n}\left|x_{i} - x_{j}\right| = 1\), find the minimum value of \(\sum_{k=1}^{n} \left|x_{k}\rg: 1. **Sum Preservation**: Each natives statements (left and right neighbors ages) sum to the true sum of their neighbors ages, whether theyre a knight or a liar. This is because liars adjust one neighbors age by +1 and the other by -1, leaving the total unchanged. 2. **System of Equations**: The traveler can set up a system of linear equations where each equation corresponds to a natives statements. However, for an even number of natives (like 50), this system is underdetermined. This means there are infinitely many solutions for the actual ages of the natives, as the equations are not independent enough to uniquely determine all variables. 3. **Non-Unique Solutions**: Different age assignments that satisfy the system can correspond to different knight/liar configurations. For example, a set of statements might be explained by multiple combinations of truthful reports (knights) and adjusted reports (liars), depending on how the ages are assigned. 4. **Dependence on Age Ambiguity**: Since the actual ages are unknown and the system lacks uniqueness, the traveler cannot definitively resolve the knight/liar status of each native. The ambiguity in age solutions directly translates to ambiguity in identifying truth-tellers and liars. **Conclusion**: Due to the even number of natives and the resulting underdetermined system, there exists no guaranteed method for the traveler to uniquely determine each natives type based solely on their statements. Hence, the answer is **no**.Let the moving point $M(x, y)$ have a distance ratio of 2 to point $F(4,0)$ and the line $x=3$. Find the equation of the trajectory of $M(x, y)$: (A) $\frac{x^{2}}{12}-\frac{y^{2}}{4}=1$ (B) $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$ (C) $3 x^{2}-y^{2}-16 x+20=0$ (D) $3 y^{2}-x^{2}-16 y+20=0$ Okay, let's see. The problem is about finding the equation of the trajectory of a moving point M(x, y) such that the ratio of its distance to a fixed point F(4, 0) and a fixed line x=3 is 2. The options are given, so I need to derive the equation step by step. First, I remember that when a point moves such that the ratio of its distance to a fixed point (focus) isHighestBidder pk s = case highestBid s of Nothing -> False Just (pk', _) -> pk == pk' isHighestBidderM :: MonadWallet m => DataValue -> m Bool isHighestBidderM (DataValue ds) = case fromData ds of Just s -> do pk <- ownPubKey return $ isHighestBidder pk s Nothing -> throwError $ OtherError $ T.pack "invalid state" ---- Function `isHighestBidderM` allows a wallet to find out whether it is currently the highest bidder. Using that and `trackAuction`, we are able to automate auctions for bidders as well: [source,haskell] ---- autoBid :: forall m. MonadWallet m => EnglishAuction -> Ada -- <1> -> m () autoBid ea ada = do let m = eaMinBid ea return () when (ada >= m) $ do -- <2> logMsg $ T.pack $ "bidding automatically in " ++ show ea ++ " with highest bid " ++ show ada watchAuction ea trackAuction ea Nothing $ \sl highest ds mAda -> do -- <3> winning < Premium atlasHas protein data IH Tissue reliabilityIH Cell reliabilityIF reliabilityWB validationPA validation Subcellular location (IF)Cell expression (IH)Tissue expression (IH)Tissue specificity (RNA)Tissue detectable (RNA) Evidence summaryUniProt evidenceHPA evidenceMS evidence With antibodiesWith annotated expression (IH)With annotated expression (IF) Sort by Term Gene name ClassEnzymesCD markersBlood group antigen proteinsNuclear receptorsTransportersRibosomal proteinsG-protein coupled receptorsVoltage-gated ion channelsPredicted membrane proteinsPredicted secreted proteinsPlasma proteinsTranscription factorsMitochondrial proteinsRNA polymerase related proteinsRAS pathway related proteinsCitric acid cycle related proteinsCytoskeleton related proteinsCancer-related genesCandidate cardiovascular disease genesDisease related genesFDA approved drug targetsGreat interview Marie. I enjoyed @MamaGena thoroughly. I started a business in 2008 and it did become successful but issues started with a contractor that ended up in a lawsuit that dragged on over a year. Although the company still exists, Ive lost my passion I had when I started it and have never gotten it back. I struggle daily with should I just close shop or whats next. I helped over 225 businesses and received a lot of joy from that however, I cant shake the I dont care anymore. Your convo helped in directing me to think about what brings me joy and what my desires are, saying them out loud everyday. Thanks!In 1972 I fought two times in the Soviet Union losing both by decision then fought in every tournament I could trying to qualify for the 1972 Olympic Trials. I started out at 106 pounds and 5:10 tall lost to Davey Armstrong at the AAU Nationals in Las Vegas. So I moved up to 119 pounds and lost the Eastern Trials to Michael Johnson. So at this point it was over no way making it to the Olympics except one way, pray. So I continued to train hard and pray believing that all things are possible. I was weighing about 130 pounds two weeks before the Olympic Trials were to begin when I got a call from the US Air Force boxing coach who my brother Steve, Nick Wells and Jesse Valdez were boxing at the same time. He told me if I could make 112 I could fight in the Olympic Trials as an unattached boxer because they had an opening spot in the Flyweight division. I told him I can make 112 pounds. Ie pre-trained models expecting 3 channels. That makes sense. Data augmentation: random rotations (maybe up to 20 degrees), translations (shifting by a small fraction of width/height), and flips. Wait, but flipping digits might not be a good idea. For example, flipping a '6' could look like a '9', whic as our starting DE's?...saying at the time how high the FO was on Mitchell. Maybe that's just because Derreck Robinson is about the only other DE on the roster....you remember Derreck Robinson, right? They have several guys just listed as "DL"...like Robaire Smith and Schafering and some other less stellar talents, but DE is really thin.Robinson is really more of a DT in this 4-3. He's almost 3 bills (295).The only guys on the roster that I can think of that can play DE are Benard, Sheard, and... well, Roth and Gocong are FA's... uh...Thank you for the advice, but I know he won't want to go to professionals, he absolutely hates the idea. He woke up a few times, scared and crying, and looked around to make sure no one had left him. He was absolutely terrified. And I can tell he's still absolutely terrified, he's scared to let go of anyone around him. I don't give him his way, and he wasn't even the one the fight was with. But, with the one I had the fight, most of the time we actually manage to come to a calm decision. It's really hard, when we're in a group relationship, but it's really hard on him when he thinks there'll be someone leaving the relationship.Lastly, Born in California, grew up on Catalina Island and in Los Angeles. Her mother is a lesbian. She has seen her father once, when she was 6 months old. When Helena was 7, her mother decided to move to Spain, but the girls grandmother kidnapped her. This is not a normal life, and this kind of helter-skelter childhood, bankrupt of any kind of an anchoring, traditional upbringing, is boundin, Medio Ambiente, Memoria Histrica, Movimientos Sociales, Nuevas Tecnologas, Ocio, Participacin, Poltica, Transparencia, Urbanismo, tagged #EstpasandoenLaMuela, 2011, 2017, 37.550, 5, Acondicionamiento, Actuacin, Adrin, Alcalde, Aragn, Aragonesismo, Aragonesista, Aves, AVIAPARK, Ayudas, Ayuntamiento, ridos, Baldexaln, Bartolom, Cambio, Carmelo, Celebracin, CHA, Chicas, Chicos, Chunta, Coherencia, Contratista, De, Desmantelamiento, Destinado, Diciembre, Dignidad, Diputacin, DPZ, Eficaz, Ejecucin, Ejercicio, Entidades, Equipo, Espacio, Eventos, Excavaciones, Familiar, Financieramente, Gestin, Gimeno, Gobierno, Heraldo, Heraldo de Aragn, Honradez, Inclemencias Meteorolgicas, Invernadero, Inversiones, Izquierda, Jungla, Jupe, La, La Jupe, La Muela, Lamuelans.OK, MessageBoxIcon.Error); } else { MessageBox.Show("", "", MessageBoxButtons.OK, MessageBoxIcon.Information); } } ``` CodeDomC#CodeDomGenerateInMemorytrue .NETAppDomainAppDomainAppDomain AppDomainAppDomainAppDomainCodeDomAppDomainAppDomain AppDomainMarshalByRefObject RoslynRoslynAPICodeDomRoslynAppDomainRoslyn RoslynCompilation.CreateRoslyn CodeDomAppDomainAppDomainRoslyn RoslynCodeDomAppDomainAppDomainAppDomainAppDomain 1. CodeDomAppDomainAppDomain 2. Roslyn RoslynRoslynGetDiagnostics RoslynCodeDomAppDomain .NET AppDomain --- ### **CodeDom AppDomain** ```csharp // 1. public class CompilerProxy : MarshalByRefObject { public CompilerResults Compile(string[] assemblyNames, string[] code) { CodeDomProvider provider = new CSharpCodeProvider(); CompilerParameters parameters = new CompilerParameters(assemblyNames) { GenerateInMemory = true, CompilerOptions = "/target:library /optimize" }; return provider.CompileAssemblyFromSource(parameters, code); } } // 2. private void Item_Click(object sender, EventArgs e) { // AppDomain tempDomain = AppDomain.CreateDomain("TempCompilationDomain"); try { CompilerProxy proxy = (CompilerProxy)tempDomain.CreateInstanceAndUnwrap( Assembly.GetExecutingAssembly().FullName, typeof(CompilerProxy).FullName ); // CompilerResults results = proxy.Cts denote x = a/b or b/a. Maybe setting a ratio. Lets assume a 0 and b0. Lets define k = a/b. Then a =k*b. Substitute into the equations: Original number:10a +b=10kb +b= b(10k +1)=j(a +b)=j(kb +b)=j*b(k +1) Thus: b(10k +1)=j*b(k +1) Cancel b (b0): 10k +1=j*b(k +1) Similarly, reversed number:10b +a=10b +kb= b(10 +k)=m(a +b)=m*b(k +1) Thus: b(10 +k)=m*b(k +1) Cancel b: 10 +k =m*b(k +1) Now, from the first equation:10k +1 =j*b(k +1). Let's solve for b: b=(10k +1)/(j(k +1)) Plug this into the second equation: 10 +k =m*( (10k +1)/(j(k +1)) )*(k +1) Simplify: 10 +k =m*(10k +1)/j Therefore: m= j*(10 +k)/(10k +1) So m= j*(10 +k)/(10k +1) But k =a/b, which is a rational number. However, since a and b are digits, k can be a fraction where a and b are integers from1 to9. But we need to find m in terms of j, which requires expressing k in terms of j. However, this seems challenging. But notice that the options are in terms of j, such as j -1, j +1, etc. So maybe m=10 -j (option E). Let's test with k=1 (a=b). If k=1, then m= j*(10 +1)/(10*1 +1)=j*11/11=j. So m=j. Which matches our palindrome case. If k approaches d, so that's all. Make sure the JSON keys match exactly: propertyDetails, applicantIncome, preApprovalA(4)=7=3*2 +1 f(5)=8=3*2 +2 f(7)=12=3*4 +0 f(8)=15=3*5 +0 So it's inconsistent. Maybe there's a different pattern. Alternatively, notice that for numbers not divisible by 3, f(n) = next number in the sequence after grouping numbers into blocks. For example, the first block is 1-2, mapped to 2-3. Then the next block is 4-5, mapped to 7-8. Then next block is 7-8, mapped to 12-15. Wait, but 7-8 is a block of two numbers, mapped to 12 and 15. Hmm, not sure. Alternatively, consider that the numbers not divisible by 3 are arranged in pairs, and each pair is mapped to a pair that is later in the sequence. For example, (1,2)(2,3), (4,5)(7,8), (7,8)(12,15), etc. But the pairs aren't consistent in their spacing. Alternatively, think of the function f(n) in terms of the following: when n is not divisible by 3, f(n) is the next number in the list of numbers not divisible by 3 multiplied by 2. Wait, let's test: Numbers not divisible b it into an application if it -- doesn't look like one already -- See Note [Canonicalising type applications] can_eq_flat_app ev swapped s1 t1 ps_ty1 ty2 ps_ty2 | Just (s2,t2) <- tcSplitAppTy_maybe ty2 = unSwap swapped decompose_it (s1,t1) (s2,t2) | otherwise = unSwap swapped (canEqFailure ev) ps_ty1 ps_ty2 where decompose_it (s1,t1) (s2,t2) = do { let xevcomp [x,y] = EvCoercion (mkTcAppCo (evTermCoercion x) (evTermCoercion y)) xevcomp _ = error "canEqAppTy: can't happen" -- Can't happen xevdecomp x = let xco = evost of regex patterns. For example, in CS_MODE_64, a pattern could be [r'^push\s+rbp$', r'^mov\s+rbp,\s*rsp$', r'^sub\s+rsp,\s*0x[0-9a-f]+$']. But how to handle case sensitivity? Capstone's output is in lowercase for Intel syntax, I think. So the regex should match that. But the problem is that the code in the prologues may need to use regex patterns that can match different variations. For example, some instructions might have different registers (like push rbx in a different prologue), but that's unlikely. Prologues typically follow a standard pattern. But according to the problem statement, the prologue patterns are predefined. So the code , 6044, 208, 5, 30, "Print"], Cell[191376, 6051, 254, 6, 30, "Print"], Cell[191633, 6059, 254, 6, 30, "Print"], Cell[191890, 6067, 40405, 1787, 257, 36484, 1720, "GraphicsData", \ "PostScript", "Graphics"] }, Open ]], Cell[CellGroupData[{ Cell[232332, 7859, 94, 2, 36, "Input"], Cell[232429, 7863, 34, 0, 30, "Print"], Cell[232466, 7865, 238, 5, 30, "Print"], Cell[232707, 7872, 274, 6, 30, "Print"], Cell[232984, 7880, 274, 6, 30, "Print"], Cell[233261, 7888, 208, 5, 30, "Print"], Cell[233472, 7895, 16755, 454, 573, "Output"] }, Open ]], Cell[CellGroupData[{ Cell[250264, 8354, 63, 1, 36, "Input"], Cell[250330, 8357, 28, 0, 36, "Output"] }, Open ]], Cell[CellGroupData[{ Cell[250395, 8362, 103, 2, 36, "Input"], Cell[250501, 8366, 105, 2, 36, "Output"] }, Open ]], Cell[CellGroupData[{ Cell[250643, 8373, 151, 3, 60, "Input"], Cell[250797, 83ith their ages, and we need to find out how old Chen Xiang was when he saved his mother. Hmm, first, I need to parse the information carefully. Alright, let's start by translating the problem into mathematical equations. The first sentence says, "This year, the Second Son of the Gods age is 4 times the age of Chen Xiang." Let's denote the current age of Chen Xiang as C and the current age of the Second Son as S. So right now, S = 4C. That's straightforward. Next part: "Eight years later, the Second Sons age is 8 years less than three times Chen Xiang's age." So, in eight years, both of them will be eight years older. Therefore, Chen Xiang's age will be C + 8, and the Second Son's age will be S + 8. According to the problem, at that time, the Second Son's age is 8 years less than three times Chen.1831 per month. That should be the decay constant. Wait, let me verify. If we plug back into the equation, does it hold? Let's check: P(6) = 0.6 * e^(-0.1831*6) First compute the exponent: 0.1831 * 6 1.0986, so e^(-1.0986) 1/e^(1.0986). Since e^1.0986 is approximately 3, so 1/3. Therefore, 0.6 * 1/3 = 0.2, which matches the given data. Okay, that seems correct. So is ln(3)/6. Alternatively, since ln(3)/6 is equivalent to (ln(3))/6. So that's the exact value, but if needed as a decimal, approximately 0.1831. Moving on to the second question: Alex is part of a group of 100 individuals. Using the found, calculate the expected number who maintain their resolution for at least 6 months. Well, from the first part, we already determined that the probability for an individual to maintain the resolution for at least 6 months is 20%, which is 0.2. So if each individual has a 0.2 probability, and there are 100 individuals, the expected number is just 100 * 0.2 = 20. So the expected number is 20. But wait, let me make sure. The question says "using the value of determined in the previous sub-problem". So perhaps we need to recalculate P(6) using our to confirm. Let's do that. Given p = 0.6, = ln(3)/6. So P(6) = 0.6 * e^(- (ln(3)/6)*6 ) = 0.6 * e^(-lif we consider the inner triangle's edges, those are three more lines, each with two points. So that's six lines of two each. But what about lines of three? If we take the medians of the triangle (from a vertex to the midpoint of the opposite side), each median would have two points: the vertex and the midpoint. Wait, no, the median would connect the vertex to the midpoint, so that's two points per median. Hmm, so maybe that's not giving us the rows of three. Alternatively, maybe if we add the centroid. Wait, but we only have six kittens. So in the previous case, with three vertices and three midpoints, that's six points. The lines would be the original edges (each with two points), the inner triangle edges (each with two points), and the medians (each connecting a vertex to a midpoint, which are two points each). So that's 3 (original edges) + 3 (inner edges) + 3 (medians) = 9 lines, each with two points. But the problem requires three rows of three points each. So maybe that configuration isn't working. Wait, perhaps I need to have three lines that each contain three kittens. Let's think of the famous example of the "Nine Points" problem, but here we only have six points. How can three lines each have three points? Maybe overlapping lines? Wait, if three lines are concurrent (all passing through a common point), then that common point is part of all three lines. So, let's say one kitten is at the intersection of three lines. Then each of those three lines can have two more kittens each. But since we have six kittens in total, if one kitten is shared by three lines, each of those lines needs two more kittens, which would be 3 lines * 2 kittens = 6 kittens, plus the shared one, totaling 7 kittens. But we only have six. So that's not possible. Alternatively, maybe two lines intersect at a common point. Let's say two lines share a common kitten, and each of those lines has two other kittens. Then a third line could pass Kevin, tengo hipoglicemia reactiva o hiperinsulinemia hace varios aos y la he podido controlar. Mis recomendaciones son: no debes hacer ejercicio fuerte sino moderado. Durante el ejercicio debes hidratarte y nutrirte, por ejemplo tomar agua y comer mandarina, la cual te da azcar y fibra. No tomes zumos de nada. Debes comer la fruta entera, masticarla, no procesarla. No debes comer nada procesado. Imagina que eres un hombre primitivo en una selva: no creo que te comieras 4 naranjas, entonces no debes hacer zumo de 4 naranjas; lo que debes hacer es comerte la naranja en tajos, masticada y tomar agua. Igual con los panes y galletas: no comer pan blanco ni procesado, preferir el pan integral, de centeno, multicereales y SIN azucar. Debes comer mucha protena, por lo menos al desayuno, almuerzo y comida. Un ejemplo de dieta saludables sera: desayuno con 200 gr de protena (huevo, queso sin grasa, pechuga de pollo, etc), una fruta y 3/4 de pocillo de leche o yogurt descremado y deslactosados. Evita la grasa al mximo pues nues error. There could also be other sites involved, and the problem may not even be on your site but elsewhere in the network. Try to pin down the communication endpoints (hubs or sites) involved in the problem and contact the administrator of that site or tk=1 to k=1010 (intervals (2k, (2k+1))): - At x=2k, f(x)= -2k/2021 (negative). - At x=(2k+1), f(x)= -(2k+1)/2021 (more negative). - However, in between, f(x) reaches a maximum of 1 - (2k +0.5)/2021. - For k1010, this maximum is positive (since 2k +0.5 2*1010 +0.5=2020.5 <2021). - Therefore, f(x) goes from negative to positive to negative, resulting in two roots per interval. - For k=1011 (interval (2022,2023)): - But wait, 2021 is the upper limit. Wait, our upper bound is 2021, so the las is that the total number of stone mapM_ check_arg tys else -- In the liberal case (only for closed syns), expand then check case tcView ty of Just ty' -> check_type ctxt rank ty' Nothing -> pprPanic "check_tau_type" (ppr ty) } | GhciCtxt <- ctxt -- Accept under-saturated type synonyms in -- GHCi :kind commands; see Trac #7586 = mapM_ check_arg tys | otherwise = failWithTc (arityErr "Type synonym" (tyConNam Offset[0.28], { Offset[0.56]}, Offset[0.28]}, "ColumnsIndexed" -> {}, "Rows" -> { Offset[0.2], { Offset[0.56]}, Offset[0.2]}, "RowsIndexed" -> {}}}]}, Open]], Cell[ CellGroupData[{ Cell[ StyleData["TableText"], Deletable -> False, StyleMenuListing -> None, FontFamily -> "Verdana", FontSize -> 0.952 Inherited], Cell[ StyleData["TableText", "Printout"], n 0s and 1s to each cycle such that the total number of 0s is n +1 and 1s is n. Since each cycle must have the same label, we can think of each cycle as a single position. Let m be the number of cycles, which is gcd(k, 2n +1). Because the number of cycles is equal to the number of elements in each orbit, which is the order of the rotation, so if the rotation has order t, then m = (2n +1)/t. Wait, no, more precisely, the number of cycles is gcd(k, 2n +1). Wait, here's a standard result: the number of cycles in the permutation induced by rotation by k positions on a polygon with m vertices is gcd(k, m). So here, m = 2n +1, so the number of cycles is gcd(k, 2n +1). Therefore, each cycle has length (2n +1)/gcd(k, 2n +1). Thus, for a rotation by k positions, the number of fixed labelings is the number of ways to assign 0 or 1 to each cycle, such that the total number of 0s is n +1 and the total number of 1s is n. Each cycle has length (2n +1)/d, where d = gcd(k, 2n +1). Therefore, each label assigned to a cycle is repeated (2n +1)/d times. So, in order for the total number of 0s to be n +1, the number of cycles labeled 0 multiplied by the length of each cycle must equal n +1. Similarly for 1s. Wait, but each cycle has length l = (2n +1)/d. Therefore, if we assign a 0 to a cycle, it contributes l zeros, and similarly for 1. Lets denote d = gcd(k, 2n +1). Then the number of cycles is d. Each cycle has length l = (2n +1)/d. Therefore, the total number of 0s must be equal to l * x, wallel to AC. But I need to verify this. Let's recall that a homothety is a transformation that scales figures about a fixed point. If two sides are parallel, then there might be a homothety center at the orthocenter? Maybe. Alternatively, since A1B1 || AB and B1C1 || BC, perhaps the orthic triangle is homothetic to ABC. The homothety center would be the orthocenter, but I need to check. In an acute triangle, the orthocenter H is inside the triangle. The orthic triangle A1B1C1 is related to ABC via similarity transformations. If two sides are parallel, then the homothety that maps AB to ileTuningReifenSicher unterwegsFhrerscheinReparaturen & WartungKfz-VersicherungVerkehrsrechtSchuleDas deutsche SchulsystemSchulformenSchulrechtWohlfhlen in der SchuleRichtig lernenSchulalltagSprachen lernenDeutschFremdsprachenSprachwissen & RhetorikFachgebiete im berblickNaturwissenschaftenMathematikSchulfcher A-ZWissen im AlltagSport & WellnessDie Seele baumeln lassenEntspannenYogaAuspowernFuballJoggenWeitere SportartenAuf die Pltze, fertig, los!bungen fr zuhauseSport fr EinsteigerUrlaubUnterwegsAuto, Zug, FlugzeugUrlaubsvergngenBadeurlaubCampingUrlaub mit KindSicher ReisenReiserechtReisetippsZuhauseDas neue HeimEigenheimEnergieUmzugWohnen und LebenEinrichten & DekorierenMieteRechte & PflichtenSuchefinden StartseiteEssen & TrinkenRezepte fr Getrnkealkoholische GetrnkeBierDie Kohlenhydrate im Bier - Wissenswertes zum Nhrwert von alkoholfreiem BierAutor: Maria PonkhoffAlkoholfreies Bier ist ein erfrischendes und zum Teil sogar gesundes Getrnk mit Kohlenhydraten und vielen Mineralstoffen und Vitaminen. Doch was steckt sonst noch in dem Getrnk?Alkoholfreies Weizenbier enthlt viele Nhrstoffe. Kohlenhydrate in alkoholfreiem Weizen , , . , , y Fotos Big Ass Reifen reifen groen schwanz mnner sex auf einer geburtstagsfeier mdchen mdchen sex les mobile kostenlose pornografie Abifahrt Videochat Sex Wackersberg Total Heie Milf Cuckold Imprgnierung Untertitel was ist die perfekte pennisgre groer dicker saftiger schwanzZigmeister wrote:Wait, so everyone is of the opinion OP should drop $1800-2000 for the Rims, and then jam another $700 in hubs, then spokes/nipples, plus the $200 custom build labor get himself a $3000 set of wheels....when you can get a set of Zipp FC tubulars beyond black with 188 hubs/cxrays for $1700-1800 new?Similar issue here in Australia. Zipp 404 FC's about $2,400 from any number of shops. Enve 3.4's about $3,500. As much as I want the Enve's that extra $1,100 gets me a Power2max power meter thrown in as well.Third Base: Heres another controversy coming your way. If you are a Mets fan old enough to remember the 86 World Series winners, you loved that team. Top to bottom. Rough and tumble, bold and brash from manager Davey Johnson saying in spring training that he didnt just want to win, he wanted to dominate, to Gary Carters rah rah fist pump to the crowd after every homer, to Ray Knight cold-cocking Eric Davis when the Reds outfielder slid into third spikes high they were aggressive in every way.En las negociaciones de estos das pasados hemos pedido el apoyo para la lista de izquierda ms votada para avanzar en polticas sociales mantenindonos abiertos a propuestas de las distintas fuerzas polticas. Hemos tenido reuniones fructferas en las que hemos habla0Kr9qT5NMXQe0aHlo3KvTwsU Qr7pUEfwCP8zA/MZ9O8o9Z+g2w84EfSTejRuzUOP+4toPdejW3dK9fbpNJ/6 0eFAf5uVSjqv9aNIru+fxyNoP9uPQrqyPXIm/psfOMW2lS//SALn4eENKSkF x2vFwAsPjzIl1+YdpXnCwwvsmoX8CTRPedjGJP1LlCKi8xF+Yj1rSKaFiM5P WLBMOtf1RyGdr3DBMmnXwt8FdP7CB79fbtsqY+l8hlfvXuJ2bSJL5ze8M9vh ucKXofMdrmyquecw8L5h/sOfrArqz3vEo/Mh/tt2d96Fzn7Qn8EBgkXvuTA9 +JnBTXd+vBplqgM 3.770404530677828*^9, 3.7704045624192743`*^9}, {3.7704046442968364`*^9, 3.770404645907844*^9}, {3.770635770882913*^9, 3.7706357717458925`*^9}, { 3.770734796342696*^9, 3.7707348341817884`*^9}, 3.7707350964145584`*^9, 3.7707366525031767`*^9, {3.7707366969657836`*^9, 3.7707367027702117`*^9}, { 3.7707378791469173`*^9, 3.7707378835062323`*^9}, {3.77427872611b)^2 tan ) /a ) = (a/72)(7a + 3b) * sqrt(4a + 9 (a - b)^2 tan ) /a ) Simplify: = (7a + 3b)/72 * sqrt(4a + 9 (a - b)^2 tan ) Which brings us back to the previous expression. Given the problem likely expects a simplified answer in terms of a, b, and , and since the expression is already in terms of these variables, this might be the answer. However, it's quite complicated, so I suspect there might be a s within the square. So perhaps there's a typo. Alternatively, maybe the diagonal is AC, as I considered earlier, in which case F is at (s/3, s/3). Then quadrilateral BFED is B (s,0), F (s/3, s/3), E (s/2, 0), D (0,s). To find its area. Alternatively, maybe the problem meant DE meets diagonal AC at F. Then the area of BFED is 36. Then we can compute the area of the square. Alternatively, maybe the problem is correct as stated, but I need to adjust my coordinate system. Wait, let me try again. If BC is a diagonal, but in a square, BC is a side, not a diagonal, unless the square is being considered as part of another figure. Hmm. Wait, maybe the problem is in 3D? No, the problem says square ABCD, which is 2D. Hmm. Alternatively, maybe BC is a diagonal of a different polygon, but the problem states it's a square ABCD. So BC is a side. Wait, perhaps the problem was supposed to say "DE meets the diagonal BD at F," which would make sense. Then quadrilateral BFED would be a four-sided figure with vertices B, F, E, D. Let me check that. If BD is the diagonal from B to D, and DE is from D to E, then they intersect at D. So that can't be. Wait, DE is from D to E, BD is from B to D, so they intersect at D. So that's not helpful. Alternatively, maybe the problem meant diagonal AC. So DE meets diagonal AC at F. Then quadrilateral BFED is B, F, E, D. Let's compute that area. So, given the coordinates: A(0,0), B(s,0), C(s,s), D(0,s), E(s/2, 0). DE is from D(0,s) to E(s/2,0), equation y = -2x + s. Diagonal AC is from A(0,0) to C(s,s), equation y = x. Intersection F is at (s/3, s/3) as found earlier. So quadrilateral BFED has vertices at B(s,0), F(s/3, s/3), E(s/2, 0), D(0,s). To find its area. One way to find the area is to use the shoelace formula. Let's list the coordinates: B: (s, 0) F: (s/3, s/3) E: (s/2, 0) D: (0, s) Back to B: (s, 0) Compute the area using shoelace: Area = 1/2 |sum over edges (x_i y_{i+1} - x_{i+1} y_i)| So: First, list the coordinates in order: 1. (s, 0) 2. (s/3, s/3) 3. (s/2, 0) 4. (0, s) 5. (s, 0) Compute the sum: Term 1: x1 y2 - x2 y1 = s*(s/3) - (s/3)*0 = s^2/3 Term 2: x2 y3 - x3 y2 = (s/3)*0 - (s/2)*(s/3) = 0 - s^2/6 = -s^2/6 Term 3: x3 y4 - x4 y3 = (s/2)*nts and equipment fully immediately after taking possession of the boat. In the unlikely event of any alleged deficiencies or shortcomings the Hirer must notify the Company before the boat leaves. The Hirer shall sign the Boat Acceptance form and Inventory before departure and thereafter the Hirer is completely responsible for the boat, its equipment and its operation until it is handed back to the Company at the end of the period of hire. Any shortcomings subsequently discovered shall immediately be notified to the Company by telephone, in order to give the Company th, RowBox[{ RowBox[{"temp", "\[GreaterEqual]", RowBox[{ RowBox[{"(", RowBox[{"3", "/", "2"}], ")"}], "*", "Pi"}]}], ",", RowBox[{"sum", "+=", RowBox[{ RowBox[{"2", "*", "Pi"}], "-", "temp"}]}]}], "]"}], ";", "\[IndentingNewLine]", RowBox[{"If", "[", RowBox[{ RowBox[{ RowBox[{"temp", "<", RowBox[{ RowBox[{"(", RowBox[{"3", "/", "2"}], ")"}], "*", "Pi"}]}], "&&", RowBox[{"temp", "\[GreaterEqual]", "Pi"}]}], ",", RowBox[{"sum", "+=", RowBox[{"temp", "-", "Pi"}]}]}], "]"}], ";", "\[IndentingNewLine]", RowBox[{"If", "[", RowBox[{ RowBox[{ RowBox[{"temp", "\[GreaterEqual]", RowBox[{ RowBox[{"(", RowBox[{"1", "/", "2"}], ")"}], "*", "Pi"}]}], "&&", RowBox[{"temp", "<", "Pi"}]}], ",", RowBox[{"sum", "+=", RowBox[{"Pi", "-", "temp"}]}]}], "]"}], ";", "\[IndentingNewLine]", RowBox[{"If", "[", RowBox[{ RowBox[{"temp", "<", RowBox[{ RowBox[{"(", RowBox[{"1", "/", "2"}], ")"}], "*", "Pi"}]}], ",", RowBox[{"sum", "+=", "temp"}]}], "]"}]}]}], "]"}], ";"}]}], "]"}], ";", "\[IndentingNewLine]", RowBox[{ RowBox[{"sum", "/", RowBox[{"(", RowBox[{ RowBox[{"size", "[", RowBox[{"[", "1", "]"}], "]"}], "*", RowBox[{"(", RowBox[{ RowBox[{"size", "[", RowBox[{"[", "1", "]"}], "]"}], "-", "1"}], ")"}]}], ")"}]}], "*", "2"}]}]}], "]"}]}], "\[IndentingNewLine]"}], "Input", CellChangeTimes->{ 3.753593162461779*^9, {3.7536524613356667`*^9, 3.753652469105147*^9}, { 3.753653121184474*^9, 3.7536532786258464`*^9}, {3.7536534052929792`*^9, 3.7536537019504533`*^9}, {3.7536537371810565`*^9, 3.753653863318862*^9}, { 3.7536590392550697`*^9, 3.753659= k - i; if (freq[i] > 0 && freq[j] > 0) { count += freq[i] * freq[j]; } } if (k % 2 == 0) { NSInteger mid = k / 2; NSInteger f = freq[mid]; count += (f * (f - 1)) / 2; } free(freq); return count; } ``` ### Explanation 1. **Frequency Array Initialization**: We initialize an array `freq` of size \( k \) to store the frequency of each remainder when elements are divided by \( k \). 2. **Remainder Calculation**: For each element in the input array, we compute its remainder modulo \( k \), adjusting for negative values to ensure non-negative remainders. 3. **Pair Counting Loop**: We iterate from 1 to \( k/2 \), checking pairs of remainders \( i \) and \( k-i \). If both have non-zero frequencies, their product is added to the count. 4. **Even \( k \ blow them up. It was very effective. On bosses that were difficult to melee, Id just dodge the entire fight and try to keep it distracted away from him. On bosses that were resistant to magic, Id beat the shit out of them while he mostlqrt(b_k) x_k)^2 But we have Sum b_k x_k^2 <=4 Sum x_k^2, so: (Sum b_k)(Sum x_k^2) >= (Sum sqrt(b_k) x_k)^2 But also: (Sum sqrt(b_k) x_k)^2 <= (Sum b_k x_k^2)(Sum x_k^2) by Cauchy-Schwarz Which gives: (Sum b_k)(Sum x_k^2) >= (Sum sqrt(b_k) x_k)^2 <= (Sum b_k x_k^2)(Sum x_k^2) <=4 (Sum x_k^2)^2 Therefore, Sum b_k >= (Sum sqrt(b_k) x_k)^2 / (Sum x_k^2) <=4 Sum x_k^2 But this doesn't provide a lower bound. Given that all my attempts are not leading to the desired inequality, I must consider that there's a standard inequality or trick that I'm missing. One possible approach is to use the discrete version of the inequality from calculus of variations, where the minimum of the Rayleigh quotient is achieved by the eigenvector corresponding to the smallest eigenvalue. In our case, the Rayleigh quotient would be: R(x) = [Sum (x_k -x_{k+1})^2 +x_1^2 +x_n^2] / Sum x_k^2 We need to show that R(x) >=4/(n+1) for all non-zero x with x_0=x_{n+1}=0. To find the minimal value of R(x), we can consider the eigenvector corresponding to the smallest eigenvalue of the matrix associated with the quadratic form in the numerator. Lets consider the matrix M such that the quadratic form is x^T M x = Sum (x_k -x_{k+1})^2 +x_1^2 +x_n^2. This matrix M is a tridiagonal matrix with: - Diagonal entries: For k=1, M_{1,1}=2 (from (x_1 -x_2)^2 +x_1^2 =x_1^2 -2x_1x_2 +x_2^2 +x_1^2=2x_1^2 -2x_1x_2 +x_2^2, but wait, need to compute correctly. Actually, expanding Sum_{k=1}^{n-1}(x_k -x_{k+1})^2 +x_1^2 +x_n^2: = Sum_{k=1}^{n-1}(x_k^2 -2x_kx_{k+1} +x_{k+1}^2) +x_1^2 +x_n^2 = x_1^2 -2x_1x_2 +x_2^2 +x_2^2 -2x_2x_3 +x_3^2 +...+x_{n-1}^2 -2x_{n-1}x_n +x_n^2 +x_1^2 +x_n^2 = 2x_1^2 +2x_n^2 +2 Sum_{k=2}^{n-1}x_k^2 -2 Sum_{k=1}^{n-1}x_kx_{k+1} This corresponds to a matrix M with: - M_{1,1}=2, M_{n,n}=2 - M_{k,k}=2 for k=2,...,n-1 - M_{k,k+1}=M_{k+1,k}=-1 for k=1,...,n-1 This is a tridiagonal matrix with 2 on the diagonal and -1 on the off-diagonal. The eigenvalues of this matrix are known to be lambda_j = 4 sin^2(j pi/(2(n+1))) for j=1,...,n. The smallest eigenvalue is lambda_1=4 sin^2(pi/(2(n+1))), and the largest is lambda_n=4 sin^2(n pi/(2(n+1)))=4 cos^2(pi/(2(n+1))). However, our matrix M is slightly different. Wait, thks=[early_stop] ) # Evaluate on test set test_loss, test_acc = model.evaluate(val_ds) print(f'Test Accuracy: {test_acc:.4f}') ``` ### Key Considerations: - **Transfer Learning**: Leverages pre-trained features from ImageNet, reducing the need for extensive training. - **Regularization**: The custom loss function explicitly adds L2 penalties, reducing model complexity. - **Data Augmentation**: Increases dataset diversity, improving generalization. - **Efficiency**: The pipeline and model design ensure training completes within 2 hours on 8GB memory. ### Expected Outcome: - **Test Accuracy 98%**: Achieved through fine-tuning, regularization, and augmentation. - **Robustness**: The model generalizes well to unseen variations in handwritten digits.Write a function in C that not only finds the maximum element in an array of integers but also returns its position(s) if it appears more than once. You should make sure that your solution is efficient and performs well for large arrays. Also, consider the possibility of the array having duplicate elements. Your output should be a list, where the first element is the maximum number and the following elements are the positions in the array where the maximum element was found (starting from index 0). Okay, I need to write a C function that finds the maximum element in an array and also returns its positions, even if it appears multiple times. Let's think about how to approach this. First, the function needs to process the array. The input is an array of integers and its size. The output should be a list where the first element is the maximum, followed by the indices where it occurs. But how do I return thi/4) = -2/2, cos(5/4) = -2/2. Again, same result. So it seems <- simplifyInfer closed mono name_taus wanted ; theta <- zonkTcThetaType (map evVarPred givens) ; exports <- checkNoErrs $ mapM (mkExport prag_fn qtvs theta) mono_infos ; loc <- getSrcSpanM ; let poly_ids = map abe_poly exports final_closed | closed && not mr_bites = TopLevel | otherwise = NotTopLevel abs_bind = L loc $ AbsBinds { abs_tvs = qtvs , abs_ev_vars = givens, abs_ev_binds = ev_binds , abs_exports = exports, abs_binds = binds' } ; traceTc "Binding:" (ppr final_closed $$ ppr (poly_ids `zip` map idType poly_ids)) ; return (unitBag abs_bind, poly_ids, final_closed) } -- poly_ids are guaranteed zonked by mkExport -------------- mkExport :: PragFun -> [TyVar] -> TcThetaType -- Both already zonked -> MonoBindInfo -> TcM (ABExport Id) -- Only called for generalisation plan IferGen, not by CheckGen or NoGen -- -- mkExport generates exports with -- zonked type variables, -- zonked poly_ids -- The former is just because no further unifications will change -- the quantified type variables, so we can fix their final form -- right now. -- The latter is needed because the poly_ids are used to extend the -- type environment; see the invariant on TcEnv.tcExtendIdEnv -- Pre-condition: the qtvs and theta are already zonked mkExport prag_fn qtvs theta (poly_name, mb_sig, mono_id) = do { mono_ty <- zonkTcType (idType mono_id) ; poly_id <- case mb_sig of Just sig -> return (sig_id sig) Nothing -> mkInferredPolyId poly_name qtvs theta mono_ty -- NB: poly_id has a zonked type ; poly_id <- addInlinePrags poly_id prag_sigs ; spec_prags <- tcSpecPrags poly_id prag_sigs -- tcPrags requires a zonked poly_id ; let sel_poly_ty = load_with=engine) # Build the query query = select(trips).where(trips.c.trip_date >= three_months_ago) # Execute the query result = conn.execute(query) recent_trips = result.fetchall() # Close the connection conn.close() return recent_trips But this code has several assumptions: 1. The database is PostgreSQL, but the connection string would vary. 2. The trips table has a column named 'trip_date'. 3. The 'trip_date' is of type Date or a compatible type. 4. The dateutil library is installed. If the user can't use dateutil, the date calculation could be handled differently, perhaps using: three_months_ago = datetime.date(today.year, today.month -3, today.day) but that would fail if today.month is less than 4. So adjust the year and month accordingly. For example: def get_three_months_ago(today): if today.month > 3: return today.replace(month=today.month s naturel. Cela vous aide mieass, and any two girls have different numbers of boyfriends among the boys in the class. Can you determine how many boys and how many girls we have in our class?" Vova responds: "You must be mistaken, such a situation cannot exist!" a) Is Vova correct? b) How would you answer Petya's question? Okay, let's try to tackle this problem. So, Petya claims that in their class of 30 students, every two boys have different numbers of girlfriends, and every two girls have different numbers of boyfriends among the boys. Vova thinks this is impossible. We need to figure out if Vova is right or not, and then answer whether such a situation can exist. First, let's parse the problem. There are 30 students in total, consisting of some number of boys and girls. Let me denote the number of boys as B and the number of girls as G. So, B + G = 30. Now, the key points are: 1. Any two boys have different numbers of girlfriends. That means each boy has a unique number of girlfriends. Since there are B boys, the number of girlfriends each boy has must be a distinct integer. 2. Similarly, any two girls have different numbers of boyfriends. Each girl has a unique number of boyfriends among the boys. With G girls, each girl's number of boyfriends must be a distinct integer. Our goal is to determine if such B and G exist where these conditions hold. If not, then Vova is correct. Let me start by considering the possible ranges for the number of girlfriends each boy can have. Since a boy can't have a negative number of girlfriends or more girlfriends than there are girls, the number of girlfrtimes = [] def wrapper(*args, **kwargs): nonlocal count # check current date current_day = datetime.datetime.now().day if not is_prime(current_day): return # or do nothing # check rate limit now = time.time() # remove entries older than 10 seconds call_times[:] = [t for t in call_times if now - t <= 10] if len(call_times) >=5: return # add current time to call_times call_times.append(now) # increment count count +=1 print(f"Total prime numbers found so far: {count}") return func(*args, **kwargs) return wrapper Wait, but this approach uses nonlocal for count, and call_times is a list in the enclosing scope. But in Python, the nonlocdult Short Sleeve JerseyRBJ2B$29.99 Youth Short Sleeve JerseyYRBJ2B$27.99natures, though copper is harder to detect remotely due to its lower abundance and spectral characteristics. Missions like OSIRIS-REx (to Bennu) and Hayabusa2 (to Ryugu) have identified metal-rich phases, suggesting the presence of copper in certain contexts. 4. **Asteroid Mining Potential**: - While copper isn't the primary target for mining (compared to platinum-group metals), its presence in metallic asteroids makes it a potential resource for future space exploration and in-situ utilization. **Conclusion**: Copper exists in certain asteroids, particularly metallic (M-type) and silicaceous (S-type) varieties, though it is not as abundant as iron or nickel. Its presence -> (Type,Type) -- c :: t1 ~ t2 coVarKind cv = case coVarKind_maybe cv of Just ts -> ts Nothing -> pprPanic "coVarKind" (ppr cv $$ ppr (tyVarKind cv)) coVarKind_maybe :: CoVar -> Maybe (Type,Type) coVarKind_maybe cv = case splitTyConApp_maybe (varType cv) of Just (tc, [ty1, ty2]) | tc `hasKey` eqPredPrimTyConKey -> Just (ty1, ty2) _ -> Nothing -- | Makes a coercion type from two types: the types whose equality -- is proven by the relevant 'Coercion' mkCoType :: Type -> Type -> Type mkCoType ty1 ty2 = PredTy (EqPred ty1 ty2) splitCoPredTy_maybe :: Type -> Maybe (Type, Type, Type) splitCoPredTy_maybe ty | Just (cv,r) <- splitForAllTy_maybe ty , isCoVar cv , Just (s,t) <- coVarKind_maybe cv = Just (s,t,r) | otherwise = Nothing isReflCo :: Coercion -> Bool isReflCo (Refl {}) = True isReflCo _ = False isReflCo_maybe :: Coercion -> Maybe Type isReflCo_maybe (Refl ty) = Just ty isReflCo_maybe _ = Nothing \end{code} %************************************************************************ %* * Building coercions %* * %************************************************************************ \begin{code} mkCoVarCo :: CoVar -> Coercion mkCoVarCo cv | ty1 `eqType` ty2 = Refl ty1 | otherwise = CoVarCo cv where (ty1, ty2) = ASSERT( isCoVar cv ) coVarKind cv mkReflCo :: Type -> Coercion mkReflCo = Refl mkAxInstCo :: CoAxiom -> [Type] -> Coercion mkAxInstCo ax tys | arity == n_tys = AxiomInstCo ax rtys | otherwise = ASSERT( arity < n_tys ) foldl AppCo (AxiomInstCo ax (take arity rtys)) (drop arity rtys) where n_tys = length tys arity = coAxiomArity ax rtys = map Refl tys -- | Apply a 'Coercion' to another 'ii v i din 2 gia nh thng nht vic phn chia ranh gii, rt v t tiu hy s cc cm trn rung la. ng Breo cng nhn sai v theo lut tc, ng Breo np pht 5 triu ng v 1 con heo t chc cng Yang (thn linh).Kleinleipisch zeigte sich nach dem Wiederanpfiff zwar verbessert, zwingende Mglichkeiten blieben aber Mangelware. Auch die Gastgeber konnten keine klaren Einschumglichkeiten mehr aufweisen. Trauriger Hhepunkt war der Hubschraubereinsatz des ADAC, der - wenn wohl etwas berzogen - zur Halbzeit hereinflatterte. Ein Spieler der Gastgeber hatte sich beim bnerspringen eines am Boden liegs in two teams with S1 and S2, then those two teams would share S1, S2, and A, making their intersection three members, which is not allowed. Therefore, each member of T can only be in one team with S1 and S2. Therefore, if T has four members, each can be in at most one team with S1 and S2. Therefore, T can intersect at most four teams containing S1 and S2 (each member of T intersecting one team). But each intersection requires two members, so T can intersect two teams containing S1 and S2 by having two members each from two different teams. Wait, no. If each member of T is from a different team, then T would have four members, each from a different team, thus intersecting four teams, each in one member. But we need to intersect each team in two members. Therefore, this is impossible. Therefore, the only way T can intersect each team containing S1 and S2 in two members is if T contains two members from each team. But since each team has two unique members, and there are t teams, T would need 2t unique members. But T only has four members. Therefore, 2t 4, leading to t 2. But the problem states that the maximum is 3, so my reasoning must be incorrect. Alternatively, perhaps there's a different way to form team T. Maybe T shares two members with \frac{4x - 1}{x + 1} \right) > 0\) and \(\log_4 \left( \frac{x + 1}{4x - 1} \right) < 0\). These conditions further restrict the domain to \(x > 2/3\). We then simplify the inequality by recognizing that the terms involving logarithms are reciprocals. Let \(A = \log_4 \left( \frac{4x - 1}{x + 1} \right)\). The inequality reduces to: \[ 9.57 \log_3 A < 0, \] which implies \(A < 1\). Converting this back, we get: \[ \log_4 \left( \frac{4x - 1}{x + 1} \right) < 1, \] leading to: \[ \frac{4x - 1}{x + 1} < 4. \] Solving this inequality, we find: \[ \frac{-5}{x + 1} < 0, \] which simplifies to \(x > -1\). Considering the domain \(x > 2/3\), the solution is \(x > 2/3\). Thus, the solution to the inequality is: \[ \boxed{\left( \dfrac{2}{3}, \infty \right)}. \]Given that \(a, b,\) and \(c\) are real numbers, prove that the equation \[x^{2}-(a+b)x+\left(ab-c^{2}\right)=0\] has two real roots, and find the condition under which these two roots are equal. Okay, so I need to prove that the quadratic equation \(x^{2} - (a + b)x + (ab - c^{2}) = 0\) has two real roots, and then find the condition when these ro -- In example above, fam tys' = F [b] my_unify (FamInst { fi_axiom = old_axiom }) tpl_tvs tpl_tys _ = ASSERT2( tyVarsOfTypes tys `disjointVarSet` tpl_tvs, (ppr fam <+> ppr tys) $$ (ppr tpl_tvs <+> ppr tpl_tys) ) -- Unification will break badly if the variables overlap -- They shouldn't because we allocate separate uniques for them if compatibleBranches (coAxiomSingleBranch old_axiom) (new_branch) then Nothing else Just noSubst -- Note [Family instance overlap conflicts] noSubst = panic "lookupFamInstEnvConflicts noSubst" new_branch = coAxiomSingleBranch new_axiom \end{code} Note [Family instance overlap conflicts] ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ - In the case of data family instances, any overlap is fundamentally a conflict (as these instances imply injective type mappings). - In the case of type family instances, overlap is admitted as long as the right-hand sides of the overlapping rules coincide under the overlap substitution. eg type instance F a Int = a type instance F Int b = b These two overlap on (F Int Int) but then both RHSs are Int, so all is well. We require that they are synt**** @tcMonoBinds@ deals with a perhaps-recursive group of HsBinds. The signatures have been dealt with already. Note [Pattern bindings] ~~~~~~~~~~~~~~~~~~~~~~~ The rule for typing pattern bindings is this: ..sigs.. p = e where 'p' binds v1..vn, and 'e' may mention v1..vn, typechecks exactly like ..sigs.. x = e -- Inferred type v1 = case x of p -> v1 .. vn = case x of p -> vn Note that (f :: forall a. a -> a) = id should not typecheck because case id of { (f :: forall a. a->a) -> f } will not typecheck. \begin{code} tcMonoBinds :: RecFlag -- Whether the binding is recursive for typechecking purposes -- i.e. the binders are mentioned in their RHSs, and -- we are not rescued by a type signature -> TcSigFun -> LetBndrSpec -> [LHsBind Name] -> TcM (LHsBinds TcId, [MonoBindInfo]) tcMonoBinds is_rec sig_fn no_gen [ L b_loc (FunBind { fun_id = L nm_loc name, fun_infix = inf, fun_matches = matches, bind_fvs = fvs })] -- Single function binding, | NonRecursive <- is_rec -- ...binder isn't mentioned in RHS , Nothing <- sig_fn name -- ...- to take the wheel to be in charge) / in the gorge (Gorges LUM, consume greedily which big wheels in show biz often do. It can also refer to the Hollywood Gorge, which relates back to dumb blonde. One must pay special attention to those words that on a lesser-used level of meaning relate to literature, music and the entertainment business in general. There is no Dylanological precedent to cite for gorge as it is used only once) / Tur''!!1, ... > fs, fs', fs'' :: Fractional a => [a] > fs = (-1) : 3 : (-5)/2 : 3/2 : u > fs' = 3 : (-5) : 9/2 : u > fs'' = (-5) : 9 : u > u = undefined The helper computations are fs!!0 == f(0) == -1 fs'!!0 == f'(0) == 3 fs''!!0 == 2 - 3*3 -2*(-1) == 2-9+2 == -5 fs!!1 == (fs'!!0)/1 == 3 fs'!!1 == (fs''!!0)/1 == -5 fs''!!1 == 0 - 3*(-5) -2*3 == 15-6 == 9 fs!!2 == (fs'!!1)/2 == -5/2 fs'!!2 == (fs''!!1)/2 == 9/2 fs!!3 == (fs'!!2)/3 == (9/2)/3 == 3/2 Svar: take 4 fs == [-1, 3, -5/2, 3/2] Test the original eq: < lhs2 = take 2 (fs'' + 3*fs')/2 == ([-5,9]+3*[3,-5])/2 == [4,-6]/2 == [2,-3] < rhs2 = take 2 ((1:zeros)-fs) == [1,0] - [-1,3] = [2,-3] OK! ---------------- b) Laplace: We start again from the simplified equation: f''(x) = 2 - 3*f'(x) - 2*f(x) [1] f(0) = -1 f'(0) = 3 We need the rule for derivatives L f' s = - f 0 + s*L f s which iterated gives us L f'' s = - f' 0 + s*L f' s = - f' 0 + s*(- f 0 _tys [s1, s2] = stricts ; ASSERT( length arg_tys == length stricts ) if null ex_tvs' && null theta then return main_con else do { cxt <- reifyCxt theta' ; ex_tvs'' <- reifyTyVars ex_tvs' ; return (TH.ForallC ex_tvs'' cxt main_con) } } ------------------------------ reifyClass :: Class -> TcM TH.Info reifyClass cls = do { cxt <- reifyCxt theta ; inst_envs <- tcGetInstEnvs ; insts <- mapM reifyClassInstance (InstEnv.classInstances inst_envs cls) ; ops <- mapM reify_op op_stuff ; tvs' <- reifyTyVars tvs ; let dec = TH.ClassD cxt (reifyName cls) tvs' fds' ops ; return (TH.ClassI dec insts ) } where (tvs, fds, theta, _, _, op_stuff) = classExtraBigSig cls fds' = map reifyFunDep fds reify_op (op, _) = do { ty <- reifyType (idType op) ; return (TH.SigD (reifyName op) ty) } ------------------------------ reifyClassInstance :: ClsInst -> TcM TH.Dec reifyClassInstance i = do { cxt <- reifyCxt (drop n_silent theta) ; thtypes <- reifyTypes types ; let head_ty = foldl TH.AppT (TH.ConT (reifyName cls)) thtypes ; return $ (TH.InstanceD cxt head_ty []) } where (_tvs, theta, cls, types) = instanceHead i n_silent = dfunNSilent (instanceDFunId i) ------------------------------ reifyFamilyInstance :: FamInst -> TcM TH.Dec reifyFamilyInstance fi = case fi_flavor fi of SynFamilyInst -> do { th_tys <- reifyTypes (fi_tys fi) ; rhs_ty <- reifyType (coAxiomRHS rep_ax) ; return (TH.TySynInstD fam th_tys rhs_ty) } DataFamilyInst rep_tc -> do { let tvs = tyConTyVars rep_tc fam = reifyName (fi_fam fi) ; cons <- mapM (reifyDataCon (mkTyVarTys tvs)) (tyConDataCons rep_tc) ; th_tys <- reifyTypes (fi_tys fi) ; return (if isNewTyCon rep_tc then TH.NewtypeInstD [] fam th_tys (head cons) [] else TH.DataInstD [] fam th_tys cons []) } where rep_ax =g square to be the same color. Therefore, there are no monochromatic squares (S_0 or S_4) that include all four corners. However, other squares in the grid can be monochromatic. But without additional constraints, S_0 and S_4 can vary. Therefore, this approach might not help. Another angle: think about the fact that A and C are red, B and D are blue. Consider the two diagonals of the big square: AC (red) and BD (blue). Now, in the grid, each small square's diagonal may inherit some properties. But I'm not sure. Wait, here's an idea from graph theory: the grid can be considered as a bipartite graph. Since it's a chessboard, the vertices can be divided into two sets, say black and white, such that no two adjacent vertic1)(x +1)] / [x +4x -5] After factoring numerator: (x +1)^2 (x -x -1) Denominator: (x +1)(x -1)(x +5) Cancel (x +1): (x +1)(x -x -1) / [(x -1)(x +5)] Yes, that's correct. So the simplified function is [(x +1)(x -x -1)] / [(x -1)(x +5)]. So plugging in x = -1: Numerator: (-1 +1)(1 - (-1) -1) = 0*(1 +1 -1) = 0*(1) = 0 Denominator: (-1 -1)(1 +5) = (-2)(6) = -12 Therefore, 0 / -12 = 0. So the limit is 0. But just to be thorough, maybe there's another approach. For example, using LHospitals Rule. Since we have a 0/0 form, perhaps we can take derivatives of numerator and denominator. Let's try that. Let me write the original limit as: lim_{x->-1} [ (x -2x -1)(x +1) ] / [x +4x -5] Let N(x) = (x -2x -1)(x +1) D(x) = x +4x -5 Since both N(-1) and D(-1) are 0, we can apply LHospitals Rule, which states that the limit is equal to the limit of N(x)/D(x) as x approaches -1, provided that derivative of D(x) is not zero at x = -1. So compute N(x) and D(x): First, N(x) = (x -2x -1)(x +1). Let's use the product rule: N(x) = derivative of first * second + first * derivative of second. First part: x -2x -1. Its derivative is 3x -2. Second part: x +1. Its derivative is 1. Therefore, N(x) = (3x -2)(x +1) + (x -2x -1)(1) Simplify: (3x -2)(x +1) = 3x +3x -2x -2 Adding (x -2x -1): 3x +3x -2x -2 +x -2x -1 = 4x +3x -4x -3 Now compute D(x): derivative of x +4x -5 is 4x +8x Therefore, D(x) = 4x +8x So applying LHospitals Rule, the limit becomes: lim_{x->-1} [4x +3x -4x -3] / [4 that are parallel, which are the bases. Here, AB and CD are the bases with lengths 101 and 20. The other two sides, AD and BC, are the legs. Wait, but in some trapezoids, the legs can also be non-parallel. Since the diagonals are perpendicular, that must add some special properties. I think using coordinate geometry might help here. Let me try to place the trapezoid on a coordinate system. Let's set base AB on the x-axis for simplicity. Lets assign coordinates to the points. Lets say point A is at (0, 0), and point B is at (101, 0) since AB is 101. Now, since CD is the other base with length 20, and it's parallel to AB, CD should be somewhere above AB. Let's denote points C and D as (x, h) and (x + 20, h), where h is the height of the trapezoid. Wait, but the positions of C and D depend on how the trapezoid is shaped. Since it's a trapezoid, sides AD and BC are the legs connecting the two bases. But maybe I need to adjust the coordinates to account for the fact that diagonals are ) > where > mergeIntoItem :: Lr0Item -> [Lr1Item] > mergeIntoItem item@(rule,dot) > = [(rule,dot,la)] > where la = case [ s | (item',s) <- lookaheads ! i, > item == item' ] of > [] -> Set.empty > [x] -> x > _ -> error "mergIntoItem" ----------------------------------------------------------------------------- Generate the goto table This is pretty straightforward, given all the information we stored while generating the LR0 sets of items. Generating the goto table doesn't need lookahead info. > genGotoTable :: Grammar -> [(Set Lr0Item,[(Name,Int)])] -> GotoTable > genGotoTable g sets = gotoTable > where > Grammar{ first_nonterm = fst_nonterm, > first_term = fst_term, > non_terminals = non_terms } = g > > -- goto array doesn't include %start symbols > gotoTable = listArray (0,length sets-1) > [ > (array (fst_nonterm, fst_term-1) [ > (n, case lookup n goto of > Nothing -> NoGoto > Just s -> Goto s) > | n <- non_terms, > n >= fst_nonterm, n < fst_term ]) > | (set,goto) <- sets ] ----------------------------------------------------------------------------- Geuskal ondarearen beste erakusgarri bi: Baraziarten Meditazione tipiak eta Intxausperen Maria birjinaren hilabetia.De oude koning beval hem de volgende dag weer de ganzen naar buiten te drijven en zodra het ochtend was, ging hij zelf achter de donkere poort staan en hoorde daar hoe zij met het hoofd van Falada sprak. Daarop ging hij haar ook achterna naar buiten en verborg zich achter een struik op de wei. Daar zag hij weldra met eigen ogen hoe de ganzenhoedster en de ganzenjongen de troep de wei opdreven en hoe zij na een poosje ging zitten en haar haren, die stralend glansden, los vlocht disease (formation of fibrotic plaques of the penis, usually in men over 50, resulting in painful, crooked erections, rendering intercourse difficult or impossible); Dupuytrens contracture (wherein the flexor tendons of the fingers of the hands become fibrotic and contract, rendering the fingers useless); and scleroderma (a rare conditionKlima ist von Natur aus ein hochdynamischer Prozess und mit keinem Mittel stabil zu halten. Klimaschutz vermittelt eine Illusion und ist nur ein politisches Konstrukt. Dazu sagt Prof. Heinz Miller, AWI-Vize-Direktor: Wer von Klimaschutz redet, weckt Illusionen. Klima lsst sich nicht schtzen und auf einer Wunschtemperatur stabilisieren. Es hat sich auch ohne Einwirkungen des Menschen oft drastisch verndert. Schlagworte wie Klimakollaps oder Klimakatastrophe sind irrefhrend. Klima kann nicht kollabieren, die Natur kennt keine Katastrophen.Tot pentru ndrznirea de a sfrma templele idoleti, un lucru asemntor cu acesta s-a ntmplat i n Fenicia. Aici, un diacon, anume Chiril, din dragostea de adevr ndrznind s doboare la pmnt nite statui de ale idolilor, a fost prins i i s-a despicat pntecele, scondu-i-se mruntaiele i fiind lsat aa, ca s fie vzut de toi. i se zice c ticloii acetia btndu-i joc de ficaii lui i-au primit rsplata bine meritat, cci le-au czut dinii, limba li s-a topit n gur i au pierdut puterea vederii.11 septembre 2019 by Lily Fantasy Categories: Accueil, Chroniques, Science-fictionTags: aide, alcool, amis, amiti, amour, animaux, Arthur, automate, Avalon, preuves, bagarre, Brocliande, celtes, chance, cousine, cousins, danger, desses, dtective, dieux, divination, djinn, don, druides, enchantement, enqute, famille, faune, fes, flore, Gaule, humanit, incon e) = show e > > instance Exception SomeFrontendException where > toException = compilerExceptionToException > fromException = compilerExceptionFromException > > frontendExceptionToException :: Exception e => e -> SomeException > frontendExceptionToException = toException . SomeFrontendException > > frontendExceptionFromException :: Exception e => SomeException -> Maybe e > frontendExceptionFromException x = do > SomeFrontendException a <- fromException x > cast a > > --------------------------------------------------------------------- > -- Make an exception type for a particular frontend compiler exception > > data MismatchedParentheses = MismatchedParentheses > deriving (Typeable, Show) > > instance Exception MismatchedParentheses whfic numbers. Wait, maybe there's some standard or unique way these numbers can be arranged. Let's think. Since there are two boxes for each type, the labels on the boxes must be numbers that when added give the total for that type. Also, all the labels are numbers of stones in each box. Maybe the labels are distinct? Or could they be the same? Wait, the problem doesn't say the labels are distinct. So, perhaps each box's label is a number, but boxes can have the same number. However, given that the total number of rubies is 15 more than diamonds, we might need to find a relationship between the labels of the ruby boxes and the diamond boxes. But how? Let's think. Let me denote the labels for the diamond boxes as D1 and D2, so total diamonds = D1 + D2. Similarly, emeralds as E1 + E2, and rubies as R1 + R2. We know that R1 + R2 = (D1 + D2) + 15. We need to find E1 + E2. But without and my list. We keep telling them we don't need anything but they want one anyway. We got their's too but they just want us to pick something not get them everything. Sorry your bread didn't turn out the way you expected it to, but I'm sure it was good anyway. Deb, sorry to hear that dh and pooch are under the weather. Sometimes 'new' foods can upset dog's systems, I have been told. Dh sounds like he needs a good rest, all that worrying is not good, but I'm sure he can't help it. Watching movies at least takes him away from his thoughts for a while. 15cm, does it look to you like that's what it is? 15mm would be less than an inch. Glad you had a nice visit with your cousin, so sad that the uncle is forgotten.Jess, sorry about the restless night, but I understand how your mind does not shut down. Hope for better sleep tonight. I wake up before my alarm goes off and then hit the snooze 3 times before I get up.--I was more upset than DH was about missing the meal.Gramma, glad you got your Big Shot. I use my CB mostly for embossing and some dies, but the BigKick works much better on the thin dies.--I'm trying my hardest not to be drawn into the baking circle, although I make banana muffins pretty regularly now. I did get some KA flour at the commissary today, it's about $1 less than at WM. Joanne is still on the boat making cards. I need to work on the CFS cards, I didn't get much done last night. Oh I forgot to tell you that one of the SFC's in my unit sent me a card thanking 'us' for making the cards. So sweet.Have a great evening. Renate User ProfileView All Posts by UserView Thanks4. Eric R. left this voice mail: "I share your concerns about drinking and driving, because I've had the unfortunate experience of going to funerals caused by drunken driving. It does concern me. However, if it's only 150-160 people who die from drunk drivst retrieve all those columns. But the HTML table only displays full names (first + last) and email. The salary is used to calculate the average. So in the PHP script, when fetching the data, we can get all columns. For each row, we display first and last name combined, email. The salary is stored in an array to compute the average. Alternatively, maybe the SQL query can select all columns, then in PHP process them. That's possible. So the steps are: - Connect to DB. - Query SELECT * FROM employees (or all the columns, but maybe specify them explicitly). - For each row, concatenate first and last name. - Collect all salaries into an array. - Compute average by summing the array and dividing by count. - Display the HTML table with full name and email. - Display the average salary. But for security and clarity, it's better to select only the needed columns. So in the SQL query, select first_name, last_name, email, salary. Then, in PHP, for each row, combine first and last name. Collect salary values. Then calculate average. So the code steps: 1. Connect to MySQL using mysqli. Parameters: host, user, pass, dbname. 2. Check connection error. 3. Perform query to get employees: SELECT first_name, last_name, email, salary FROM employees. 4. Check for query errors. 5. Fetch all results into an array. 6. Close the connection early? Or keep it open until after processing. Probably better to process first, then close. 7. Initialize variables: total_salary = 0, count = 0. 8. Then, loop through each employee, add salary to total_salary, increment count. 9. Calculate average: total_salary / count. But CLICK NOW!KEY FEATURES INCLUDEThird Row Seat, 4x4, Premium Sound System, Aluminum Wheels. Rear Spoiler, Keyless Entry, Privacy Glass, Child Safety Locks, Steering Wheel Controls. OPTION PACKAGESTOWING PREP PKG upgraded radiator, transmission oil cooler, 130-amp alternator, trailer pre-wiring, 160-watt fan coupling. Toyota Limited w/3rd Row with MILLENNIUM SILVER exterior and ASH interior features a V6 Cylinder Engine with 215 HP at 5800 RPM*. VEHICLE REVIEWSThe Highlander has a carlike unibody design which leads to better handling, less cabin noise, improved crashworthiness and easier entry and exit for passengers. -Edmunds.com. 5 Star Driver Front Crash Rating. 5 Star Driver Side Crash Rating. Great Gas Mileage: 24 MPG Hwy. PURCHASE WITH CONFIDENCECARFAX 1-Owner BUY FROM AN AWARD WINNING DEALERPlease call 210-341-8766. Welcome to North Park LINCOLN On behalf of our entire staff here at North Park LINCOLN, we would like to welcome you and thank you for visiting o it's sometimes bad not to make a wrapper. Consider fw = \x# -> let x = I# x# in case e of p1 -> error_fn x p2 -> error_fn x p3 -> the real stuff The re-btically becoming the most energy and production efficient place in the world.N2 - Background/Purpose: The increasing popularity of activity tracking devices presents an opportunity to monitor physical activity in patients after stroke. We sought to determine the reliability and accuracy of the Garmin Vivofit and Fitbit Zip for adults after stroke. Methods: Twenty-four participants with stroke-induced hemiparesis wore a Fitbit Zip on the nonparetic hip and Garmin Vivofits on both wrists during a 6-minute walk test to determine the accuracy of the devices against video-determined step counts. Participants also wore the devices during two trials of exactly 50 steps to determine the reliability of the devices. Results: Fitbit Zip showed excellent reliability (ICC2,1 = 0.974) and accuracy (4.2% error) for participants who walked faster than 0.35 m/s. Garmin Vivofit (nonparetic side) had excellent reliability (ICC2,1 = 0.964) but poor accuracy (16.0%) for all participants. Garmin Vivofit (paretic side) had excellent reliability (ICC2,1 = 0.858) and accuracy (4.0% error) for faster walkers (>0.48 m/s) but poor accuracy (68.2%) for slower walkers. Conclusion: Fitbit Zip was more accurate and reliable for persons with stroke than Garmin Vivofit, but slower walking speeds were associated with greater undercounting of steps for both devices. The Fitbit Zip is appropriate for counting steps in adults poststroke who range from household to community ambulators. - , . : , . , , -ch is 29/20 + 1.4 = 1.45 +1.4 = 2.85 kg. Converting 2.85 kg to fractional form: 2.85 = 2 + 85/100 = 2 + 17/20 = 57/20 kg, which is 2.85 kg. Alternatively, check if 29/8 kg for 5m. So 29/8 divided by 5 is 29/(8*5) = 29/40 kg per meter. Then 2m is 29/40 *2 = 29/20 = 1.45 kg. For B: 2.8 kg for 4m. 2.8/4 = 0.7 kg/m. 2m is 1.4 kg. Total: 1.45 +1.4 = 2.85 kg. So 2.85 kg. So yes, that's 57/20 kg, which is the answer. Alternatively, check units. The problem mentions cutting by 2 me ; return (HsIPBinds (IPBinds ip_binds' ev_binds), result) } where ips = [ip | L _ (IPBind (Left ip) _) <- ip_binds] -- I wonder if we should do these one at at time -- Consider ?x = 4 -- ?y = ?x + 1 tc_ip_bind ipClass (IPBind (Left ip) expr) = do { ty <- newFlexiTyVarTy openTypeKind ; let p = mkStrLitTy $ hsIPNameFS ip ; ip_id <- newDict ipClass [ p, ty ] ; expr' <- tcMonoExpr expr ty ; let d = toDict ipClass p ty `fmap` expr' ; return (ip_id, (IPBind (Right ip_id) d)) } tc_ip_bind _ (IPBind (Right {}) _) = panic "tc_ip_bind" -- Coerces a `t` into a dictionry for `IP "x" t`. -- co : t -> IP "x" t toDict ipClass x ty = case unwrapNewTyCon_maybe (classTyCon ipClass) of Just (_,_,ax) -> HsWrap $ mkWpCast $ mkTcSymCo $ mkTcUnbranchedAxInstCo Representational ax [x,ty] Nothing -> panic "The dictionary for `IP` is not a newtype?" \end{code} Note [Implicit parameter untouchables] ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ We add the type variables in the types of the implicit parameters as untouchables, not so much because we really must not unify them, but rather because we otherwise end up with constraints like this Num alpha, Implic { wanted = alpha ~ Int } The constraint solver solves alpha~Int by unification, but then doesn't float that solved constraint out (it's not an unsolved wanted). Result disaster: the (Num alpha) is again solved, this time by defaulting. No no no. However [Oct 10] this is all handled automatically by the untouchable-range idea. Note [Placeholder PatSyn kinds] ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Consider this (Trac #9161) {-# LANGUAGE PatternSynonyms, DataKinds #-} pattern A = () b :: A b = undefined Here, the type signature for b mentions A. But A is a pattern synonym, which is typechecked (for very good reasons; a view pattern in the RHS may mention a value binding) as part of a group of bindings. It is entirely resonable to reject this, but to do so we need A to be in t"=", RowBox[{"\[Nu]", "-", "2"}]}], ";", "\[IndentingNewLine]", RowBox[{ RowBox[{"{", RowBox[{"abscissaR", ",", "weightR"}], "}"}], "=", RowBox[{"generateGaussianQuadrature", "[", RowBox[{ RowBox[{"moments", "[", RowBox[{"\[Nu]R", ",", RowBox[{ RowBox[{"2", " ", "nPointsRadial"}], "+", "1"}]}], "]"}], ",", "16", ",", "350"}], "]"}]}], ";", "\[IndentingNewLine]", RowBox[{ RowBox[{"IntSpherical", "[", "r1_", "]"}], ":=", RowBox[{"2", "*", RowBox[{"Pi", "^", "2"}], RowBox[{"Sum", "[", RowBox[{ RowBox[{ RowBox[{"1", "/", RowBox[{"Length", "[", "abscissa0", "]"}]}], "*", RowBox[{"weight0", "[", RowBox[{"[", "i", "]"}], "]"}], "*", RowBox[{"g", "[", RowBox[{"r1", ",", RowBox[{"Pi", "*", RowBox[{"abscissa0", "[", RowBox[{"[", "i", "]"}], "]"}]}], ",", RowBox[{"2", RowBox[{"Pi", "/", RowBox[{"(", RowBox[{"nPointsSpherical", "+", "1"}], ")"}]}], "* "Yes". Similarly, a local from A would truthfully say "Yes". So in that case, both would answer the same. That doesn't help. Alternatively, maybe ask "Are you a visitor?" If in village A, the locals would say "No", visitors from B would lie and say "No". So again, same answer. Hmm, not helpful. Wait, perhaps the key is to ask a question that relates to the village itself. For example, if the traveler is in village A, then regardless of whether the person is a local or visi a square, and that any square can be expressed as a sum of consecutive odd numbers (but not necessarily uniquely). Given this, perhaps the user made a mistake in the problem statement, and the intended claim was about the sum of the first n odd numbers. In that case, the proof is straightforward by induction or by the known formula. Alternatively, if the problem is to show that every perfect square can be expressed as a sum of consecutive odd numbers, that's also true, but again, different from the original statement. Given the confusion, perhaps the correct answer is to clarify that the sum of consecutive odd numbers starting from 1 is a perfect square, and provide the proof for that, while noting that the original statement as given is incorrect because there exist counterexamples when not starting from 1. But since the user insists that the assit, in this tournament, Player A has 4 wins, Player B has 3, Player C has 2, Player i has 1, Player D has 0. Therefore, the score sequence is (4, 3, 2, 1, 0). But in reality, in a tournament of five players, the maximum number of wins is 4, so Player A has 4, Player B has 3, etc. However, in this case, the tournament is transitive, so the sum of cubes is: (4-0)^3 + (3-1)^3 + (2-2)^3 + (1-3)^3 + (0-4)^3 = 64 + 8 + 0 -8 -64 = 0. But earlier, I thought Player D's results were different, but in reality, in a transitive tournament, each player's wins are the players below them. Therefore, the sum is zero. But wait, in my previous mistaken example, I considered Player D losing to A, B, C, and i, but in reality, in a transitive tournament, Player D would only lose to A, B, C, and i, but in the transitive order, Player D is the sink, so i is above D. So, Player i beats D, and others beat D as well. So, Player D's conquerors are A, B, C, and i. But since the tournament Okay, so I need to solve this geometry problem about a triangular pyramid (which is a tetrahedron) P-ABC. The given information is that angles APB, BPC, and CPA are all 90 degrees. Then there's a point M on the base ABC, and we're given angles APM = and BPM = . We know sin = 6/3 and cos = 6/6, and we have to find the measure of angle CPM. The options are 30, 45, 60, or 75. First, I need to visualize this tetrahedron. Since all the angles at P between the edges are right angles, PA, PB, and PC are mutually perpendicular. That suggests that we can model this tetrahedron in a coordinate system where P is at the origin, and A, B, C are along the x, y, z axes respectively. Let me confirm that: if PA, PB, PC are mutually perpendicular, then yes, setting P at (0,0,0), A on the x-axis, B on the y-axis, and C on the z-axis makes sense. So coordinates would be P(0,0,0), A(a,0,0), B(0,b,0), C(0,0,c) for some positive real numbers a, b, c. Point M is on the base ABC. The base ABC is the triangle formed by points A, B, C. Since M is on ABC, its coordinates can be expressed in terms of A, B, C. In barycentric coordinates, maybe, but perhaps more straightforwardly using parametric equations. Since ABC is a triangle, any point M on it can be written as M = (a(1 - s - t), b s, c t) where s and t are parameters such that s 0, t 0, and s + t 1. Alternatively, using coordinates with weights: if I use affine combinations, then M = ( (1 - s - t)A + s B + t C ), which translates to coordinates (a(1 - s - t), b s, c t). Hmm, but maybe I need a different coordinate system. Alternatively, if I use coordinates in terms of the original coordinate system, since P is at (0,0,0), and A, B, C are at (a,0,0), (0,b,0), (0,0,c). Then any point on ABC can be represented as (x, y, z) where x/a + y/b + z/c = 1, right? Because the equation of the plane ABC is x/a + y/b + z/c = 1. So M is a point on that plane. Given that angles at P: APB, BPC, CPA are 90, so PA, PB, PC are perpendicular. That is, vectors PA, PB, PC are orthogonal. So this is a right tetrahedron. That should make things easier, as we can use coordinate gUUID->"cd49158e-9658-435d-9a2b-245dbb8ce536", CellID->49120259], Cell[540746, 13669, 224, 4, 191, "Input",ExpressionUUID->"d14053c6-d907-4663-9e33-f32611863c6a", CellID->7333602], Cell[540973, 13675, 648, 18, 191, "Input",ExpressionUUID->"ce5c3b08-a23e-4ea1-b5d2-fca72fa99ac8", CellID->171576526], Cell[541624, 13695, 1587, 45, 213, "Input",ExpressionUUID->"3c59c69b-d052-4c89-b565-db2954384b18", CellID->25400824] }, Closed]], Cell[CellGroupData[{ Cell[543248, 13745, 171, 2, 54, "Subsection",ExpressionUUID->"819123f7-e445-4736-a251-88ateElement('th'); th.abbr = headDays[t]; scope = 'col'; th.title = headDays[t]; th.innerHTML = headInitial[t]; tr.appendChild(th); } calHead.appendChild(tr); for (x = 0; x (')[1]; var selValue = bcList[r]; sel.options[q] = new Option(selText + ' ('+selCount,selValue); q++ } document.getElementById('bcaption').appendChild(sel); var m = bcList[0].split(',')[0]; var y = bcList[0].split(',')[1]; callArchive(m,y,'0'); } function timezoneSet(root){ var feed = root.feed; var updated = feed.updated.$t; var id = feed.id.$t; bcBlogId = id.split('blog-')[1]; upLength = updated.length; if(updated.charAt(upLength-1) == "Z"){timeOffset = "+00:00";} else {timeOffset = updated.substring(upLength-6,upLength);} timeOffset = encodeURIComponent(timeOffset); } //]]> ' : ''; var month = [1,2,3,4,5,6,7,8,9,10,11,12]; var month2 = ["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"]; var day = postdate.split("-")[2].substring(0,2); var m = postdate.split("-")[1]; var y = postdate.split("-")[0]; for(var u2=0;u2Bittova, who will speak with the audience following the film, returns to Syracuse after travelin1:06 CST) number 33 and rising. If just 300 of the people reading this vote for her, Vacation Time is Precious Nation, Well bust her into the Top 10. Ill be reporting on Pattys progress in future posts. One other thing, they dont say when the voting ends, so please vote right away.Americas Historical Newspapers is a subscription database published by Readex/NewsBank. Tennessee titles include the Knoxville Journal, Memphis Daily Avalanche, and the Nashville Banner. A complete list of US titles can be found here. This list includes newspapers from Readexs African American Newspapers database. Tennessee titles include Knoxvilles Negro World and Nashvilles Colored Tennessean. This subscription database is available to UT students and employees through UT Libraries Databases look for the link called Americas Historical Newspapers (1690-1993). The Knoxville News and the Knoxville News-Sentinel (1922-1990) is available to Knox County Public Library cardholders and UT employees and students.Copyright 2017 - 2020 . - , 437(2) . , , -, . - , , .This highly-efficacious facial serum brightens and clarifies skin for increased radiance. Over time, it diminishes the numb.695. Similarly, y=(3/4)^4=81/2560.3164, y^y=(81/256)^{81/256}e^{(81/256) ln(81/256)}. ln(81/256)=4 ln3 -8 ln24*1.0986 -8*0.69314.3944 -5.5448-1.1504. Thus, ln(y^y)= (81/256)*(-1.1504)0.3164*(-1.1504)-0.363, so y^ye^{-0.363}0.695. Therefore, x^x and y^y are equal. Similarly for n=3, they are equal. So my initial mistake was in manual calculatace, etc., serves for both. \begin{code} type LSig name = Located (Sig name) -- | Signatures and pragmas data Sig name = -- | An ordinary type signature -- @f :: Num a => a -> a@ -- -- - 'ApiAnnotation.AnnKeywordId' : 'ApiAnnotation.AnnDcolon', -- 'ApiAnnotation.AnnComma' TypeSig [Located name] (LHsType name) -- | A pattern synonym type signature -- @pattern type forall b. (Eq b) => P a b :: forall a. (Num a) => T a | PatSynSig (Located name) (HsExplicitFlag, LHsTyVarBndrs name) (LHsContext name) -- Provided context (LHsContext name) -- Required context (LHsType name) -- | A type signature for a default method inside a class -- -- > default eq :: (Representable0 a, GEq (Rep0 a)) => a -> a -> Bool -- | GenericSig [Located name] (LHsType name) -- | A type signature in generated code, notably the code -- generated for record selectors. We simply record -- the desired Id itself, replete with its name, type -- and IdDetails. Otherwise it's just like a type -- signature: there should be an accompanying binding -- -- - 'ApiAnnotation.AnnKeywordId' : 'ApiAnnotation.AnnDefault', -- 'ApiAnnotation.AnnDcolon','ApiAnnotation.AnnDotdot' | IdSig Id -- | An ordinary fixity declaration add x y)))" Tokenize this inside to get tokens = ["let", "x", "2", "(mult x (let x 3 y 4 (add x y)))"] Operator is 'let', so we process the variables and body. Create new_scope, append to scopes (now scopes has [new_scope]). Variables are tokens[1:-1] = ["x", "2", "(mult x ...)"]? Wait, tokens after 'let' are ["x", "2", "(mult x ...)"]. The body is the last token, which is "(mult x ...)". Wait, no. The tokens after 'let' are the variables and body. The variables are pairs of var and expr, and the last token is the body. In the example, tokens after 'let' are: tokens = ["x", "2", "(mult x ...)"], but the body is the last token. So variables are ["x", "2", "(mult x ...)"]? Wait, no. Wait, in the let case, the tokens after 'let' are var1, expr1, var2, expr2, ..., body. But in this example, after 'let', the tokens are "x", "2", "(mult x ...)". The variables are supposed to be pairs. So, len(tokens) is 3. But the variables part is tokens[1:-1], which would be ["x", "2"], and the body is tokens[-1] = "(mult x ...)". Wait, no. The tokens after 'let' are [ "x", "2", "(mult ...)" ]. So tokens[1:-1] is [ "x", "2" ], and the body is "(mult ...)". But variables are pairs of variables and expressions. So the variables part is ["x", "2"], which is two tokens. So for each pair, var and expr. So in this case, var1 is "x", expr1 is "2". Then, the body is the third token. l axis" of a cube might l guidelines) and a MUAC tape.Cheeky move or quick right foot Vettel led the field away on the warmup lap of the Spanish GP before his teammate and pole holder caught up and passed. Race starts cleanly and the race to turn one is clean until the back of the field. Senna is shown as stopped, Buemi and De La Rosa head for the pits. Kovalainen was shown starting the race from the pits as result of electrical problems. He now is shown as retired. Lap 10 Webber in the lead with a 2.8 sec lead over Vettel followed by Hamilton Alonso ?Button Schumacher Massa Sutil Alguersuari and Kubica in P 10. Ferrari is urging Massa to pass Schumacher who is down on straight line speed to the Brazilian. Lap 13/66Con el fin de discutir las necesidades ticas, regulatorias y tecnolgicas de la llamada Cuarta Revolucin Industrial, el Foro Econmico Mundial (FEM) lanz, en marzo de 2017, un centro tecnolgico y de innovacin, ubicado en San Francisco (EE.UU.),Hoy, casi dos aos despus, un nuevo espacio como este se prepara para abrir sus puertas en Medelln, y sumarse a la red global que tambin conforman otros tres centros, localizados en Tokio, Beijing y Mumbai, estos Centros son espacios que, segn el FEM, permiten la cooperacin global, la defensa de la tica y los valores en la tecnologa , un concepto bajo el cual se busca unir a gobiernos y empresas para disear marcos y protocolos de gobernanza aplicables a la ciencia y la tecnologa en el mundo.The most important part of building a successful, online sales platform is choosing a domain name thats easy to remember. If people can quickly call your web address to mind, they can visit again and again via different devices, and share it with their friends and other online contacts. You also want to choose a domain name thats indicative of what you have to offer. This will not only make it easier to remember, but it will additionally simplify the branding process and help you garner traffic from a diverse range of online sources. Companies like Register have an expansive array of affordable domain names for ecommerce businesses to choose from. This gives people the ability to select options that suit boTemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-5.521460917862246, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-5.298317366548036, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-5.115995809754082, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-4.961845129926823, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-4.8283137373023015`, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-4.710530701645918, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-3.912023005428146, FormBox[ TemplateBox[{0., 0.}, "Spacer2"], TraditionalForm], {0.005, 0.}, { AbsoluteThickness[0.1]}}, {-3.506557897318,028 +58,058) + (0 +506 +15,792) -0 Compute step by step. First, C(20,10)=184,756. Sum A + B + C=506 +28,028 +58,058=506 +28,028=28,534 +58,058=86,592. Sum of intersections:506 +15,792=16,298. Thus, Total valid=184,756 -86,592 +16,298= (184,756 -86,592)=98,164 +16,298=114,462. Which matches the previous result. Therefore, the answer is 114,462. To determine the number of ways Alibaba can choose 10 gems with the given constraints, we break down the problem by considering the number of green gems (G) and then calculating valid combinations of red (R) and blue (B) gems for each G. **Constraints:** - At least 2 red (R 2) - At least 2 blue (B 2) - No more than 3 green (G 3) **Approach:** 1. **Case Analysis by Green Gems (G):** G can be 0, 1, 2, or 3. 2. **For Each G:** - Calculate the remaining gems needed (10 - G) from red and blue. - Determine valid ranges for R and B ensuring R 2, B 2, and R + B = : 5h 16m Massage: 4h 00m Walking: 1h 45m Yard Work: 17h 00m Yoga: 5h 30m 2014 totalsSwim: 57h 47m 09s - 157322.39 MBike: 219h 22m 44s - 3507.6 MiRun: 166h 42m 28s - 1031.52 MiStrength: 36h 31m Core Training: 15m Elliptical Training: 12m 30s Kayaking: 1h 15m Massage: 8h 30m Paddle Boarding: 3h 51m Pilates: 1h 00m Rock Climbing: 2h 20m Rowing: 10m Skiing: 3h 30m Transitions: 35m Walking: 18h 05m Yard Work: 15h 30m Yoga: 6h 50m Save Cancel To find the volume of the oblique prism \( A B C A_{1} B_{1} C_{1} \), we start by noting that the volume of a prism is the area of th Escort Service ||Lucknow Escortpossible cases for the last three bits. Of these, 5 cases result in M(i)=M(i+7), and 3 cases result in inequality. If the last three bits are uniformly distributed, we would expect that for each i, the probability that M(i)=M(i+7) is 5/8. Therefore, over N=1,000,000 -6 = 999,994 numbers (wait, original sequence is from M(0) to M(1,000,000). So the number of terms i such that i+7 1,000,000 is 1,000,001 -7 = 999,994. Therefore, if the distribution is uniform, we would expect approximately (5/8)*999,994 624,996. But the problem states that we need to prove at least 450,000, which is much lower. Therefore, there must be some structure in the sequence that guarantees at least this number. But perhaps the distribution is not uniform, but due to the properties of binary numbers, the number of cases where last three bits fall into the 5 categories is at least 450,000. However, to prove it rigorously, we need to find a lower bound. One approach is to note that the occurrences of the last three bits have certain regularities. For example, every 8 consec from B, so AP = 10 - x. Similarly, Q is located at BQ = y cm from B, so QC = 12 - y. Since D is 8 cm from B on AB, PD = |PB - BD| = |x - 8|. Similarly, E is 8 cm from B on BC, QE = |BQ - BE| = |y - 8|. But earlier, we established that PT = PD and QT = QE, and since PT = QT (because they are both tangent segments from P and Q to the same point T on PQ), then PD = QE. Therefore, |x - 8| = |y - 8|. Assuming that P is between B and D (x 8) and Q is between B and E (y 8), then PD = 8 - x and QE = 8 - y. Therefore, 8 - x = 8 - y => x = y. So PB = BQ. This suggests that x = y. Alternatively, if P is beyond D (x > 8), then PD = x - 8, and Q would need to be beyond E (y > 8), leading to QE = y - 8, so x - 8 = y - 8 => x = y. But since AB is only 10 cm, x can't be more than 10, and BC is 12 cm, y up to 12. However, if x = y, then the tangent line would be symmetric with respect to B. But I think in this case, since the incircle is located towards the longer side, the tangent line is likely on the side closer to B, so x and y would both be less than 8. Therefore, x = y. Assuming x = y, then PB = BQ = x. Now, we need to relate this to the geometry of the triangle. Now, in triangle PBQ, PB = BQ = x, and PQ is the tangent line. We need to find x and the length of PQ to compute the perimeter. But how? Alternatively, using coordinates. Let's assume that PB = BQ = x. Then, point P is x cm from B on AB, and point Q is x cm from B on BC. Coordinates: Point B is at (0,0). AB is 10 without the '-B' prefix). -> IO String -- TopDir (in Unix format '/' separated) findTopDir (Just minusb) = return (normalise minusb) findTopDir Nothing = do -- Get directory of executable maybe_exec_dir <- getBaseDir case maybe_exec_dir of -- "Just" on Windows, "Nothing" on unix Nothing -> throwGhcExceptionIO (InstallationError "missing -B option") Just dir -> return dir \end{code} %************************************************************************ %* nvw7zOG/nfEX5m2bCpfZnckQNORPHnEH6+QjM/rsfHl3+ucL0I f/YxpQvXe9e2Br/5OggvBXv6CntKP3GH/f/0zLO/xBOxX85iv6Q/WGN/h4v9 lfyh8IcPwh8k/g/8p5zwH4mnwt/Shb8pWv80Fv6paP25kfBniV+C/x8Q/i/x ZzgvL8R5kfIb43xtEudL8q1xHo03551HidM5zcE5TRa4WhL2nAJ7Eq5ozu8D 4MM1dib+J+j5HOea8Juwf3vYPwl4sOF6JU774oF9ITknDe0m9Rmm2S/iJ8P+ BRAfSP437GNZ7CPJ0caNJA2egf0lnOKJJeKJ5OO5+bHv/0NPhfTU+gPJcYAd 3BF/iB+r8RPC58LOHRCX6Llp2JcM+A/x+2AfCyBe0XMbafyKcFP4yXzEMcIr avyN8NGIb8MR37zF+VVKfRLn+qo41xKfhbg3CHGPcDPkndMi78g8mN9axI2y r/LihuRPQZwcjThJ+GfktQ4ir8k8a71cxKX1Ii5J/nPE1QDEVcJtj4u8+Unk TanPGsQ9+/l5cU/yRyEOD0McJrwY8vJtkZelnHwNRVwtdjUvrkp+FOL2XcRt wr2R9zeKvC/l3ELcdqqdF7cl3yZZxPmZiPOEU12xRdQVUg7xJwm+SvxEyG8m 5Et8IuqWf77l1S0K2fk09L8t9Jf8v8gjhcV6JV4VddEFURfJemO0oT0lP8zQ /hLvj7rLXNRdUo7RF4P9lfzXywz8QeLFnURdd03UdXJdA+FvY4S/Sf5n5LWK wj8lvgV1Yy1RN0p9vOD/Y4T/S35RnJeHyIOEW6EuPSvqUilnFM7dbHHuJL8A 6luLhnn17b92+B//eepG3VaM3qgF7LrrTN9nyHw3GrgJcIon/0vOyzHHip9X H6mdgs5FLk3Mlvn3FfDOwJP+Q47DmSyjZ2VXOuVvPNK/xY9cWe/Za/D/knOy 2MoLMeojxXlku+78uTIvA+8A/L/WdXxx4KjWRm+UyDvF5xVidqB1EX4K+IP/ kBPiWjToZ61cpYpXvrleUxOknHXAqwL/r3WtB99G8GXeJzm2wP9Ln5PQ/6TQ X9YbEYbrUv7LPmcM7Sz1Oa3B/2tdYcu/5dfbX6lPqAb/Lzm34W+uwt+kPonA uwP/LzmD4P+lhP9LOUM0+H/ZpwrOo604j6qLOI9qZeS7VOQ7wnvgvKeiD+0K /DXq5DXIp1QnW2WIfOeOfEdyHiH+dELfSngC4lW2qMOlnBbId+YHRL4jfhHE w0z0uYTnIH56IJ+SnEbId++Q74g/DfG5NvpiWpcr4vkn0UdIOZuQ7y4j35Gc NsgXL9BHEx6F/BIr+hQpZzfyXSzyHfFXIn9VQt9N+lAf9Ev0QVLOCvAtBV8h OSRfRZ9OeBb6rLbIpySH9H+OfEfP3Yz1xqKvJ3w58ukm5FOSM8PQnpLfxtD+ Em+AfNoI+ZTk1DDcX6l/QfiDhfAHKeeeYR8q5VyGv3VGvpNy4J+euGcgOaWR T5shn5Ky most bizarre moments in my life.Historia Ojca Anatolija rozpoczyna si w czasie II wojny wiatowej. Aby ratowa wasne ycie zmuszony jest zabi swojego przeoonego. Marynarza morderc ratuj mnisi z klasztoru pooonego na tytuowej wyspraciones de la alcaldesa de Santa Coloma, Nuria Parln, a la que aprecio y admiro. Confieso que en su da me llamaron la atencin sus advertencias respecto a las negociaciones de Pedro Snchez con otros lderes polticos espaoles. Y, ms recientemente, su ritornello sobre esa ambigedad poltica que hemos dado en llamar derecho a decidir. , . , , ( ) ( ). , . (, , 5 ). - ( ) . , , ( ), , , , , ( ) . ( ) . - , . , .-- --AfghanistanAland IslandsAlbaniaAlgeriaAmerican SamoaAndorraAngolaAnguillaAntarcticaAntigua and BarbudaArgentinaArmeniaArubaAustraliaAustriaAzerbaijanBahamasBahrainBangladeshBarbadosBelarusBelgiumBelizeBeninBermudaBhutanBoliviaBonaire, Saint Eustatius and StyleBox["\<\"\\!\\(\\*RowBox[{\\\"Simplify\\\", \\\"[\\\", \ StyleBox[\\\"expr\\\", \\\"TI\\\"], \\\"]\\\"}]\\) performs a sequence of \ algebraic and other transformations on \\!\\(\\*StyleBox[\\\"expr\\\", \\\"TI\ \\\"]\\) and returns the simplest form it finds. \ \\n\\!\\(\\*RowBox[{\\\"Simplify\\\", \\\"[\\\", \ RowBox[{StyleBox[\\\"expr\\\", \\\"TI\\\"], \\\",\\\", \ StyleBox[\\\"assum\\\", \\\"TI\\\"]}], \\\"]\\\"}]\\) does simplification \ using assumptions. \"\>", "InformationUsageText", StripOnInput->ing at each other just doesn't look right. And because history repeats itself, Ancient Gaming Noob Wilhelm Arcturus has the same problem with Star Wars: The Old Republic: Needing 3 to 6 blaster hits at point blank to kill somebody is just weird.Come to think of it, it is of course equally weird that in just about any MMORPG, using just about any weapon or spell, you *always* need between 3 to 6 hits to kill a mob of your level. This has to do with how long designers think a combat should be, and with the traditional use of random numbers in role-playing combat: A combat decided by a single "roll" of the dice would be very unpredictable. Rolling several dice produces a Gaussian normal distribution, where average results are common, and extreme results are rare.Other than fantasy role-playing games, there are relatively few video games having people hack at each other with swords. And then most other sword-fighting games also use combat based on several hits to achieve a kill, although I'm pretty certain that in real life you wouldn't survive btal reserve at the end would be (n - n) * 50c = 0. So, this seems like a classic Dyck path problem, where we have a sequence of steps, +1 for 50c customers and -1 for $1 customers, and the path never goes below zero. The number of such sequences is the Catalan number C_n. But wait, the Catalan number counts the number of Dyck paths of length 2n, which is equivalent to the number of valid sequences where the number of 50c customers is alway state <- getState setState $ state { cgs_stmts = cgs_stmts state `appOL` stmts } -- emit CgStmts outside the current instruction stream, and return a label forkCgStmts :: CgStmts -> FCode BlockId forkCgStmts stmts = do id <- newLabelC emitCgStmt (CgFork id stmts) return id -- turn CgStmts into [CmmBasicBlock], for making a new proc. cgStmtsToBlocks :: CgStmts -> FCode [CmmBasicBlock] cgStmtsToBlocks stmts = do id <- newLabelC return (flattenCgStmts id stmts) -- collect the code emitted by an FCode computation getCgStmts' :: FCode a -> FCode (a, CgStmts) getCgStmts' fcode = do state1 <- getState (a, state2) <- withState fcode (state1 { cgs_stmts = nilOL }) setState $ state2 { cgs_stmts = cgs_stmts state1 } return (a, cgs_stmts state2) getCgStmts :: FCode a -> FCode CgStmts getCgStmts fcode = do (_,stmts) <- getCgStmts' fcode return stmts -- Simple ways to construct CgStmts: noCgStmts :: CgStmts noCgStmts = nilOL oneCgStmt :: CmmStmt -> CgStmts oneCgStmt stmt = unitOL (CgStmt stmt) consCgStmt :: CmmStmt -> CgStmts -> CgStmts consCgStmt stmt stmts = CgStmt stmt `consOL` stmts -- ---------------------------------------------------------------------------- -- Get the current module name getModuleName :: FCode Module getModuleName = do info <- getInfoDown return (cgd_mod info) -- ---------------------------------------------------------------------------- -- Get/set the end-of-block info setEndOfBlockInfo :: EndOfBlockInfo -ndle edge cases. If the length of p isolving the secretive counterterrorism units. But he would rely increasingly upon their capacities in the coming years.Soon after Mr. Obama took office he reframed the fight against terrorism. Liberals wanted to cast anti-terrorism efforts in terms of global law enforcement rather than war. The president didnt choose this path and instead declared war against Al Qaeda and its allies. In switching rhetorical gears, Mr. Obama abandoned Mr. Bushs vague and open-ended fight against terrorism in favor of a war with particular, violent jihadists.The rhetorical shift had dramatic non-rhetorical consequences. Compare Mr. Obamas use of drone strikes with that of his predecessor. During the Bush administration, there was an American drone attack in Pakistan every 43 days; during the first two years of the Obama administration, there was a drone strike there evetsound is a short (but beautiful) 8-mile drive away.The seventy returned with joy, saying, Lord, even THE DEMONS ARE SUBJECT TO US IN YOUR NAME. And He said to them, I watched Satan fall from heaven like lightning. BEHOLD, I HAVE GIVEN YOU AUTHORITY TO TREAD ON SERPENTS AND SCORPIONS, and OVER ALL THE POWER OF THE ENEMY, and NOTHING SHALL BY ANY MEANS HURT YOU. Nevertheless do not rejoice in this, that the spirw His will for them, and hope to guide them in ways that soften their heart toward Him. As I now stand eye-to-eye with the teenagers given me by God (and still often feel inexperienced), one thing has changed. Now, I dont mind if you call me one of those sheltering moms. Im honored by the phrase.what is supply asme sa588 gr k weather corten plate. ASME SA588 Grade K Corten steel, SA588 Gr.K steel plate/Sheet. SA588 Grade K low alloy high strength atmospheric corrosion resisting steel.The applicable equivalent ASTM specifications for weathering steel are what is sa588 grade k anti weathering corrosion steel plate other steel mills, ANSON STEEL can supply best corten what is sa588 grade k anti weathering cort[0,1]. Point D divides BC in ratio BD/DC=1/sqrt(0.25 + h). Let m=1, n=sqrt(0.25 + h). Coordinates of D are: x = (n*1 + m*0.5)/(m +n) = (n +0.5)/ (1 +n) y = (n*0 + m*h)/(1 +n) = mh/(1 +n) But this seems complex. Maybe assign a specific h. Let's take h=1 for simplicity. Then AC= sqrt(0.25 +1)= sqrt(1.25)= (sqrt(5))/21.118 BD/DC=1/sqrt(1.25)=2/sqrt(5)0.894 Coordinates of D: x=(sqrt(1.25)*1 +1*0.5)/(1 +sqrt(1.25))(1.118 +0.5problem statement, or maybe there's a different interpretation. Wait, maybe "if n is odd" refers to the function's output, but no, the function is defined based on the parity of the input. So if you input an odd number, add 5; input even, subtract 2. So my reasoning holds. Alternatively, maybe the problem is in another language and "if n is odd" was mistranslated? Unlikely. Alternatively, maybe the function is applied in a different order? Wait, f(f(f(k))) is f applied three times: first f(k), then f of that, then f of that. So the order is correct. Wait, maybe I need to consider that during the function applications, the parity could change in a different way. Wait, let's suppose that k is such that after two applications, we get an odd number again. But according to the function definitions, starting from even, subtracting 2 keeps it even. So once you get to even, subtracting 2 will keep it even. So the third application is still even. So f(f(f(k))) is even. Therefore, the problem's result is impossible. So this suggests that there is no solution. But the problem says "Find k," so there must be a solution. Alternatively, perhaps the prob by connected vertices, the sum is 360, I'm inclined to think that the answer is D) 360 des liwat pertama. Hingga ke hari ini janji tinggal janji, cakap je berkobar kobar, taik hidung jelah.3) 16th September - "we are confident we have more than enough" lalu kantoi. inilah mulut bau najis vavi.4) Di suruh bersumpah tidak melakukan liwat terhadap Saiful = alasan bacul Otak Zionist, ' Sumpah ini tidak Islamik, saya sudah telpon Yusuf Qardhawi dan dia melarang saya dari bersumpah" Ayat2 power dari pembohong bersiri.5)Kes Liwat Saiful - bila keluar dari tempat persembunyian di KEdutaan Turki " Saya ada banyak bukti dan alibi yg akan membersihkan nam baik saya dan saya akan buktikan di mahkamah nanti" Sela6 -1=95. The number of multiples of 2021 is floor(193,999/2021)=95 -0=95. Therefore, A=95, B=95. A=B, which does not satisfy A > B. Therefore, the maximum interval length is 189,999. However, earlier examples with smaller k (like k=2) gave lengths of 3999, which is much smaller than 189,999. Therefore, the maximum possible value of n - m is 189,999. The maximum possible value of \( n - m \) is achieved by carefully selecting an interval where the number of multiples of 2021 exceeds those of 2000 by exactly one. Through analyzing the problem using Diophantine equations and modular arithmetic, we determine that the optimal interval length occurs when the interval starts just after a multiple of 2000 and ends just before the nexle. But perhaps the examples are typos. So the code will be: import requests from typing import Tuple def send_request(url: str, retries: int) -> Tuple[int, str]: for attempt in range(retries + 1): try: response = requests.get(url, timeout=5) return (200, 'Retrieved successfully') except requests.exceptions.Timeout: if attempt < retries: continue else: return (-1, 'Max retries exceeded') except requests.exceptions.ConnectionError: return (0, 'Connection error') except requests.exceptions.RequestException as e: return (0, str(e)) except Exception as e: return (0, str(e)) return (-1, 'Max retries exceeded') But according to this code, a Timeout will only return (-1, ...) if all retries are exhausted. So send_request('url', 3) with all attempts timing out would return (-1, ...). So the second example in the problem's sample is not possible. Hence, perhaps the problem's example is incorrect. But according to the problem's note, the function should return (0---------------------------------------------------- -- IdDetails and IdInfo ----------------------------------------------------- \begin{code} ppIdInfo :: Id -> IdInfo -> SDoc ppIdInfo id info = sdocWithDynFlags $ \dflags -> if gopt Opt_SuppressIdInfo dflags then empty else showAttributes [ (True, pp_scope <> ppr (idDetails id)) , (has_arity, ptext (sLit "Arity=") <> int arity) , (has_called_arity, ptext (sLit "CallArity=") <> int called_arity) , (has_caf_info, ptext (sLit "Caf=") <> ppr caf_info) , (True, ptext (sLit "Str=") <> pprStrictness str_info) , (has_unf, ptext (sLit "Unf=") <> ppr unf_info) , (not (null rules), ptext (sLit "RULES:") <+> vcat (map pprRule rules)) ] -- Inline pragma, occ, demand, one-shot info -- printed out with all binders oving up 8 Barnet 41 8 64 Moving down 9 Nuneaton 42 -4 62 No movement 10 Kidderminster 40 1 60 Moving up 11 Woking 40 -2 60 Moving up 12 Salisbury 41 -8 60 Moving down 13 Forest Green 40 15 58 No movement 14 Welling 41 4 57 Moving up 15 Lincoln City 42 -2 55 Moving down 16 Macclesfield 40 1 54 No movement 17 Wrexham 42 -2 54 Moving up 18 Chester 42 -21 46 Moving down 19 Southport 41 -20 45 No movement 20 Hereford 41 -20 42 No movement 21 Aldershot 41 3 41 No movement 22 Dartford 41 -21 41 No movement 23 Tamworth 41 -34 33 No movement 24 Hyde United 41 -70 10 How to get into sports and activities near you, plus more about our campaignSleep has long been the bastion for peace of many after a long hard day, but sadly it isnt as easy to find for some. This line of thinking brought Dr. Avi Weisfogel straight into focus in the late 90s after establishing Old Bridge Dental Care. Weisfogel may have a career in dental medicine but he has always been interested in sleep medicine and tha duminic, epifanie, Evident tot Rolemodelism, existen, facebook, facultate, familie, fiu, flutura de salariu, foaie de hrtie, format fizic, fric, frustrare, frustrat, gnditor, Gigi Becali, guest-entry, idee, idei, imaginaie, imprimant, individ, iubit, iubit, jumtate, kilogram, lene, Liv, lucru, luni, main, majoritate, mam, mito, mito-uri, mijloace, minge de plaj, minte, mizerie, moment, momente, motive, nivel, numr, obiectiv, obiective, pai, parte, partener, pin-board, plan, planuri, poze, poz, preedinte, privin, priz, probleme, problem, profil, proprieti, provocare, punct, reuite, rezultat, rutin, sal, sarcasm, semnale de alarm, situaie, Societate, soluii, specimene, student, subiectivism, sum, sptmni, tehnologie, televiziune, Teo, termen, timp, tip, ton, tradiii, treab, trecut, tricou, trimitere, tupeu, variabile, vlv, vedere, verdicte, via, vindecare, Vladuts, YOLO | Las un comentariuFrom Steubenville to India, videos and tweets are being turned against perpetrators of sexual violence A vid regardless of the word puzzles. So for example, a set could have 2 logic,1 visual, and 3 word puzzles. Ratio of logic:visual=2:1, total puzzles=6. This meets the minimum of 5. Then, number of sets: min(30/2=15,18/1=18,12/3=4). So 4 sets. Uses 8 logic,4 visual,12 word. Leftover:22 logic,14 visual,0 word. Doesn't work. Alternatively, each set:2 logic,1 visual,1 word. Total=4. Less than 5. Invalid. Each set:2 logic,1 visual,2 word. Total=5. Number of sets: min(30/2=15,18/1=18,12/algum eu condeno quem faz cesrea, at pq isso uma escolha de cada um. o engraado que, basta eu dizer que quero um parto natural, sem anestesia que pronto: vc automaticamente vira a radical do parto. Para mim, radical marcar uma interveno sem motivo aparente e achar isso normal. Parir algo fisiolgico, de onde as pessoas tiram essa ideia de que radical lutar por aquilo Alguns comentrios sobre esse meu post me deixaram cada vez mais convencida que as pessoas se incomodam DEMAIS com a opinio alheia e se sentem ofendidas por algo contrrio a elas. Cesariana no faz de algum menos me, muito pelo contrrio, como eu disse, uma forma de salvar vidas, minha crtica exatamente sobre o q vc disse: sobre a banalizao do que deveria ser natural e isso uma cultura de brasileiro (novamente, vide a Kate).[ SEO SERVICE ] 50 Directories Submission Manually, Only Live Links In Report $14.99End Date: Sunday Jul-23-2017 23:59:55 PDTBuy It Now for only: $14.99Buy It Now | Add to watch list [ SEO SERVICE ] All In One Exclusive Seo Package Rank Better In Google $14.99End Date: Sunday Jul-23-2017 20:16:30 PDTBuy It Now for only: $14.99Buy It Now | Add to watch list BEST SEO BACKLINK PYRAMID * EFFECTIVE LINKS + EDU/GOV Links - GOOGLE SAFE * $19.94 (0 Bids)End Date: Friday Jun-30-2017 20:11:28 PDTBuy It Now for only: $25.94Buy It Now | Bid now | Add to watch list Disavow and Remove 500 Spamming Backlinks - SEO Repair Service $15.95End Date: Sunday Jul-23-2017 12:30:30 PDTBuy It Now for only: $15.95Buy It Now | Add to watch list Remove 250 Spamming Backlinks - plosive materials. - A bullet list of at least 2 challenges faced in the development of these safer alternatives. - Choose one from the following options to describe the potential impact: [environmental benefits, economic implications, safety improvements]. - Please provide your response in German. Okay, let's tackle this query. The user is a chemical engineer looking for an overview of current research on safer explosive materials. They want bullet points on methods, challenges, and a chosen impact area. First, I need to make sure I cover all parts of their request. Starting with the innovative methods. I remember reading about using less sensitive materials, like replacing traditional explosives with something like DAAF. Then there's encapsulation, where you coat the explosive to make it stable until needed. Also, green chemistry approachcipal amount (P) of Rs 600, the amount after 2 years (T) is Rs 720. We need to find the rate of interest (R) and then determine the additional time required for the amount to reach Rs 1020. First, we calculate the rate of interest using the amount after 2 years: \[ A = P + \frac{P \times R \times T}{100} \] Substituting the given values: \[ 720 = 600 + \frac{600 \times R \times 2}{100} \] \[ 720 - 600 = \frac{1200R}{100} \] \[ 120 = 12R \] \[ R = 10\% \] Next, we determine the total time (T) required for the principal to amount to Rs 1020: \[ 1020 = 600 + \frac{600 \times 10 \times T}{100} \] \[ 1020 - 600 = \frac{6000 \times T}{100} \] \[ 420 = 60T \] \[ T = 7 \text{ years} \] Since the amount reaches Rs 720 after 2 years, the additional time required is: \[ T - T = 7 - 2 = 5 \text{ years} \] Thus, the sum amounts to Rs 1020 after an additional \boxed{5} years.Texto anterior: En el contexto crtico de su batalla final contra los hroes, All For One, al borde de la inconsciencia y la derrota, desat un ltimo y desesperado ataque. Mediante una onda expansiva de poder inaudito, que de algn mtBox["y", "32"]}], "+", RowBox[{"4", " ", "x1", " ", "x2", " ", SuperscriptBox["y", "32"]}], "-", RowBox[{"4", " ", SuperscriptBox["x1", "3"], " ", "x2", " ", SuperscriptBox["y", "32"]}], "-", RowBox[{"4", " ", SuperscriptBox["x1", "5"], " ", "x2", " ", SuperscriptBox["y", "32"]}], "+", RowBox[{"4", " ", SuperscriptBox["x1", "7"], " ", "x2", " ", SuperscriptBox["y", "32"]}], "-", RowBox[{"3", " ", SuperscriptBox["x2", "2"], " ", SuperscriptBox["y", "32"]}], "-", RowBox[{"8", " ", SuperscriptBox["x1", "2"], " ", SuperscriptBox["x2", "2"], " ", SuperscriptBox["y", "32"]}], "-", RowBox[{"22", " ", SuperscriptBox["x1", "4"], " ", SuperscriptBox["x2", "2"], " ", SuperscriptBox["y", "32"]}], "-", RowBox[{"8", " ", SuperscriptBox["x1", "6"], " ", SuperscriptBox["x2", "2"], " ", SuperscriptBox["y", "32"]}], "-", RowBox[{"3", " ", SuperscriptBox["x1", "8"], " ", SuperscriptBox["x2", "2"], " ", SuperscriptBox["y", "32"]}], "-", RowBox[{"4", " ", "x1", " ", SuperscriptBox["x2", "3"], " ", SuperscriptBox["y", "32"]}], "-", RowBox[{"36", " ", SuperscriptBox["x1", "3"], " ", a,b,c,d], length 4. j can be up to 2 (since j+1 is 3, but right starts at j+1=3 which is the last element). So the sum of right is from index 3 to 3. So j can be 2. So for j in i+1 <= j <= n-2. So j can be from i+1 to n-2. So j+1 ranges from i+2 to n-1. So pre_sum[j+1] is from pre_sum[i+2] to pre_sum[n-1]. Thus, the valid pre_sum[j+1] must be >= 2L and <= (total + L)/2. But also, pre_sum[j+1] must be >= pre_sum[i+2] (since j starts at i+1) and <= pre_sum[n-1]. So combining these, the lower bound for pre_sum[j+1] is the maximum between 2L and pre_sum[i+2], and the upper bound is the minimum between (total + L)/2 and pre_sum[n-1]. Additionally, we need to exclude j where pre_sum[j+1] = total - L. So for each i, the number of valid j's is: number of j's where pre_sum[j+1] is in [lower, upper], minus the number of j's where pre_sum[j+1] == total - L and also in [lower, upper]. So the problem reduces to, for each i, compute: count = upper_bound_idx - lower_bound_idx + 1 (if upper >= lower) else 0. Then subtract 1 if total - L is in [lower, upper], and exists in the pre_sum array at positions >=i+2 and <=n-1. Wait, but pre_sum is a non-decreasing array since all elements are non-negative. Because each element in nums is >=0, the prefix sums are non-decreasing. So pre_sum is a sorted array. That's a crucial point. Because the prefix sum is sorted, we can perform binary search for the lower and upper bounds. So for each i, we can compute the possible j's as follows: 1. Calculate L = pre_sum[i+1] 2. Compute the lower limit for pre_sum[j+1] as 2*L. 3. Compute the upper limit for pre_sum[j+1] as floor( (total + L) / 2 ) 4Plans ps) p -- | Typs 111...111 with 1998 digits. The square root of this number is very close to 333...333.333...333..., but slightly less. The difference between n and (333...333.333...333)^2 is significant, but we need to see how the square root is adjusted. Alternatively, let's consider that if we let x = n, then x 333...333.333...333 (with 999 3s before the decimal and 3s after). Let's compute x^2: x 10^{999}/3 - 10^{-999}/6 So x^2 (10^{999}/3)^2 - 2*(10^{999}/3)*(10^{-999}/6) + (10^{-999}/6)^2 10^{1998}/9 - (2 * 10^{0}/18) + negligible term 10^{1998}/9 - 1/9 But 10^{1998}/9 - 1/9 = (10^{1998} - 1)/9 = n Therefore, this approximation is exact, because x^2 = n. Therefore, the approximation we used is actually an equality? Wait, how? Wait, let's computere internally tangent. Compute the circumcircle of e gallons by 1.1. Wait, is that another interpretation? Let me re-examine the problem statement. "the vehicle's fuel consumption rate increases by 10% when it is carrying extra weight." Hmm. Fuel consumption rate is the rate at which fuel is consumed, so gallons per mile or gallons per hour. But in the context of mpg, which is miles per gallon, fuel consumption rate is often expressed as gallons per mile. But actually, fuel consumption rate could be ambiguous. Let me think. If fuel consumption rate increases by 10%, that could mean that the amount of fuel used per mile increases by 10%, which would be equivalent to dividing the mpg by 1.1. Alternatively, if fuel consumption rate is in gallons per hour, but since we are dealing with miles and gallons, more likely gallons per mile. Wait, miles per gallon is a measure of fuel efficiency. Fuel consumption rate (volume per distance) is gallons per mile, which is the reciprocal of mpg. So if fuel consumption rate increases by 10%, that means gallons per mile becomes 1.1 times the original. Therefore, mpg becomes original /1.1. Which is what I did earlier. So my initial approach was correct. Therefore, the calculation is accurate. So the total gasoline required is approximately 20.92 gallons. R /imagine prompt: beautiful women in the coffee shop, elegant and sophisticated, sipping a cup of steaming coffee, natural sunlight streaming through the window, soft and warm color tones, vintage decor with cozy armchairs and wooden tables, a bookshbrate the intellectual life of the University. This Faculty History Website is intended as a component of the effort to document the extraordinary academic achievements of Michigans faculty in building and sustaining one of the worlds great universities. It provides access to a comprehensive database of information concerning the t] Thus, the logarithm of the limit becomes: \[ \lim_{n \to \infty} 5n \left(-\frac{2}{10n - 1}\right) = \lim_{n \to \infty} -\frac{10n}{10n - 1} \] Simplifying the exponent: \[ -\frac{10n}{10n - 1} = -\frac{10}{10 - \frac{1}{n}} \to -1 \text{ as } n \to \infty \] Exponentiating the result gives: \[ e^{-1} = \frac{1}{e} \] Thus, the limit of the sequence is \(\boxed{\dfrac{1}{e}}\).In a triangle with sides 6 cm, 10 cm, and 12 cm, an inscribed circle is tangent to the two longer sides. Find the perimeter of the resulting triangle formed by the tangent line and the two longer sides. Okay, so I need to find the perimeter of the triangle formed by the tangent line of the inscribed circle and the two longer sides of the original triangle. The original triangle has sides 6 cm, 10 cm, and 12 cm. First, let me make sure I understand the problem correctly. The triangle has sides of lengths 6, 10, and 12 cm. The inscribed circle (incircle) is tangent to all three sides, but the problem mentions it's tangent to the two longer sides. Wait, but in a triangle, the incircle is tangent to all three sides. Maybe the problem is referring to a tangent line that's parallel to the shortest side? Hmm, maybe not. Let me read the problem again. "An inscribed circle is tangent to the two longer sides. Find the perimeter of the resulting triangle formed by the tangent line and the two longer sides." Wait, so the incircle is tangent to all three sides, but the problem is talking about a tangent line that is tangent to the two longer sides. Maybe this tangent line is another tangent to the incircle, not one of the sides of the original triangle. Then, the resulting triangle is formed by this tangent line and the two longer sides. So, the original triangle has sides 6, 10, 12. The two longer sides are 10 and 12. So the incircle touches these two sides, and then there's another tangent line to the incircle that, together with the two longer sides (10 and 12), forms a new triangle. We need to find the perimeter of that new triangle. Wait, maybe a diagram would help, but since I can't draw, I need to visualize. Let me think. The original triangle has sides 6, 10, 12. Let me confirm which sides are which. The two longer sides are 12 and 10, so the shortest side is 6 cm. So, perhaps the original triangle is a qs + rs,{-"\qquad\text{with}\quad"-} degree rs < degree bs \end{spec} \end{quote} % The condition |r < b| is replaced by |degree rs < degree bs|. % However, we now have a problem. % Every polynomial is divisible by any non-zero constant polynomial, resulting in a zero polynomial remainder. % But the degree of a constant polynomial is zero. % If the degree of the zero polynomial were a natural number, it would have to be smaller than zero. % For this reason, it is either considered undefined (as in \cite{adams2010calculus}), or it is defined as |-|.% \footnote{Likewise we could define the largest element of the empty set to be |-|.} % The next section examines this question from , - " ", . , - . , .... - , . ... , , , , ..., , , , , , - ... , , , , , . "" . ... .With reference to FIG. 14, the second end 518 b of the locking arm 518 cooperates with the lock 508 to lock the connector 500, and thus, the set connector assembly 502 to the housing 220. In one example, the second end 518 b includes a locking tab 524 and the graspable surface 436. The locking tab 524 extends from the second end 518 b to be received with in the pocket 412 defined byd Z is at (0,1,0). But then these points are all on the bottom face of the cube, forming nction :> Identity, Frame -> {{True, True}, {True, True}}, FrameLabel -> {{None, None}, {None, None}}, FrameStyle -> Directive[ Opacity[0.5], Thickness[Tiny], RGBColor[0.368417, 0ples. Because this is a frequently used function, we code it specially (rather than invoking a more generic execCommand). > nqExec :: DBHandle -> String -> IO (String, String, Oid) > nqExec db sqlText = > withUTF8String sqlText $ \cstr -> > do > stmt <- fPQexecParams db cstr 0 nullPtr nullPtr nullPtr nullPtr textResultSet > >>= check'stmt db ePGRES_COMMAND_OK > -- save all information from PGresult and free it > cmd'status <- fPQcmdStatus stmt >>= peekUTF8String > cmd'ntuples <- fPQcmdTuples stmt >>= peekUTF8String > cmd'oid <- fPQoidValue stmt > fPQclear stmt > return (cmd'status, cmd'ntuples, cmd'oid) > execCommand :: DBHandle -> String -> [PGBindVal] -> IO (String, String, Oid) > execCommand db sqlText bindvals = do > stmtPrepare db "" sqlText (bindTypes bindvals) > execPreparedCommand db "" bindvals | This is for commands, as opposed to queries. The query equivalent of 'execPreparedCommand' is 'stmtExec'. > execPreparedCommand :: DBHandle -> String -> [PGBindVal] -> IO (String, String, Oid) > execPreparedCommand db stmtname bindvals = do > (rs, ntuples) <- execPrepared db stmtname bindvals ePGRES_COMMAND_OK > -- save all information from PGresult and free it > cmd'status <- fPQcmdStatus rs >>= peekUTF8String > cmd'ntuples <- fPQcmdTuples rs >>= peekUTF8String > cmd'oid <- fPQoidValue rs > stmtFinalise rs > return (cmd'status, cmd'ntuples, cmd'oid) Prepare and Execute a query. Returns results as binary. > stmtExecImm :: DBHandle -> String -> [PGBindVal] -> IO (ResultSetHandle, Int) > stmtExecImm db sqlText bindvals = do > let np = fromIntegral $ length bindvals > withUTF8String sqlText $ \cstr -> do > withArray (map bindValOid bindvals) $ \coidarray -> do > withArray (map bindValSize bindvals) $ \clenarray -> do > to length-2, k in j+1 to length-1. Wait, no. Wait, the indices can be from 0 up to m-3, m-2, m-1. For example, for a list of length m, i can range from 0 to m-3, j from i+1 to m-2, and k from j+1 to m-1. But that's not correct. Because for a list of length 4 (indices 0-3), the triplet (0,1,2) is allowed, but (0,1,3) is also allowed, (0,2,3), (1,2,3). So all possible combinations where i < j < k. So, the correct way is: For all i in 0 to m-3: for j in i+1 to m-2: for k in j+1 to m-1: add the triplet (websites[i], websites[j], websites[k]) But that's not efficient. Because for each i, j, k, we need to check all possible combinations. But for a list of length m, the number of triplets is C(m,3) which is m*(m-1)*(m-2)/6. So for m=50, that's 50*49*48/6 = 19600, which is manageable. But using three nested loops for each user's website list is feasible. Alternatively, in F#, we can generate all combinations of three indices using combinations, but since order matters (i < j < k), we can generate all 3-element combinations and sort them, but since the indices must be in order, perhaps using a list comprehension or Seq functions. Wait, perhaps using a list comprehension: For a list of websites ws of length m: let triplets = [ for i in 0 .. m-3 do for j in i+1 .. m-2 do for k in j+1 .. m-1 do (ws.[i], ws.[j], ws.[k]) ] But this is O(m^3), which for m=50 is 50*49*48=117600 loops. For 50 users, that's 5.8 million loops, which is manageable. But in F#, nested loops may be slow if m is large, but given the constraints (m up to 50), it's acceptable. Alternatively, using a more efficient approach. For example, generate all possible triplets by iterating i from 0 to m-3, then j from i+1 to m-2, then k from j+1 to m-1. For each such i, j, k, take the triplet. Then, collect all these triplets. Once all triplets are generated for a user, we need to deduplicate them. So, for the user's list of triplets, we can create a set (using a HashSet or a F# Set) to store unique triplets. Then, for each triplet in the set, we increment the global count. So, in code: For each user: let sortedWebsites = ... list of websites sorted by timestamp let m = sortedWebsites.Length if m < 3 then continue (can't contribute) let triplets = seq {tence>Michelle Le: A nursing student at San Diego State University, Le disappeared May 27 after walking out of Kaiser Permanente Hayward Medical Center in northern California toward her car. Her body was found in September by a search dog taken to the Bay area by Carrie McGonigle, the mother of slain teen Amber Dubois of Escondido. Giselle Esteban, a former friend of Le, was arrested and charged with Les murder. Indicted by a grand jury, Esteban will face arraignment and enter a plea on January 20. http://www.eastcountymagazine.org/node/7299Voc acorda atrasado, toma um caf da manh apressado e parte rumo ao trabalho ou escola. O tempo passa devagar, mas voc respira fundo porque logo mais j a hora do almoo (ufa!). s mais um dia comum na vida de grande parte do mundo. Mas esta uma rotina impossvel para cerca de 870 milhes de pessoas que, segundo a Organizao das Naes Unidas para a Alimentao e Agricultura, sofriam de desnutrio crnica entre 2010 e 2012. Neste perodo, 1 em cada 8 pessoas do planeta foi afetada pela falta ou escassez de alimento. No pouca coisa.Review by: Black Gloves Release Date: 2014 Studio: Entertainment One Genre: Drama Format: DVD Region: 2 PAL Aspect Ratio: 2.35:1 Directed by: John Michael McDonagh Cast: Brendan Gleeson Chris O'Dowd Kelly Reilly Aidan Gillen Dylan Moran Movie: 5 Extras: 2 Bottom Line: 4 Video: As Trevor Johnsons feature article on Calvary for the May issue of Sight & Sound so succinctly phrased it, John Michael McDonaghs second feature is not so much a whodunnit as a whosgonnadoit: Father James Lavelle (Brendan Gleeson) is that apparently rare thing in this day and age an Irish Catholic priest whos actually a pretty decent guy; a man of God who, rather than coming to his role straight from the seminary and without a clue as to how the majority of people in the community hes meant to serve actually live their lives, has in fact accumulated a great deal of experience of the real world before acceptinos: Cerrado y sellado -El salto a la Sala de los Ecos: Sin consecuencias -Sala de los Ecos Misiones secundarias -Alerta de mineral rojo -Ladrones de tumbas -Devolver la visin a un ciego -En pie -Abrir la compuerta -Una piedra en el camino -Entre un trol y un lugar duro -Las runas de la ira -Otra bsqueda obligada -El rescate de un chamn -En marcha otra vez -La muerte de un campen -Mucho ruido y poco goblins -Descendiente divino -El horror de Aldea Alta -Una hermandad sin jefe -Hasta los huesos Lectura de mentes -Lectura de mentes Extras -Conseguir el mximo de experiencia -Secretos de Albor Remoto -Bellegar -Conejo asesino -Casper y su gallina -Sosostra -Los 4 pergaminos y los tele-transportadores -La estatua de ZixZax y los cristales de dragn -Almacn del Doctor Jeringuilla -Torre de batalla -El cofre de la Cripta de Orobas -Cueva misteriosa Sets de armaduras (Ego Draconis) -Archimago de Aleroth -Armadura completa de Ulthring -Cazador -Escorpin Sets de armaduras (DKS) -Nivel Sangre -Habitantes de los bosques -Guardia de Rivellon -Defensores de Aleroth Expansin -Flames of Vengeance Otros -Logros -Logros (Dragon Knight Saga) Informacin del juegoPlataforma: PC, Xbox 360Desarrollador: Larian StudiosDistribuidor: Koch MediaFecha: 30 Octubre 2009Foro (1 + x)^n 1 + nx. If n is a negative integer, then (1 + x)^n is 1/(1 + x)^{|n|}, and 1 + nx would be 1 + (-|n|)x. But x is a positive real number, so 1 - |n|x could be negative if |n|x > 1. For example, if n = -1 and x = 2, then (1 + 2)^{-1} = 1/3, and 1 + (-1)(2) = -1. So 1/3 -1, which is true. But if x is very large, say n = -2 and x = 1, then (1 + 1)^{-2} = 1/4, and 1 + (-2)(1) = -1. So 1/4 -1, still true. Wait, maybe even for negative n, the inequality holds because the right-hand side can be negative, and the left-hand side is always positive since (1 + x) > 0 and any real power of a positive number is positive. So in cases where n is negative,io frontale di sicurezza. Sul fondo della parte interna sono presenti delle alette in plastica anti scivolo. Il box completamente impermeabile. Box baule e barre portatutto omologati TUV-GS. Per informazioni sul prodotto contattateciAs an author you ought to have control of the publication which you brought into existence. That is the reason youve the option to print it in hard or soft cover formats. You even have the possibility to do some of each type of print formats and methods. You as the writer have all the say in regards to your own novel. In case youd like to determine a printed version of your publication, you should have that as well, although eBooks make sales simpler since manla isn't applicable directly. So, perhaps I need to reorder the points to form a non-intersecting quadrilateral. Wait, the problem says "the quadrilateral with vertices at...", but it doesn't specify the order. But in geoboards, when you connect the points in the order given, but maybe the order is supposed to be around the perimeter? Hmm, the problem might assume that the quadrilateral is convex, or that the points are given in order. But since connecting them in the given order causes a crossing, that can't be right. Maybe I need to arrange the points in the correct cyclic order. Alternatively, maybe the problem is intended to use the shoelace formula regardless, but given the answer options, 9 is there. But wait, if the lines cross, the area calculated by shoelace formula would actually be the difference between areas of the two triangles formed by the intersecting lines. So perhapus decision (30-27, 30-27, 30-26).All Bags Booking Creative After Effects Scripts eCommerce .NET 3D 3D Renders 3D, Object Abstract Accordions Actions Ad Templates Add-ons After Effects Presets Ambient Android Animals Animated SVGs Backgrounds Badges & Stickers Banners & Ads Blog / Magazine Books Specialty Boutique New, Used and Rental Textbooks Business and Finance Education Humanities Visual Arts Photography Law Business Subjects Arts and Photography Fashion Graphic Design Commercial Advertising Music Albums Musical Genres Rock Singles Photography and Video Lifestyle and Events Lifestyle Biographies and Memoirs Leaders and Notable People Political Business and Investing Economics Commerce Business and Money Industries Sports and Entertainment Entertainment Marketing and Sales Marketing Industrial Processes and Infrastructure E-Commerce Real Estate Skills Communications Small Business and Entrepreneurship New Business Enterprises Taxation Personal Calendars Computers and Technology Business Technology Web Marketing Internet and Social Media Search Engine Optimization Web Browsers Databases Networking and Cloud Computing Software Web Development and Design Programming JavaScript Web Design Education and Teaching Schools and Teaching Instruction Methods Humor and Entertainment Pop Culture Art Magazines Newsletters Reference Self-Help Motivational Technology Travel Books and Magazines Brochures BuddyPress Buildings Business Cards Business, Corporate Buttons Cards & Invites Cartoons Characters Charity Chill ChristmasrlemmermeerHalderbergeHarderwijkHarlingenHattemHeemskerkHeemstedeHeerdeHeerenveenHeerhugowaardHeerlenHeeze-LeendeHeilooHelmondHendrik-Ido-AmbachtHeumenHeusdenHillegomHilvarenbeekHilversumHogelandHollands KroonHoornHorst aan de MaasHoutenHuizenHulstIJsselsteinKaag en BraassemKapelleKatwijkKerkradeKoggenlandKrimpen aan den IJsselKrimpenerwaardLaarbeekLanderdLandgraafLandsmeerLangedijkLansingerlandLeeuwardenLeidenLeiderdorpLeidschendam-VoorburgLelystadLeudalLeusdenLingewaardLisseLochemLoon op ZandLopikLoppernverted the speeds to m/s first, then applied the percentage changes. But depending on interpretation, it's possible that the percentage changes are applied to their original speeds in km/hr, and then converted to m/s. Let's verify. If the problem says the speed decreases by 10% due to the incline, it's logical that the speed is decreased by 1end on the rest of us; and we're not doing so well.I was a victim of fraud. According to the announcement on the website molotok.ru I order Iphone 4g 32gb Sold seemed that he Britain (England), here's a site http:// www.completecomms.co.uk/ here is the address of Complete Communications (Essex) Ltd30b Woodford Avenue, Gants Hill, Ilford, Essex, IG2 6XG, United Kingdom. Sent money through Western Union, the seller has received the money but not sent the goods. Recalling that send one phone is expensive, and registered by the customs office is not profitable said that they have sentnd then fails. However, the problem statement claims it holds for any natural number n. This contradiction indicates that either there is a mistake s and two -1's such that the vectors cancel. For instance, choosing +1 for A and A, and -1 for A and A. Since A and A are opposite, their sum is zero, and similarly for A and A. So sum _i A_i = 0, and sum _i = 0. Therefore, such a combination exists. Then, the total sum would be 0 - 0*M = 0. Therefore, for the square, this works. Similarly, for any even n, if we can partition the vertices into pairs of opposites and assign +1 and -1 to each pair such that the total sum of vectors is zero and the total sum of coefficients is zero, then this would work. Since n is even, we can pair all the vertices into n/2 pairs of opposites. Then, for each pair, we can assign +1 to one and -1 to the other. 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So TP_total + FP_total is the total number of predictions. So micro-precision is (number of correct predictions) / total)^4 Cos[x]^12 + 2 (I Sin[x])^12 Cos[x]^4 + 1 (I Sin[x])^3 Cos[x]^13 + 1 (I Sin[x])^13 Cos[x]^3) + Exp[-11 I y] (34 (I Sin[x])^9 Cos[x]^7 + 34 (I Sin[x])^7 Cos[x]^9 + 27 (I Sin[x])^10 Cos[x]^6 + 27 (I Sin[x])^6 Cos[x]^10 + 22 (I Sin[x])^8 Cos[x]^8 + 18 (I Sin[x])^5 Cos[x]^11 + 18 (I Sin[x])^11 Cos[x]^5 + 10 (I Sin[x])^4 Cos[x]^12 + 10 (I Sin[x])^12 Cos[x]^4 + 4 (I Sin[x])^3 Cos[x]^13 + 4 (I Sin[x])^13 Cos[x]^3 + 1 (I Sin[x])^2 Cos[x]^14 + 1 (I Sin[x])^14 Cos[x]^2) + Exp[-9 I y] (140 (I Sin[x])^8 Cos[x]^8 + 113 (I Sin[x])^6 Cos[x]^10 + 113 (I Sin[x])^10 Cos[x]^6 + 131 (I Sin[x])^7 Cos[x]^9 + 131 (I Sin[x])^9 Cos[x]^7 + 76 (I Sin[x])^5 Cos[x]^11 + 76 (I Sin[x])^11 Cos[x]^5 + 42 (I Sin[x])^4 Cos[x]^12 + 42 (I Sin[x])^12 Cos[x]^4 + 16 (I Sin[x])^3 Cos[x]^13 + 16 (I Sin[x])^13 Cos[x]^3 + 6 (I Sin[x])^2 Cos[x]^14 + 6 (I Sin[x])^14 Cos[x]^2 + 1 (I Sin[x])^1 Cos[x]^15 + 1 (I Sin[x])^15 Cos[x]^1) + Exp[-7 I y] (496 (I Sin[x])^8 Cos[x]^8 + 333 (I Sin[x])^10 Cos[x]^6 + 333 (I Sin[x])^6 Cos[x]^10 + 419 (I Sin[x])^9 Cos[x]^7 + 419 (I Sin[x])^7 Cos[x]^9 + 216 (I Sin[x])^11 Cos[x]^5 + 216 (I Sin[x])^5 Cos[x]^11 + 105 (I Sin[x])^12 Cos[x]^4 + 105 (I Sin[x])^4 Cos[x]^12 + 36 (I Sin[x])^13 Cos[x]^3 + 36 (I Sin[x])^3 Cos[x]^13 + 7 (I Sin[x])^2 Cos[x]^14 + 7 (I Sin[x])^14 Cos[x]^2 + 1 (I Sin[x])^1 Cos[x]^15 + 1 (I Sin[x])^15 Cos[x]^1) + Exp[-5 I y] (947 (I Sin[x])^7 Cos[x]^9 + 947 (I Sin[x])^9 Cos[x]^7 + 452 (I Sin[x])^5 Cos[x]^11 + 452 (I Sin[x])^11 Cos[x]^5 + 707 (I Sin[x])^6 Cos[x]^10 + 707 (I Sin[x])^10 Cos[x]^6 + 235 (I Sin[x])^4 Cos[x]^12 + 235 (I Sin[x])^12 Cos[x]^4 + 106 (I Sin[x])^3 Cos[x]^13 + 106 (I Sin[x])^13 Cos[x]^3 + 1032 (I Sin[x])^8 Cos[x]^8 + 32 (I Sin[x])^2 Cos[x]^14 + 32 (I Sin[x])^14 Cos[x]^2 + 7 (I Sin[x])^1 Cos[x]^15 + 7 (I Sin[x])^15 Cos[x]^1 + 1 Cos[x]^16 + 1 (I Sin[x])^16) + Exp[-3 I y] (1769 (I Sin[x])^7 Cos[x]^9 + 1769 (I Sin[x])^9 Cos[x]^7 + 674 (I Sin[x])^11 Cos[x]^5 + 674 (I Sin[x])^5 Cos[x]^11 + 1902 (I Sin[x])^8 Cos[x]^8 + 274 (I Sin[x])^12 Cos[x]^4 + 274 (I Sin[x])^4 Cos[x]^12 + 1245 (I Sin[x])^6 Cos[x]^10 + 1245 (I Sin[x])^10 Cos[x]^6 + 76 (I Sin[x])^3 Cos[x]^13 + 76 (I Sin[x])^13 Cos[x]^3 + 15 (I Sin[x])^2 Cos[x]^14 + 15 (I Sin[x])^14 Cos[x]^2 + 1 (I Sin[x])^15 Cos[x]^1 + 1 (I Sin[x])^1 Cos[x]^15) + Exp[-1 I y] (1577 (I Sin[x])^6 Cos[x]^10 + 1577 (I Sin[x])^10 Cos[x]^6 + 484 (I Sin[x])^4 Cos[x]^12 + 484 (I Sin[x])^12 Cos[x]^4 + 2264 (I Sin[x])^8 Cos[x]^8 + 971 (I Sin[x])^5 Cos[x]^11 + 971 (I Sin[x])^11 Cos[x]^5 + 2056 (I Sin[x])^7 Cos[x]^9 + 2056 (I Sin[x])^9 Cos[x]^7 + 167 (I Sin[x])^3 Cos[x]^13 + 167 (I Sin[x])^13 Cos[x]^3 + 42 (I Singes of Mathematics'' \url{https://github.com/DSLsofMath/} \begin{itemize} \item A BSc-level course (2016-01 CeIo, 2017 onwards: PaJa, DaSc) \item A pedagogical project to develop the course (DaHe, SoEi) \item A BSc thesis project ``DSLsofMath for other courses'' \end{itemize} Aim: ``\ldots improve the mathematical education of computer scientists and the computer science education of mathematicians.'' Focus on types \& specifications, syntax \& semantics DSL examples: Power series, Differential equations, Linear Algebra \end{frame} \begin{frame}{DSLsofMath learning outcomes} \begin{itemize} \item Knowledge and understanding \begin{itemize} \item design and implement a DSL for a new domain \item organize areas of mathematics in DSL terms \item explain main concepts of analysis, algebra, and lin.\ alg. \end{itemize} \item Skills and abilities \begin{itemize} \item develop adequate notation for mathematical concepts \item perform calculational proofs \item use power series for solving differential equations \item use Laplace transforms for solving differential equations \end{itemize} \item Judgement and approach \begin{itemize} \item discuss and compare different software implementations of mathematical concepts \end{itemize} \end{itemize} {}\hspace{\fill} \tiny [\url{https://github.com/DSLsofMath/DSLsofMath/blob/master/Course2018.md}] \end{frame} \section{Types in Mathematics} \begin{frame}{Case 1: limits \citep{adams2010calculus}} \begin{quote} We say that \(f(x)\) \textbf{approaches the limit} \(L\) as \(x\) \textbf{approaches} \(a\), and we write \vspace{-0.5cm} \[\lim_{x\to a} f(x) = L,\] if the following condition is satisfied:\\ for every number \(\epsilon > 0\) there exisrixSizes' > neededWeights = fmap (\(a,b) -> a*b) matrixSizes > totalNeededWeights = sum neededWeights > availableWeights = length weights > weights' = take totalNeededWeights weights > partitions = fromList $ partition' (toList neededWeights) (toList weights') > mkMatrix (_,j) values = matrix j values > in if availableWeights >= totalNeededWeights > then Right $ zipWith mkMatrix matrixSizes partitions > else Left $ "Was not given enough weights." Breaking down networks ---------------------- > serialize :: Network -> Seq Double > serialize = Sequence.fromList . concat . concat . toLists > toLists :: Network -> Seq [[Double]] > toLists = fmap Matrix.toLists Using a network --------------- > evaluate :: Network -> Vector Double -> Vector Double > evaluate network input = > let accumulator last next = Matrix.cmap sigmoid (last #> (next `mappend` vector [1])) > in foldr accumulator input network > error :: Network -> TrainingSet -> Seq Double > error network training = > let xs = fmap fst training > ys = fmap snd training > actual = fmap (evaluate network) xs > differences = zipWith (-) actual ys > in fmap norm_2 differences \documentstyle{article} \begin{document} \section{Introduction} This is a trivial program that prints the first 20 factorials. \begin{code} module TLHSLaTeX where -- >>> prod -- [(1,1),(2,2),(3,6)] prod = [ (n, product [1..n]) | n <- [1..3]] \end{code} \end{document} % % (c) The University of Glasgow 2006 % (c) The GRASP/AQUA Project, Glasgow University, 1992-1998 % Matching guarded right-hand-sides (GRHSs) \begin{code} {-# OPTIONS -fno-warn-incomplete-patterns #-} -- The above warning supression flag is a temporary kludge. -- While working on this module you are encouraged to remove it and fix -- any warnings in the module. See -- http://hackage.haskell.org/trac/ghc/wiki/Commentary/CodingStyle#Warnings -- for details module DsGRHSs ( dsGuarded, dsGRHSs ) where #include "HsVersions.h" import {-# SOURCE #-} DsExpr ( dsLExpr, dsLocalBinds ) import {-# SOURCE #-} Match ( matchSinglePat ) import HsSyn import MkCore import CoreSyn import Var import Type import DsMonad import DsUtils import TysWiredIn import PrelNames import Name import SrcLoc import Outputable \end{code} @dsGuarded@ is used for both @case@ expressions and pattern bindings. It desugars: \begin{verbatim} | g1 -> e1 ... | gn -> en where binds \end{verbatim} producing an expression with a runtime error in the corner if necessary. The type argument gives the type of the @ei@. \begin{code} dsGuarded :: GRHSs Id -> Type -> DsM CoreExpr dsGuarded grhss rhs_ty = do match_result <- dsGRHSs PatBindRhs [] grhss rhs_ty error_expr <- mkErrorAppDs nON_EXHAUSTIVE_GUARDS_ERROR_ID rhs_ty empty extractMatchResult match_result error_expr \end{code} In contrast, @dsGRHSs@ produces a @MatchResultan 180 degrees. Let's see. The quadrilateral is almost a triangle with three points near the origin and one far away. Connecting A to B to C to D to A. The edges AB is from (0,0) to (0.1,0), BC is from (0.1,0) to (0,0.1), CD is from (0,0.1) to (100,0), and DA is from (100,0) back to (0,0). The angle at D is between CD and DA. CD is from (0,0.1) to (100,0), which has direction (100, -0.1), and DA is from (100,0) to (0,0), direction (-100,0). The angle at D is between vectors (-100, 0.1) and (-100,0). The angle between these two vectors is very small, so less than 180. Similarly, other angles are likely less than 180, so the quadrilateral is convex. But the diagonals AC and BD do not intersect. Wait, that can't be. In a convex quadrilateral, the two diagonals must intersect inside the quadrilateral. Wait, maybe I made a mistake in the definition. Wait, in a convex quadrilateral, the order of the vertices matters. The quadrilateral should be A, B, C, D connected in order, forming a convex shape. If we connect A(0,0), B(0.1,0), C(0,0.1), D(95 L .54802 .66036 L .54812 .66642 L .54938 .67282 L .55183 .67922 L .55547 .68532 L .5602 .6908 L .56589 .69538 L .57235 .69883 L .57936 .70099 L .58665 .70176 L .59395 .70109 L .60099 .69905 L .6075 .69575 L .61327 .69136 L .61809 .68614 L .62182 .68036 L .62437 .67435 L .62573 .66842 L .62592 .6629 L .62506 .65809 L .62329 .65425 L .62081 .65159 L .61787 .65026 L .61472 .65035 L .61165 .65186 L .60892 .65473 L .60678 .6588 L Mistroke .60546 .66389 L .60514 .66973 L .60594 .67603 L .60792 .68247 L .6111 .68872 L .61542 .69448 L .62076 .69944 L .62695 .70336 L .63377 .70604 L .64098 .70736 L .64831 .70725 L .65548 .70573 L .66223 .7029 L .66831 .69891 L .we need: 1. For all x in (1, 2), (x - a)^2 < 1. 2. For all x not in (1, 2), (x - a)^2 1. So first, let's handle the first condition: For all x in (1, 2), |x - a| < 1. So this means that the distance between x and a is less than 1 for all x in (1, 2). Therefore, a must be within distance 1 of every point in (1, 2). So the maximum distance between a and any x in (1, 2) must be less than 1. What's the maximum distance from a to the interval (1, 2)? If a is inside (1, 2), the maximum distance is max(a - 1, 2 - a). To ensure that this maximum distance is less than 1, we need that both a - 1 < 1 and 2 - a < 1. So a < 2 and a > 1. Therefore, a must be in (1, 2). Waitds in the category of endofunctors author: Oleg Grenrus --- I define free `Monad). Since n is odd, n-1 is even, so gcd(2, n-1)=2. Therefore, ord_n(2) divides 2. So ord_n(2)=1 or 2. If ord_n(2)=1, then 21 mod n, which implies n divides 1, impossible. If ord_n(2)=2, then 21 mod n, so n divides 3. Hence, n=3. So the only prime solution is n=3. Now, suppose n is composite. Let n=pq where p and q are primes (not necessarily distinct). Let's consider n=p, a square of a prime. Let's test n=9, which we did earlier. 2 +1=513. 513/81=6.333, not integer. So n=9 doesn't work. Similarly, n=25: 2 +1=33,554,433. Divided by 625 gives remainder 58, as before. So primes squared don't work. What about n being a product of two different primes? Let's say n=pq where p3 is a solution. Then n must be a Wieferich prime or something similar? Wait, Wieferich primes are p retracement level, the buyers didn't give up and stopped further declines, which resultedi); count++; break; } else if (i == 1) { s.erase(0, 1); count++; } } } return count; } ``` Okay, I need to port this C++ code to C#. Let me look at the original function. It's called maxOperations and takes a string s. The goal is to count the maximum number of operations where in each step, we remove the largest possible prefix that's duplicated. So, for example, if 5060*(10A/11) +5060*3000=5060*(21A/23) +5060*(21x/20) Compute each term: 5060/11=460, so 460*10A=4600A 5060*3000=15,180,000 5060/23=220, so 220*21A=4620A 5060/20=253, so 253*21x=5313x Thus, equation becomes: 4600A +15,180,000=4620A +5313x Subtract 4600A: 15,180,000=20A +5313x Thus, 20A +5313x=15,180,000 Now, from equation 1: x=(0.21A -6930)/1.21 Let me write 0.21 as 21/100 and 1.21 as 121/100, so: x=(21A/100 -6930)/(121/100)= (21A -693,000)/121 Thus, x=(21A -693,000)/121 Plug this into equation 2: 20A +5313*(21A -693,000)/121=15,180,000 Multiply numerator and denominator: 5313*(21A -693,000)=21*5313A -5313*693,000 Compute 21*5313=111,573 5313*693,000=5313*693*1000. Compute 5313*693: First, 5000*693=3,465,000 313*693=216, 309 Wait, 300*693=207,900 13*693=9,009 So 207,900 +9,009=216,909 Thus, total 5313*693=3,465,000 +216,909=3,681,909 Therefore, 5313*693,000=3,681,909,000 So equation becomes: 20A + (111,573A -3,681,909,000)/121=15,180,000 Multiply all terms by 121 to eliminate denominator: 20A*121 +111,573A -3,681,909,000=15,180,000*121 Compute each term: 20*121=2,420, so 2,420A 15,180,000*121=1,836, 180,000*100=15,180,000*100=1,518,000,000; 15,180,000*20=303,600,000; 15,180,000*1=15,180,000. So total 1,518,000,000 +303,600,000 +15,180,000=1,836,780,000 So equation: 2,420A +111,d anyone else. If the memory is still accessible then your familiarity with the common patterns of human experience helps you access it more easily by narrowing the search. Rather than trying to fill in the blank by searching your mind aimlessly, your awareness of the most likely outcomes allows you to search your mind specifically for traces of any previously encoded memory that happens to match one of those probable scenarios. Again, if a previously encoded memory trace is accessible, the chances are good youll succeed in remembering it. (And not just "remembering" it.)Many aspects of an animal or plant can be correctly called adaptations, though there are always some features whose function remains in doubt. By using the term adaptation for the evolutionary process, and adaptive trait for the bodily part or function the product, one. 1er cycle Sur la piste de la vie: adaptation Fiche dactivits Voil science! 2 Droulement suite Ensuite, demandez aux lves sils croient que les animaux sadaptent leur environnement. numrez quelques animaux et discutez globalement de leurs adaptations. The increased exposure that grassland animals face is a daunting challenge. Not only do animals living in these flat, featureless areas have to cope with bad weather, they also have few places to hide from predators. While some grassland species have evolved physical adaptations, such. Animal Adaptation F. First, I remember that stock prediction is a time series problem, so maybe models like ARIMA or LSTM could work. But the user mentioned machine learning, so perhaps they want something like regression models or neural networks. Wait, the user specified using historical data. So I need to get historical stock prices. Maybe using Yahoo Finance with pandas-datareader or yfinance library. Then, data preprocessing is necessary. Features like Open, High, Low, Close, Volume are typical. Maybe calculate moving averages or other technical indicators as features. But how do I structure the data for a machine learning model? Time series prediction usually uses past data points to predict future ones. So I need to create lag features. For example, use the past 7 days' closing prices to predict the next dayere because the triangles in a convex polygon have geometric intersections, allowing points to cover multiple hyperedges (triangles) beyond what is not significantly related to ventilatory efficiency. Peak Vo2, ventilatory inefficiency, and LAVI are associated with an increased risk of major events in the short-term follow-up."When they start hearing you get defensive of course then they start asking themselves rightly, 'Why is he so defensive, maybe I should probe on this more?'" Costolo remembers. He said he had to learn not to think of the questions as a challenge to his authority. He came to understand that investors wanted to learn whether, as the company leader, he was thoughtful about the architecture. They probed to understand whether his team had designed the product to be extensible instead of just a point solution.Maka mulailah dari sekarang untuk membangkitkan energi positif dan pikiran positif untuk melawan penyakit anda. Karena dengan diawali dengan sugesti yang baik, maka herbal kamipun akan bekerja secara maksimal. Kemudian pertanyaan berikutnya adalah, Apakah produk kami sudah ada yang membuktikan? Ternyata dari hasil penelitian diatas, produk kami bekerja sangat baik dan tentunya semua kesembuhan yang berasal dari Tuhan. Berikut adalah sms dari pasien yang sudah mengkonsumsi Ziirzak De Natur yang kebetulan kami abadikan supaya anda juga tersupport bahwa kanker itu bisa dilawan !!!trova Porto di Golfo Aranci - arrivo traghetti in Sardegna Golfo Aranci nasce nella parte Nord Est dellisola di Sardegna, a nord di Olbia.Ci sono diverse interpretazioni sul significato del suo nome, la leggenda vuole che il nome sia dovuto ad una nave carica di arance che naufrag nel suo mare, ma potrebbe anche provenire da Ranci che nel dialetto del posto significa ir (book) + $2 (toy) = **$8** **Calculation**: $12 + $7 = $19 $19 $6 = $13 $13 $2 = $11 $11 + $10 = **$21** **Answer**: Emily now has **$21**.There are 6 students, among which 3 can sing, 2 can dance; 1 can both sing and dance. Now, if we need to select 2 students who can sing and 1 who can dance for a cultural performance, there are a total of $\boxed{15}$ ways to make the selection. Okay, so I need to figure out how many ways there are to select 2 students who can sing and 1 who cut JS, the hamburger menu won't toggle, so in CSS, maybe hide the hamburger icon and show the menu items as a block. The table won't load, so a noscript message is needed. The iframe links would need to be regular links, but with JS, they prevent default and load into the iframe. Putting it all together. Let me start writing the HTML structure, then CSS, then JS. Wait, the navigation links need to load content into the iframe. So each nav link should have a data attribute with the URL. When clicked, JS changes the iframe's src. Without JS, those links would navigate the page, but with JS, they load into the iframe. So, in the HTML, the href could be "#" and prevent default, but without JS, they won't do anything. Hmm, maybe the href should point to the actual page, and with JS, we intercept the click and load it into the iframe. That way, without JS, clicking the link would take the user to the page, but with JS, it stays within the iframe. That's better for graceful degradation. For the JSON endpoint, I'll simulate it with a placeholder URL. The user can replace it with a real endpoint. The data structure should be an array of objects with properties like id, name, email, etc. Pagination: calculate the number of pages based on the data leore than 200 students. Starting in 1921, the city's program for deaf students was housed at T.T. Minor. Notes Handwritten on negative: T.T. Minor School .Handwritten on sleeve: Seattle - schools - Minor, T.T. (overcrowding) .Caption by MOHAI staff .Date photograph was filed at the Seattle Post-Intelligencer (date of photograph and file da-1|=1989 Subtract 2: 1989 -2 =1987 Subtract 1988: |1987 -1988|=1 Subtract 3: |1 -3|=2 Subtract 1987: |2 -1987|=1985 Subtract 4: |1985 -4|=1981 Subtract 1986: |1981 -1986|=5 Subtract 5: |5 -5|=0 (but 5 is already used?) Wait, this seems too convoluted. Maybe there's a pattern here. If we alternate between subtracting small and large numbers, we might be able to create a kind of "sawtooth" pattern where we decrease a bit, then increase by a large amount. Let's test this. Let's take n=6. Let me try with x=1990, x=1, x=2, x=1989, x=3, x=1988. Compute step by step: 1. |1990 -1| =1989 2. |1989 -2|=1987 3. |1987 -1989|=2 4. |2 -3|=1 5. |1 -1988|=1987 6. |1987 -4|=1983 (Wait, but x=1988, so step 5 should be |1 -1988|=1987, then step 6 would subtract x=4? Wait, no. Wait, in this ordering, after x=3, x=1988. So step 5: |1 -3|=2, then step 6: |2 -1988|=1986. But in my previous calculation I messed up. Wait, let's retraogether to make the vineyard what it is today. After years of searching for the perfect land, the Fields family decided on a place along Woodbridge Road, in the heart of the Lodi appellation. With the help of Ryan Sherman, their winemaker, and the constant care of the Fields family, including Russ, Melinda, Van, Zack, and Mia, they collected their first harvest at the end of 2007.EnlacesEnlaces hacia fuera bokra.net www.bokra.netnmlsconsumeraccess.org NMLS Consumer Accessconsumerfinance.gov CFPB > Consumer Financial Protection Bureaumycardstatement.com Credit Card StatementServidoresLocalizacin del Servidor Jack Henry & Associates Inc.MissouriMonettEstados Unidos 36.918272, -93.928335Est alojado por Jack Henry & Associates Inc (Missouri, Monett,) usando el Goldleaf servidor web. El entorno del lenguaje de programacin es ASP.NET. ns2.nwinternet.com, y ns1.nwinternet.com son tus servidores de nombre DNS.IP: 74.200.56.83Desarrollado por: ASP.NETServidor web: GoldleafCodificacin: iso-8859-1El tiempo de carga del sitio es 1718 milisegundos, ms lento que el 62% de otros sitios medidos.Configuracin del Servidor Content-Type:text/html Last-Modified:-- Accept-Ranges:bytes ETag:"a0a92a8d914fd01:0" X-Powered-By:ASP.NET Server:Goldleaf Web Server Date:-- Content-Length:17834I'd seen this supernatural drama before but I'd forgotten most of what it was about so I watched it againcamente, y esos lugares a los que normalmente volvemos con mayor frecuencia cuanto ms lejos estn de nosotros se cuentan entre nuestras posesiones ms valiosas.United StatesCanadaUnited KingdomAfghanistanAlbaniaAlgeriaAmerican SamoaAndorraAngolaAnguillaAntarcticaAntigua and BarbudaArgentinaArmeniaArubaAustraliaAustriaAzerbaijanBahamasBahrainBangladeshBarbadosBelarusBelgiumBelizeBeninBermudaBhutanBoliviaBosnia and HerzegowinaBotswanaBouvet IslandBrazilBritish Indian Ocean TerritoryBrunei DarussalamBulgariaBurkina FasoBurundiCambodiaCameroonCape VerdeCayman IslandsCentral African RepublicChadChileChinaChristmas IslandCocos (Keeling) IslandsColombiaComorosCongoCook IslandsCosta RicaCote D\IvoireCroatiaCubaCyprusCzech RepublicDenmarkDjiboutiDominicaDominican RepublicEast TimorEcuadorEgyptEl SalvadorEquatorial GuineaEritreaEstoniaEthiopiaFalkland Islands (Malvinas)Faroe IslandsFijiFinlandFranceFranceFrench GuianaFrench PolynesiaF (L loc $ HsCmdTop cmd' cmd_stk res_ty names') } ---------------------------------------- tcCmd :: CmdEnv -> LHsCmd Name -> CmdType -> TcM (LHsCmd TcId) -- The main recursive function tcCmd env (L loc cmd) res_ty = setSrcSpan loc $ do { cmd' <- tc_cmd env cmd res_ty ; return (L loc cmd') } tc_cmd :: CmdEnv -> HsCmd Name -> CmdType -> TcM (HsCmd TcId) tc_cmd env (HsCmdPar cmd) res_ty = do { cmd' <- tcCmd env cmd res_ty ; return (HsCmdPar cmd') } tc_cmd env (HsCmdLet binds (L body_loc body)) res_ty = do { (binds', body') <- tcLocalBinds binds $ setSrcSpan body_loc $ tc_cmd env body res_ty ; return (HsCmdLet binds' (L body_loc body')) } tc_cmd env in_cmd@(HsCmdCase scrut matches) (stk, res_ty) = addErrCtxt (cmdCPH5B9Zn4X+tnGuD13L/8/6ZOyda5B6 /I+tiXLs4zkWnir1y5dR/0TBY//zr9/4jzaoUx6N/7dlDRTamv8rYb0VGt2m wTQ9uTJyb2h60wQ7Vs7N5LH3/7ymDLbe8qSx3ySYZuj/fE2Wcv9bzfoU9EO6 q9yb+7qW52XNHZoqnVAOz70515TBcqOs+UI/fDs26oGnST33ph594LnYs/Fb tEHuz3RpP/xO2kAz9L/MJgTTJ+Vq3h+bCcG0SjlCD3Qt32j6XMNfpZyu6H9w zJDyLYLpnP7vxsyxoYeJm6SPmTOZNx6eeqwR82DX/q/HdGnP5sE35rwzpg3+ x9ksqZNOaY0SuuWWORdNlf9Z/eakdB3rbrzVNd6rb+CuYjO1h1Y3pocNmp5m TiwWzTya3iMekeYjplJsJT2MZiNekJ4jVpENfUv8oHK6jnhGsY9DPQmmQdGE xEHSi8TZibej5YjFE5NHbxCjZw6A/4sBFCM41EhgWgptxrH0GfqHGEGaCW1A XB0NgW6hTnqI2EA29BLzBvXz+4mR1B56lfg7v502YC5vLm0eLQ7UHN783bxY OT+qeE/zw1W6Nkdnw49qvqic/9Oc0hx4GLsKmwuL+9NOeob5onkj/6f5lnkX 36b5ivkVvyUejU8PYzlhvFosnmNX7VrMpvne6l2Lr3SseZmYSnWu0bW5Fxs+ ZLGB5lTmU8ZhMSjGZTEd4j8eDV8We4RXiwkSw2TeID7R/Mfcx9xLPfylOD4b PF8skmPxbjGM5pnmmOZDjuXDNC9UzkfKZyU2CA8VsyN2B5ekh9PNzYnEGYl/ GnJtGIcXZyTeaMijYdoUHqoefNW73xjwTHiuc+HAtF7jsbH4z8F3h9+J6cE5 xfwBox[{"(*", " ", RowBox[{"Random", " ", "vector", " ", "function"}], " ", "*)"}], "\[IndentingNewLine]", RowBox[{ RowBox[{"randomVec", " ", ":=", " ", RowBox[{"RandomReal", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{"-", "1"}], ",", "1"}], "}"}], ",", "3"}], "]"}]}], ";"}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{"randomVec", " ", ":=", " ", RowBox[{"RandomReal", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{"-", "1"}], ",", "1"}], "}"}], ",", "2"}], "]"}]}], ";"}]}], "\[IndentingNewLine]", "]"}], "\[IndentingNewLine]", "\[IndentingNewLine]", RowBox[{"(*", " ", "Functionst the One Piece: Unlimited Cruise Ep. 2 [Wii] Trailer off WP (150mb)More articles about One Piece: Unlimited Cruise Ep. 2Please enable JavaScript to view the comments powered by Disqus.blog comments powered by DisqusExample 3. H and W are U.S. citizens. H resides in State T and W is a bona fide resident of the Virgin Islands. For 2008, H and W prepare a joint Form 1040, U.S. Individual Income Tax Return, reporting total adjusted gross income of $75x, of which $40x is attributable to compensation that W received for services performed in the Virgin Islands and $35x to compensation that H received for services performed in State T. Pursuant to section 932(d) and paragraph (d) of this section, because W would have the greater adjusted gross income if computed separately, H and W must file their joint Form 1040 with the Virgin Islands as required by section 932(c) and paragraph (c)(1) of this section. H and W may claim a tax credit on such return for income tax withheld during 2008 and paid to the IRS.Posted on March 7, 2015March 7, 2015 by captainakiraareportingforduty I always suspected that he wasnt my real dad, but it hurt when she confirmed it. I get angry and depressed most days, I feel broken and abandoned by a man Ive never known and a man that I looked up to. I shall admit though that Ive said harsh things because I cant manage my anger probably. Yet, why cant I be angry? Why do I get introuble for saying the truth? He isnt * %************************************************************************ Warn about functions like toInteger, fromIntegral, that convert between one type and another when the to- and from- types are the same. Then it's probably (albeit not definitely) the identity \begin{code} warnAboutIdentities :: CoreExpr -> CoreExpr -> DsM () warnAboutIdentities (Var v) wrapped_fun | idName v `elem` conversionNames , let fun_ty = exprType wrapped_fun , Just (arg_ty, res_ty) <- splitFunTy_maybe fun_ty , arg_ty `eqType` res_ty -- So we are converting ty -> ty = warnDs (vcat [ ptext (sLit "Call of") <+> ppr v <+> dcolon <+> ppr fun_ty , nest 2 $ ptext (sLit "can probably be omitted") , parens (ptext (sLit "Use -fno-warn-identities to suppress this messsage)")) ]) warnAboutIdentities _ _ = return () conversionNames :: [Name] conversionNames = [ toIntegerName, toRationalName , fromIntegralName, realToFracName ] -- We can't easily add fromIntegerName, fromRationalName, -- becuase they are generated by literals \end{code} %************************************************************************ %* * \subsection{Errors and contexts} %* * %************************************************************************ \begin{code} -- Warn about certain types of values discarded in monadic bindings (#3263) warnDiscardedDoBindings :: LHsExpr Id -> Type -> DsM () warnDiscardedDoBindings rhs rhs_ty | Just (m_ty, elt_ty) <- tcSplitAppTy_maybe rhs_ty = do { -- Warn about discarding non-() things in 'monadic' binding ; warn_unused <- woptDs Opt_WarnUnusedDoBind ; if warn_unused && not (isUnitTy elt_ty) then warnDs (unusedMonadBind rhs elt_ty) else -- Warn about discarding m a things in 'monadic' binding of the same type, -- but only if we didn't already warn due to Opt_WarnUnusedDoBind do { warn_wrong <- woptDs Opt_WarnWrongDoBind ; case tcSplit, "2"]}], "150"], "+", RowBox[{ FractionBox["1", "75"], " ", SqrtBox[ FractionBox["2", "3"]], " ", SuperscriptBox["h", "2"], " ", SuperscriptBox["\[Mu]", "2"]}]}], ")"}], " ", RowBox[{"(", RowBox[{"U", "+", RowBox[{ FractionBox["1", "2"], " ", RowBox[{"(", n entfernen und so das Opt-out-Plug-in aktivieren: [Piwik iFrame].See if any keys under HKLM\SYSTEM\CurrentControlSet\Services\Browser\Parameters\ auto or backup browser 2. As almost all 24183933 ora-12012 oracle_ocm_config_dir reconnects some recovery scenarios IsDomainMaster True ? Are you using an operating system that required the error ora 12520 limited. That HSF will work fine The card error into a PCI slot? You need logitech's,there are lots of anywhere is the a run command I can do? My cpu temp has been prod ora-12012 both maintain and ora-20001: yellow or red flag. If they're always long, execute on Computer Browser Service and of was recognized without problems, and I installed the supplied driver.The Special went featured on the couple ebook David Bellamy\'s Watercolour Landscape Course of the century, while the coast found on the reasons of the f to mine) shall perchance have sometimg for the fire of doom to cleanse,' 130.2 . achd ni Maolroin imargu asrubart frissim n `it was not, however, Maelruain that had told him to do this,' 129.10 . isem ro ba anmchara d-si, bi s immorro oc a thothlugud-som `he was her ike terms: (1 -G)x^2 + (-3 +6G)x +15 Thus: Hx +7H = (1 -G)x^2 + (-3 +6G)x +15 Therefore, coefficients: For x^2: 1 -G =0 - **Digital wallets**: expecting a straightforward answer, it's probable that all members of the couples are students. So each couple is two students, hence 60*2=120 students. Then the remaining 123-120=3 students came on their own. Therefore, the answer should be 3 students attended on their own. But just to make sure, let's rephrase the problem in equations. Let S be the number of students who attended on their own. Each couple has 2 students, so 60 couples have 60*2=120 students. Total students is S + 120 = 123. Solving for S gives S = 123 - 120 = 3. Yep, that checks out. I think that's solid. The key assumption here is that all members of the couples are students, which seems reasonable given the problem statement. Unless there's a trick or a different interpretation, 3 is the right answer. **Final Answer** The number of students who attended on their own is \boxed{3}. Chalktown High School had their prom with 123 students attending. We need to determine how many students attended on their own given that 60 couples came to the prom. 1. Each couple consists of 2 students. Therefore, the number of students in couples is calculated as: \[ 60 \text{ couples} \times 2 \text{ students/couple} = 120 \text{ students} \] 2. The total number of students is the sum of students in couples and students who attended on their own.obtuse. So if D is below and to the left of A, that would make the angle at A obtuse. So p would be negative, and D is at (p, -1) with p negative. Similarly, point C would be at (q, -1). Let's see. From equation 3: q - p = s or q - p = -s. But since CD is a side, length |q - p| = s. If D is at (p, -1) with p negative, then C is at (q, -1). If q - p = s, then q = p + s. Since p is negative, q could be positive or negative depending on s. Let's try to work through the equations. From equation 1: p = -sqrt(s^2 -1) (since p is negative) From equation 2: q = s sqrt(s^2 -1). Let's check which sign to take. If we take the positive sign: q = s + sqrt(s^2 -1). Then, from equation 3: q - p = [s + sqrt(s^2 -1)] - (-sqrt(s^2 -1)) = s + 2 sqrt(s^2 -1). This would equal s only if 2 sqrt(s^2 -1) = 0, which is impossible. So this case is invalid. If we take the negative sign: q = s - sqrt(s^2 -1). Then, q - p = [s - sqrt(s^2 -1)] - (-sqrt(s^2 -1)) = s - sqrt(s^2 -1) + sqrt(s^2 -1) = s. So that works. Therefore, q = s - sqrt(s^2 -1) Therefore, we have: - p = -sqrt(s^2 -1) - q = s - sqrt(s^2 -1) - CD length: q - p = [s - sqrt(s^2 -1)] - [-sqrt(s^2 -1)] = s - sqrt(s^2 -1) + sqrt(s^2 -1) = s, which is correct. So coordinates: - A: (0,0) - B: (s,0) - D: (-sqrt(s^2 -1), -1) - C: (s - sqrt(s^2 -1), -1) Now, the next part is that a point K is taken on the extension of side AD beyond point D. So AD goes from A(0,0) to D(-sqrt(s^2 -1), -1). Extending beyond D would go further in the same direction. So parametrize AD: starting at A(0,0), moving towards D, which is direction vector (-sqrt(s^2 -1), -1). Thrst few numbers: Number1: Group1 (odd square) Number2: Group1 Number3: Group2 (closer to4) Number4: Group2 (exact square) Number5: Group2 (closer to4) Number6: Group2 (closer to4) Number7: Group1 (closer to9) Number8: Group1 Number9: Group1 Number10: Group1 Number11: Group1 Number12: Group1 Number13: Group2 Number14: Group2 Number15: Group2 So up to15, the counts are: Group1:1,2,7,8,9,10,11,12 8 numbers Group2:3,4,5,6,13,14,157 numbers But the sums are equal (60 each). So even though Group1 has more numbers, the numbers in Group2 are larger on average. But since each interval balances the sums, the total sums are equal. Therefore, the conclusion is that the sums om the very own corporate ranks, individuals totally committed to use the instrument of military expenditures in order to provide the high profits that are needed to make the very rich of America even richer. In this respect, Bill Clinton had fallen short of expectations, and corporate America could never forgive his original sin, namely, that he had managed to have himself. The structure would be: - app.py (Flask backend) - templates/ - index.html (form) - result.html (result display) In app.py: from flask import Flask, render_template, request app = Flas the union of two coprime sets. For example, all primes and numbers coprime to all primes. But numbers coprime to all primes are just 1, which we can't include. Alternatively, think of the problem in terms of graph colorings. Each number is a node, and edges represent gcd conditions. But this might not help. Wait, let's revisit the earlier idea where if S consists of all primes, it satisfies both conditions (assuming a and b are distinct). The size is 28. The question is, can we include more numbers without violating the conditions? Suppose we include a prime p and a number that is a multiple of p, say p*q where q is another prime. Then, for a = p and b = p*q, condition (ii) requires a c' such that gcd(p, c') >1 and gcd(p*q, c') =1. But gcd(p*q, c') =1 implies c' is not divisible by p or q. However, gcd(p, c') >1 implies c' is divisible by p. Contradiction. Therefore, suchakes the grid, the current queue, visited, and the current area. It returns the final area and the updated visited. But in Elixir, functions can't modify variables outside their scope, so we need to pass state around. Alternatively, use a recursive approach. Here's a possible outline: defmodule Solution do def max_area_of_island(grid) do m = length(grid) if m == 0, do: 0, else: n = length(Enum.at(grid, 0)) max_area = 0 visited = MapSet.new() for i <- 0..m-1, j <- 0..n-1, reduce: {max_area, visited} do {current_max, visited_acc} -> if Enum.at(Enum.at(grid, i), j) == 1 and not MapSet.member?(visited_acc, {i, j}) do {area, new_visited} = bfs(grid, i, j, visited_acc, m, n) {max(current_max, area), new_visited} else {current_max, visited_acc} end end |> elem(0) end defp bfs(grid, i, j, visited, m, n) do queue = [{i, j}] visited = MapSet.put(visited, {i, j}) bfs_loop(grid, queue, visited, m, n, 0) end defp bfs_loop(grid, queue, visited, m, n, area) do case queue do [] -> {area, visited} [{x, y} | rest_queue] -> current_area = area + 1 neighbors = for {dx, dy} <- [{0, 1}, {1, 0}, {0, -1}, {-1, 0}], do: {x + dx, y + dy} valid_neighbors = Enum.filter(neighbors, fn {nx, ny} -> nx >= 0 and nx < m and ny >= 0 and ny < n and Enum.at(Enum.at(grid, nx), ny) == 1 and not MapSet.member?(visited, {nx, ny}) end) new_visited = Enum.reduce(valid_neighbors, visited, fn {nx, ny}, acc -> MapSet.put(acc, {nx, ny}) end) new_queue = rest_queue ++ valid_neighbors bfs_loop(grid, new_queue, new_visited, m, n, current_area) end end end Wait, but in the code above,things. So the definition can be subjective based on political views. I should also consider legal definitions. The UN has struggled to agree on a single definition, but there are common elements: violent acts intended to create fear, perpetrated for ideological goals, and targeting non-combatants. Psychological impact is a big part of it, aiming to influence a wider audience beyond the immediate victims. Another point is the use of media. Terrorist acts are often designed to get media coverage to spread their message and amplify fear. So the publicity aspect is crucial. Also, terrorism can be domestic or international. Domestic terrorism happens within a country's borders without foreign involvement, while international terrorism crosses borders. I should also mention the motivations: political, religious, ideological, or social. Groups like ISIS, Al-Qaeda, or domestic groups like the Oklahoma City bombers. Each has different motivations but use similar tactics. Counter-terrorism measures include intelligence gathering, military action, law enforcement, and addressing root causes like poverty or political grievances. But there's debate over the effectiveness of different counter-terrorism strategies. Some argue that military responses can exacerbate the problem, while others focus on prevention and community engagement. Also, the impact on human rights and civil liberties when governments implement strict security measures. Wait, did I miss anything? Maybe the difference between terrorism and insurgency. Insurgencies aim to overthrow a government or create a separate state through prolonged violence, which can include guerrilla warfare. Terrorism is a tactic used within insurgencies or by other groups. So terrorism is a tactic, not necessarily an ideology itself. Also, state-sponsored terrorism, where governments support terrorist groups as proxies. That adds another layer, complicating international relations. Examples might include alleged support by some states for groups like Hezbollah or Hamas. In summary, terrorism is complex with multiple dimensions: use of violence for coercion, targeting non-combatants, ideological motives, psychological impact, and the aim to influence politics or society. Definitions can vary, but these elements are commonly cess (t-1) p exitcode | otherwise -> loop chan hProcess t p exitcode readerProc :: Chan BuildMessage -> Handle -> (String -> String) -> IO () readerProc chan hdl filter_fn = (do str <- hGetContents hdl loop (linesPlatform (filter_fn str)) Nothing) `finally` writeChan chan EOF -- ToDo: check errors more carefully -- ToDo: in the future, the filter should be implemented as -- a stream transformer. where loop [] Nothing = return () loop [] (Just err) = writeChan chan err loop (l:ls) in_err = case in_err of Just err@(BuildError srcLoc msg) ows. Wait, no: first row is black, second white, third black, fourth white, fifth black, sixth white, seventh black. Therefore, black-starting rows: 1,3,5,7 (4 rows), and white-starting rows: 2,4,6 (3 rows). Each black-starting row has 6 black squares, each white-starting row has 6 black squares as well (since they start with white, the even positions are black). Wait, no: Wait, in a white-starting row, the first square is white, then black, etc. So positions 1: white, 2: black, 3: white, ..., 12: black if even. So in a white-starting row, there are 6 white and 6 black squares. Similarly, in a black-starting row, 6 black and 6 white. Therefore, each row has 6 of each color. Therefore, over 7 rows, total black squares = 7*6 =42, white squares=7*6=42. Therefore, equal. Therefore, the maximum independent set is 42. Therefore, 25 red squares can be placed on a single color class, no two adjacent, each with 0 red neighbors. Therefore, satisfying the condition. Therefore, the answer should be yes. Therefore, the final answer is \boxed{Yes}. To determine if it's possible for each of the 25 red squares on a $7 \times 12$ chessboard to be adjacent to an even number of red squares, we can use the following reasoning: 1. **Chessboard Structure**: A $7 \times 12$ chessboard has 84 squares, which can be divided into tw arranging mber of scenes were Jack is taken advantage of and often times brutally beaten. As Jack is allowed to participate in the business, he starts to think to highly and eventually puts himself, the woman he is courting, Bertha Minnix (Mia Wasikowska), his best friends that acts as his right hand man, Cricket Pate (Dane DeHaan) and the entire operation in danger. Experience, anger and a personal assault on Cricket eventually ends up teaching Jack how to be a Bondurant and he steps up in the end of the film to try and save his reputation.When we reported you some ways U2 could have prevented No Line On The Horizon from making its way onto the Internet, we didn't think we'd have to consider the obvious: Make Sure Your Label Doesn't Leak It.As it turns out, the leak has been traced to Universal Music Group's Australian arm, which was offering high-quality downloads for purchase from their digital music site Getmusic.au. A-ha!A huge mistake, for sure, and one which wasn't caught for a few hours - enough time for the album to be all over the Internet.Undoubtedly, U2 weren't thrilled by the gaffe, but they've turned a negative into a positive by streaming the entire record on their MySpace page.By the way, we've heard from some readers who have had problems accessing the tracks on U2's MySpace page. Check out the image below:On the left, you'll see thumbnails of U2 album and single covers. Click on the No Line On The Horizon thumb and the playlist will come up, as shown. But NOT if your IP address is in the UK.For everyone else? Have a listen and read MusicRadar's exclusive track-by-track review and blog on the album. Then let us know what you think. , ( ) -. , . , , . , 19 . Interfax: . : () , SANA .Goldman recently reached out to other banks to get them involved, the sources said. - ReutersSHOW ALLCompaniesRantao appointed as Independent Media Press OmbudsmanUnions warn of second strike if Robertson wage talks failStrange weather dents global wine productionMarmitegate shows strain on consumer industryBHP needs 21 000 women to angles of the triangle. For example, the touch point on side a is located at coordinates (r, 0) if we align the triangle such that side a is horizontal. But this might not capture the entire picture. Alternatively, use trilinear coordinates. The touch points have trilinear coordinates (0, s - c, s - b), etc. But converting trilinear coordinates to cartesian might be tedious. Alternatively, use vector geometry. Lets denote the incenter as I. The touch points can be expressed as I plus vectors in the direction of the angle bisectors, scaled by the inradius r. However, without knowing the specific angles, this might not help. Alternatively, recall that the area of the contact triangle can be expressed as r^2 * (cot(A/2) + cot(B/2) + cot(C/2))/2. Wait, but cot(A/2) = (s - a)/r, since in a triangle, cot(A/2) = (s - a)/r. Therefore, cot(A/2) + cot(B/2) + cot(C/2) = (s - a)/r + (s - b)/r + (s - c)/r = (3s - (a + b + c))/r = (3s - 2s)/r = s/r. Therefore, the area of the contact triangle is r^2 * (s/r)/2 = (r s)/2. Wait, but that would mean the area of the contact triangle is (r s)/2, and the area of the original triangle is r s. Then the ratio would be 2. But in the equilateral triangle case, this is not true, since the contact triangle (which is the medial triangle in that case) has area 1/4 of the original, so ratio 4. Therefore, this formula must be incorrect. Hmm, what's wrong here? Wait, if cot(A/2) + cot(B/2) + cot(C/2) = s/r, then the formula for the area of the contact triangle would be r^2 * (s/r)/2 = (r s)/2. But in +x_n^2 - (4/(n+1)) Sum x_k^2 We need to show that Q(x) >=0 for all x. To do this, we can diagonalize Q(x) or find its eigenvalues. Suppose we look for the minimal e , co_ax_branches = FirstBranch branch } where branch = CoAxBranch { cab_loc = getSrcSpan name , cab_tvs = tvs , cab_lhs = mkTyVarTys tvs , cab_roles = roles , cab_rhs = rhs_ty , cab_incomps = [] } mkPiCos :: Role -> [Var] -> Coercion -> Coercion mkPiCos r vs co = foldr (mkPiCo r) co vs mkPiCo :: Role -> Var -> Coercion -> Coercion mkPiCo r v co | isTyVar v = mkForAllCo v co | otherwise = mkFunCo r (mkReflCo r (varType v)) co -- The first coercion *must* be Nominal. mkCoCast :: Coercion -> Coercion -> Coercion -- (mkCoCast (c :: s1 ~# t1) (g :: (s1 ~# t1) ~# (s2 ~# t2) mkCoCast c g = mkSymCo g1 `mkTransCo` c `mkTransCo` g2 where -- g :: (s1 ~# s2) ~# (t1 ~# t2) -- g1 :: s1 ~# t1 -- g2 :: s2 ~# t2 [_reflk, g1, g2] = decomposeCo 3 g -- Remember, (~#) :: forall k. k -> k -> * -- so it takes *three* arguments, not two \end{code} %************************************************************************ %* * Newtypes %* * %************************************************************************ \begin{code} -- | If @co :: T ts ~ rep_ty@ then: -- -- > instNewTyCon_maybe T ts = Just (rep_ty, co) -- -- Checks for a newtype, and for being saturated instNewTyCon_maybe :: TyCon -> [Type] -> Maybe (Type, Coercion) instNewTyCon_maybe tc tys | Just (tvs, ty, co_tc) <- unwrapNewTyCon_maybe tc -- Check for newtype , tys `lengthIs` tyConArity tc -- Check saturated = Just (substTyWith tvs tys ty, mkUnbranchedAxInstCo Representational co_tc tys) | otherwise = Nothing topNormaliseNewType_maybe :: Type -> Maybe (Coercion, Type) -- ^ Sometimes we want to look through a @newtype@ and get its associated coercion. -- This function strips off @newtype@ layers enough to reveal something that isn't -- a @newtype@. Specifically, here's the invariant: -- -- > topNormaliseNewType_maybe rec_nts ty = Just (co, ty') -- -- then (a) @co : ty0 ~ ty'@. -- (b) ty' is not a newtype. -- -- The function returns @Nothing@ for non-@newtypes@, -- or unsaturated applications topNormaliseNewType_maybe ty = go initRecTc Nothing ty where go rec_nts mb_cve, the problem might not require us to consider that, as it's only asking about a. So, the answer is \boxed{20}. **Final Answer** The number of days a alone can finish the job is \boxed{20}. Given that \(a\) and \(b\) together can finish a work in 40 days, their combined work rate is \(\frac{1}{40}\) per day. They worked together for 10 days, completing \(10 \times \frac{1}{40} = \frac{1}{4}\) of the work. The remaining work is \(1 - \frac{1}{4} = \frac{3}{4}\). After \(b\) left, \(a\) finished the remaining \(\frac{3}{4}\) of the work in 15 days. Let \(a\)'s work rate be \(\frac{1}{a}\) per day. The work done by \(a\) in 15 days is \(15 \times \frac{1}{a} = \frac{3}{4}\). Solving for \(a\): \[ 15 \times \frac{1}{a} = \frac{3}{4} \] \[ \frac{15}{a} = \frac{3}{4} \] \[ a = \frac{15 \times 4}{3} = 20 \] Thus, \(a\) alone can finish the job in \(\boxed{20}\) days.How does the behavior of an animal affect the transmission of diseases from one animal to another? Okay, so I need to figure out how an animal's behavior affects disease transmission among animals. Let me start by breaking down the question. The user is asking about the relationship between animal behavior and the spread of diseases. I remember from biology class that behaviors like social interactions, mating, feeding, and migration can play roles. Let me think of some examples. First, social animals. Animals that live in groups, like wolves or meerkats, probably have higher disease transmission because they're close to each other. Diseases can spread through direct contact, like grooming or fighting. Then there's mating behavior. If animals mate with multiple partners, that could spread sexually transmitted infections. Birds or mammals with promiscuous mating systems might have higher STI rates. Feeding behavior is another aspect. Scavengers, like hyenas or vultures, that eat dead animals might get diseases from carcasses. Also, if animals share feeding areas, they might come into contact with each other's saliva or feces, which can transmit pathogens. Forthat for a polynomial of degree d, the maximum number of distinct integer solutions to p(x) = 0 or p(x) = c is d + k, where k is some constant. But I need a better approach. Wait, here's a crucial observation. For the polynomial p(x) with integer coefficients, the equations p(x) = 0 and p(x) = -6 cannot both have many roots be a bit messy, but maybe manageable. Alternatively, perhaps we can find a homogeneous recurrence modulo13. Let me check if the nonhomogeneous term n*2^{n} is congruent to 0 modulo13 for certain n. If that term is 0 mod13, then the recurrence would be homogeneous, which might be easier to handle. But n*2^{n} mod13: For this term to be 0 mod13, either n0 mod13 or 2^{n} 0 mod13. However, 2 and 13 are coprime, so 2^{n} is never 0 mod13. Therefore, the nonhomogeneous term is 0 mod13 only when n0 mod13. For other n, the term is non-zero. So maybe except when n is a multiple of13, we have an inhomogeneous recurrence. However, the problem is about n=1989,1990,1991. Let's see what these are modulo13 and modulo other numbers. First, compute 1989,1990,1991 modulo13. Let's compute 1989 13. 13*153=1989, since 13*150=1950, 13*3=39, 1950+39=1989. Therefore, 19890 mod13. Similarly, 1990=1989+11 mod13, and 19912 mod13. So f(1989)f(0)0 mod13, but wait, f(0)=0. Wait, but 1989 is a large n. Wait, perhaps the recurrence's terms can be related modulo13? Wait, perhaps we can use induction. Maybe if we can show that f(n) 0 mod13 for all n0,1,2 mod13, but that might not be the case. Let me check the base cases. f(0)=0, f(1)=0. Then f(2)=4^{2} f(1) -16^{1} f(0) +0*2^{0}=16*0 -16*0 +0=0. So f(2)=0. Then f(3)=4^{3} f(2)-16^{2} f(1) +1*2^{1}=64*0 -256*0 +1*2=2. So f(3)=22 mod13. So f(3)2 mod13. Therefore, not 0. So the function isn't always 0 mod13. Therefore, the zeros are only at n=0,1,2. Then f(3)=2, which is 2 mod13. Therefore, the problem is specifically about n=1989,1990,1991. So maybe there's some periodicity modulo13 or modulo some multiple of13, such that f(n) 0 mod13 for n0,1,2 mod13. Let's check that. Wait, let's compute a few more terms modulo13. Let's start computing f(n) mod13 up to n=6 or so to see if there's a pattern. Given: f(0)=0 f(1)=0 f(2)=4^{2} f(1) -16^{1} f(0) +0*2^{0} =0-0+0=0 f(3)=4^{3} f(2)-16^{2} f(1)+1*2^{1}=64*0 -256*0 +2=2 f(4)=4^{4} f(3)-16^{3} f(2)+2*2^{4}=256*2 -4096*0 +2*16=512 +0 +32=544 Compute 544 mod13: 13*41=533, so 544-533=11. So f(4)=11 mod13. f(5)=4^{5} f(4)-16^{4} f(3)+3*2^{9} First compute 4^{5} mod13.7939 L .33783 .67667 L .33098 .67509 L Mistroke .32395 .67472 L .31696 .6756 L .31024 .67771 L .304 .68096 L .29842 .68521 L .29368 .69029 L .28988 .69598 L .28711 .70202 L .2854 .70816 L .28472 .71413 L .28501 .71967 L .28616 .72453 L .28802 .72852 L .2904 .73145 L .29311 .73321 L .29591 .73373 L .29858 .73301 L .3009 .73109 L .30267 .72807 L .3037 .7241 L .30385 .7194 L .30303 .71417 L .30116 .70869 L .29823 .70321 L .29429 .698 L .28942 .69332 L .28374 .6894 L .27741 .68642 L .27064 .68455 L .26363 .68389 L .25661 .68447 L .2498 .68628 L .24343 .68927 L .23768 .69329 L .23272 .69819 L .22869 .70375 L .22566 .70974 L .22369 .71588 L .22277 .72192 L .22283 .72758 L .22379 .73263 L .22549 .73684 L .22777 .74004 L .23042 .74209 L .23321 .74292 L .23594 .74249 L .23836 .74085 L .24028 .73808 L .2415 .73433 L .24188 .72978 L Mistroke .2413 .72466 L .23968 .71922 L .23701 .71371 L .23331 .70842 L .22865 .70359 L .22315 .69946 L .21697 .69624 L .21028 .69409 Lhe dependency on x may not show up in the loop_breaker_edges (see note [Choosing loop breakers} above). A normal "strong" loop breaker has IAmLoopBreaker False. So pWith4Equal msg z as bs cs ds zipWith4Equal _ _ [] [] [] [] = [] zipWith4Equal msg _ _ _ _ _ = panic ("zipWith4Equal: unequal lists:"++msg) #endif \end{code} \begin{code} -- | 'zipLazy' is a kind of 'zip' that is lazy in the second list (observe the ~) zipLazy :: [a] -> [b] -> [(a,b)] zipLazy [] _ = [] zipLazy (x:xs) ~(y:ys) = (x,y) : zipLazy xs ys \end{code} \begin{code} stretchZipWith :: (a -> Bool) -> b -> (a->b->c) -> [a] -> [b] -> [c] -- ^ @stretchZipWith p z f xs ys@ stretches @ys@ by inserting @z@ in -- the places where @p@ returns @True@ stretchZipWith _ _ _ [] _ = [] stretchZipWith p z f (x:xs) ys | p x = f x z : stretchZipWith p z f xs ys | otherwise = case ys of [] -> [] (y:ys) -> f x y : stretchZipWith p z f xs ys \end{code} \begin{code} mapFst :: (a->c) -> [(a,b)] -> [(c,b)cCvKVmt96WHXhicFVi3Yvo+DSFyJrcnW7cCD1Upi7PgVNoo4v OSjWhYEv1kmLXqbg7r2BdeH8Tvx4XeIo9wYFHcUf333P7sR9dTNWSTEU9Awv mTOnvRPbuwwNZLIp+DhgDTv9Yydu3/wlNqGbgi/1jLIb0jrRTyP+9x4WBd8S dt4zLzrRN2kr5M1SsOBMaNe+W534PUJ78adlVBzwoP+dPNiJG6v6S713UHFc gROpaNeJJf7azJE9VOTTFu2xsujEYNN/HPX9VJR2gtKETZ1oc0yd2+pFxVVL 9odUqHTic/3HL2wvUXHDb59trJWd+OFhx6XQICqa7H6ZaybSiXnXJuXlI6m4 e95nXx8eA0uqLzyPeEpFx5JajWdjDNz8TislO5GKnsazKUNUBk41NhaYZFPx sbZbbMx7Brq1Tt90raZi4uhFux/JDDylLcQq/kPFjMx7C3ufMTBNoffVcAsV UeX7TcMbDBxPTFW/2UXFqs6G7ccCGOhb9SN3vJ+KjYmDs/fPMnB3mmrHulEq Dq5Y6Uffz8C3uu8vNU6R+Tdr64jv b is just a*b. So if a*b is a square, then1*a*b = a*b is a square, which is forbidden. Therefore, if we include1, we must ensure that the product of any two numbers in M is not a square. Therefore, in this case, M must be such that it's a set where no two elements multiply to a square, and also no three elements multiply to a square. But if we include1, we have to satisfy both conditions: no two elements multiplying to a square (because1*a*b = a*b would be forbidden), and also no three elements (including non-1 elements) multiplying to a square. Therefore, including1 imposes a stricter condition. Thus, including1 might limit the size of M. Let's explore this possibility. Suppose we include1 in M. Then, we must ensure that: 1. No two elements in M multiply to a square (since1*a*b = a*b would be a square). 2. No three elements imy wonderful manager who was kind enough to let us play with his iPhone, at times. Here us refers to me and two other ardent fans of the game. We used to take turns playing with the phone. I do remember one Saturday night at work ,when the three of us were playing and creating some commotion and a few others, around 4 of them also joined with us for the tournament. It was fun, playing amidst deadlines at 2 am!",{id:!a.aanFeatures.r?a.sTableId+"_processing":null,"class":a.oClasses.sProcessing}).html(a.oLanguage.sProcessing).insertBefore(a.nTable)[0]}function C(a,b){a.oFeatures.bProcessing&&h(a.aanFeatures.r).css("display",b?"block":"none");w(a,null,"processing",[a,b])}function rb(a){var b=h(a.nTable);b.attr("role", "grid");var c=a.oScroll;if(""===c.sX&&""===c.sY)return a.nTable;var d=c.sX,e=c.sY,f=a.oClasses,g=b.children("caption"),i=g.length?g[0]._captionSide:null,j=h(b[0].cloneNode(!1)),n=h(b[0].cloneNode(!1)),l=b.children("tfoot");cns its maximum and minimum. Due to the tetrahedron's symmetry, the maximum and minimum areas will occur along directions that are symmetric with respect to the tetrahedron's geometry. We already saw that projecting onto a plane parallel to a face gives the face area, which is a candidate for maximum. Projecting onto a plane perpendicular to a face's normal gives a different shape, possibly with a smaller area. But earlier, when we projected onto the plane perpendicular to (1,1,1), we got an area equal to the face area, which is confusing. Maybe this is because of the specific coordinates chosen. Alternatively, let's use a different coordinate system where the calculations are clearer. Let's take a regular tetrahedron with vertices at (1,1,1), (-1,-1,1), (-1,1,-1), (1,-1,-1), edge length 22, as before. The normals to the faces are vectors like (1, -1, -1), etc. The area of each face is 23. When we project onto a plane perpendicular to (1,1,1), the area came out as 23, same as the face area. But that seems contradictory. Unless projecting along that direction somehow also captures a face's area, but that doesn't make sense because the face is not perpendicular to (1,1,1). Wait, perhaps the mistake is in the projection. When we project the tetrahedron onto a plane perpendicular to (1,1,1), we are essentially squishing the tetrahedron along that direction. The resulting projection's area being equal to a face's area might be a coincidence due to the symmetry. Alternatively, maybe the maximum area is achieved not only when projecting parallel to a face but also in other symmetric directions. Wait, let's calculate the area when projecting onto a plane perpendicular to an edge direction. For example, take the edge from (1,1,1) to (-1,-1,1). The midpoint of this edge is (0,0,1). The direction vector of the edge is (-2, -2, 0). A plane perpendicular to this edge would have a normal vector (-2, -2, 0), or simplified to (1,1,0). Lets project the tetrahedron onto a plane with normal (1,1,0). Compute the projected area. First, find the projections of the four vertices onto this plane. Using the same projection method: for each vertex **p**, its projection onto the plane is **p** - (**p** **n** / ||**n**||) **n**, where **n** = (1,1,0). ||**n**|| = 2. 1. Vertex (1,1,1): **p** **n** = 1*1 + 1*1 + 1*0 = 2 Projection: (1,1,1) - (2/2)(1,1,0) = (1,1,1) - (1,1,0) = (0,0,1) 2. Vertex (-1,-1,1): **p** **n** = (-1)(1) + (-1)(1) + 1*0 = -2 Projection: (-1,-1,1) - (-2/2)(1,1,0) = (-1,-1,1) + (1,1,0) = (0,0,1) 3. Vertex (-1,1,-1): **p** **n** = (-1)(1) + 1*1 + (-1)*0 = 0 Projection: (-1,1,-1) - (0/2)(1,1,0) = (-1,1,-1) 4. Vertex (1,-1,-1): **p** **n** = 1*1 + (-1)(1) + (-1)*0 = 0 Projection: (1,-1,-1) - (0/2)(1,1,0) = (1,-1,-1) So the projected points are (0,0,1), (0,0,1), (-1,1,-1), and (1,-1,-1). But since we are projecting onto a plane, we need to express these in 2D coordinates on the plane. The plane has normal (1,1,0), so we can choose an orthonormal basis within the plane. Let's take **u** = (1,-1,0)/sqrt(2) and **v** = (0,0,1). Wait, but **v** is not in the plane. The plane's normal is (1,1,0), so vectors in the plane must satisfy x + y = 0. Wait, no. Wait, the plane is perpendicular to (1,1,0), so a vector **p**ath Lovely Morning Recap Brazen Debacle Everything Okay? Disappeared Horndog Doing Better College Public School Staying Flake The Last Word Hugs Chill Blubbering Big Deal Over the Moon Frowny Important Message Schedules Ball Is That a No? Processing Faith Harmful Outnumbered Actually Dating Ask Rain Delay Special Next> Latest>>{"_RUDA-10A_10":{"ThumbnailURL":"https://img.chooseacottage.co.uk/property/rud/95/ruda-10a_10.jpg","ImageURL":"https://img.chooseacottage.co.uk/property/rud/480/ruda-10a_10.jpg","ImageAltText":"Surf shop","DisplayOrder":"1"},"_RUDA-10B_11":{"ThumbnailURL":"https://img.chooseacottage.co.uk/property/rud/95/ruda-10b_11.jpg","ImageURL":"https://img.chooseacottage.co.uk/property/rud/480/ruda-10b_11.jpg","ImageAltText":"","DisplayOrder":"2"},"_RUDA-11A_11":{"ThumbnailURL":"https://img.chooseacottage.co.uk/property/rud/95/ruda-11a_11.jpg","ImageURL":"https://img.chooseacottage.co.uk/property/rud/480/ruda-11a_11.jpg","ImageAltText":"Croyde Bay","DisplayOrder":"3"},"_RUDA-01A_15":{"ThumbnailURL":"https://img.chooseacottage.co.uk/property/rud/95/ruda-01a_15.jpg","ImageURL":"https://img.chooseacottage.co.uk/property/rud/480/ruda-01a_15.jpg","ImageAltText":"The park setting","DisplayOrder":"4"},"_RUDA-01B_13":{"ThumbnailURL":"https://img.chooseacottage.co.uk/property/rud/95/ruda-01b_13.jpg","ImageURL":"https://img.chooseacottage.co.uk/property/rud/480/ruda-01b_13.jpg","ImageAltText":"Croyde Bay beach\u003cbr /\u003e","DisplayOrder":"5"}}Atmosphere. From physical conditions and dress code to receptivity to new ideas, think through all aspects of atmosphere to ensure everyone is comfortable and understands that the retreat offers a safe space to share and exchange concepts and viewpoints. The right atmosphere will encourage openness and spark creativity. Also consider timing. After a long day, you may want to cap things off with social events that run into late evening. However, be sure to build in breaks and offer staff time to recharge. dpaBeate Zschpe dpaMnchen - Im NSU-Prozess hat das Gericht in einer zhen Vernehmung versucht, das rechtsextreme Umfeld des spteren Terrorlast one of the entire split) contributing their cost. Wait, this is getting complicated. Alternatively, perhaps the DP[i] represents the minimal cost to split the first i words (0..i-1) into lines, where each line (except the last line of the entire split) is considered. So when we compute DP[i], it's possible that the last line of the split is i-1, but if that's part of the entire split, then its cost is added only if it's not the last line of the entire split. But this seems too vague. Let me think again. Let's model dp[i] as the minimal cost to split the first i words (0.= 0; return 0; } ``` Compiled with `gcc -g test.c -o test -fno-stack-protector`. The disassembly shows loops that modify memory in chunks of 4096 bytes and then some more. The user is confused because the total allocated stack space doesn't add up to 4,000,004 bytes (since 1,000,000 ints are 4,000,000 bytes, plus 4 bytes for 'num'), and there's an OR instruction that doesn't seem to do anything. Let me start by recalling that the stack is typically managed in pages. The stack grows downwards, and the operating system allocates stack memory in pages (usually 4KB each). Accessing memory beyond the current stack allocation can cause a page fault, which the OS handles by expanding t The traveler will **not** always be able to determine who is a knight and who is a liar. Here's the reasoning: 1. **Sum Preservation**: Each natives statements (left and right neighbors ages) sum to the true sum of their neighbors ages, whether theyre a knight or a liar. This is because liars adjust one neighbors age by +1 and the other by -1, leaving the total unchanged. 2. **System of Equations**: The traveler can set up a system of linear equations where each equation corresponds to a natives statements. However, for an even number of natives (like 50), this system is underdetermined. This means there are infinitely many solutions for the actual ages of the natives, as the equations are not independent enough to uniquely determine all variables. 3. **Non-Unique Solutions**: Different age assignments that satisfy the system can correspond to different knight/liar configurations. For example, a set of statements might be explained by multiple combinations of truth all of them. Otherwise, if there's between k and 2k-1 characters left, reverse the first k and leave the rest. Let me think with the examples. First example: "abcdefg", k=2. The output is "bacdfeg". Let's break it down. The string length is 7. The 2, in our case, x_{51} = 51/101. Then, 1 - x_{51} = 1 - 51/101 = 50/101 = x_{50}? Wait, x_{50} = 50/101, but 1 - x_{51} = 50/101 = x_{50 + 1}? Wait, x_{50} = 50/101, but 1 - x_{51} = 50/101, so the pair for x_{51} would be x_{50}, but x_{50} is already paired with x_{52}? Wait, maybe my earlier pairing was incorrect. Wait, let's think again. If i ranges from 1 to 101, then x_i = i/101. Then, the term paired with x_i is x_{102 - i} = (102 - i)/101. So, for example, i=1: x_1=1/101, paired with x_{101}=101/101=1. Then, i=2: x_2=2/101 paired with x_{100}=100/101=1 - 1/101. Wait, but 1 - x_i = 1 - i/101 = (101 - i)/101, which is x_{101 - i +1} = x_{102 - i}. Wait, no, 101 - i +1 = 102 - i. So, 1 - x_i = x_{102 - i -1}? Wait, maybe not. Let me check with i=1: 1 - x_1 = 1 - 1/101 = 100/101 = x_{100}, but x_{102 - 1} = x_{101} = 101/101 =1. So, actually, 1 - x_i is x_{101 - i +1}? Wait, 101 - i +1 = 102 -i. So, 1 - x_i = (101 - i)/101 = x_{101 - i}, but if x_{101 - i} is for index 101 - i. Wait, but if i ranges from 1 to 101, then 101 - i ranges from 100 down to 0. But x_0 isn't defined here. So maybe my initial thought is slightly off. Wait, perhaps 1 - x_i = x_{101 - i +1}? Let's check for i=1: 1 - x_1 = 1 - 1/101 = 100/101. Then, x_{101 -1 +1} = x_{101} = 101/101 =1. Not equal. So that's not right. Alternatively, maybe x_{102 -i}? For i=1: x_{102 -1}=x_{101}=1. 1 - x_1 =100/101. Not equal. So perhaps 1 - x_i is not in the set of x_j's. Wait, except when x_i =0.5, but 0.5 is 50.5/101, which is not an integer. So, actually, there is no term where x_j =1 - x_i, unless perhaps j=102 -i. But 1 - x_i = (102 -i)/101 - 1/101. Wait, let's compute 1 - x_i: 1 - x_i = 1 - i/101 = (101 - i)/101. So, if we have a term x_j where j =101 -i, then x_j = (101 -i)/101 =1 - x_i. But the index j=101 -i. However, in the summation, we are summing from i=1 to 101. So j=101 -i would range from 100 down to 0. But when i=101, j=0, which. For example, if you set \( G_E \) to its minimum (zero), then \( P_E \) would need to be 115.5, but since that's impossible, you have to increase \( G_E \) until \( P_E \) is feasible. Alternatively, if you set \( P_E \) to its minimum possible (theoretical zero), then \( G_E \) would need to be 231, which is also impossible. But given that the problem states "find the minimum value", maybe we are to provide the equation or express the relationship. But since this is a math problem, likely expecting numerical answers, perhaps there is a misunderstanding here. Wait, perhaps there is a typo, and the problem is actually asking for the minimum value of \( P_E(i) \) OR \( G_E(i) \), but as written, it's AND. Alternatively, maybe it's asking for the minimum combined value? But again, unclear. Alternatively, maybe the problem wants the minimum value of \( R(i) \), but no, it's given that \( R(i) \geq 1 \), and wants the min values of \( P_E \) and \( G_E \). Hmm. Wait, let's consider the possibility that the question is actually asking for the minimum values such that both \( P_E \) and \( G_E \) are minimized, but in a way that their combination meets the requirement. This is similar to optimization where you minimize a function subject to a constraint. If we consider minimizing a function like \( P_E + G_E \), subject to \( P_E + 0.5 G_E \geq 115.5 \). But since ecking, I found that the solution involves using the Law of Sines in the sub-triangles formed by the trisectors and setting up a system of equations leading to a cubic equation in cos . The key steps are: 1. Let angle BAC = 3. 2. The trisectors divide BC into BD = d, DE = e, EC = f. 3. Using the Law of Sines in triangles ABD, ADE, and AEC, and recognizing that the angles at D and E relate to the angles of the main triangle. 4. Derive a system of equations that ultimately reduces to a cubic equation in cos . 5. Solve the cubic equation to find , then multiply by 3 to get angle A. Following this method, let's attempt it for case I: d=1, e=1, f=2. Lets denote angle BAC = 3. Applying the Law of Sines to triangles ABD, ADE, and AEC. For triangle ABD: AB / sin(angle ADB) = BD / sin AB = (BD * sin(angle ADB)) / sin = (1 * sin(angle ADB)) / sin For triangle ADE: AD / sin(angle AED) = DE / sin AD = (DE * sin(angle AED)) / sin = (1 * sin(angle AED)) / sin For triangle AEC: AC / sin(angle AEC) = EC / sin AC = (2 * sin(angle AEC)) / sin In triangle ABC, angles at B and C are and respectively, with + = 180 - 3. Additionally, angles at D and E: In triangle ABD, angle ADB = 180 - - In triangle ADE, angle AED = 180 - angle AEC - In triangle AEC, angle AEC = 180 - - But angle AED in triangle ADE is also related to angle ADB and angle AEC through the straight line BC. This is getting too complex. Perhaps another approach is needed. Given the time I've spent without success, I think I need to look up tnstruct such a set. Alternatively, use the fact that the maximum size is related to the size of a Sidon set in the vector space, but again, I'm not sure. Alternatively, think greedily: start adding vectors one by one, ensuring that adding a new vector doesn't create a triple that sums to zero. That is, when adding a new vector v, check that there are no two existing vectors u, w such that u + w = v. Because if there are, then u + w + v = 0. Therefore, the problem reduces to selecting a subset of vectors such that no vector is the sum of any two others. This is equivalent to the set being sum-free in the additive sense, except that instead of avoiding a + b = c in integers, here it's in GF(2). In coding theory, such a set is called a "code with no three codewords summing to zero," but I need to think of it constructively. Alternatively, consider that each vector added restricts the future vectors that can be added. For example, once we add vectors u and v, we cannot add the vector u + v. So the problem is similar to building a set where the sum of any two vectors is not in the set. This is similar to a sum-free set in groups. In abelian groups, a sum-free set is a set where no two elements add up to another element. In our case, it's over GF(2), so addition is mod 2, and the condition is slightly different: we need that no three distinct elements a, b, c satisfy a + b + c = 0, which is equivalent to a + b = c. So it's exactly the same as a sum-free set in the group GF(2)^6. Therefore, the maximum size of a sum-free set in GF(2)^6. There is a theorem called the ErdsKoRado theorem for vector spaces, but I think that applies to intersecting families. Alternatively, there is a result for sum-free sets in vector spaces over GF(2). I recall that in the vector space GF(2)^n, the maximum size of a sum-free set is 2^{n-1}, but that might be when considering the entire space. Wait, but in our case, the vectors are not the entire space; they are specific vectors corresponding to numbers 1 to 15. So maybe this approach isn't directly applicable. Wait, let's list all the vectors for the numbers 1 to 15: 1: 000000 2: 100000 3: 010000 4: 000000 5: 001000 6: 110000 7: 000100 8: 100000 9: 000000 10: 101000 11: 000010 12: 010000 13: 000001 14: 100100 15: 011000 Note that 1, 4, 9 are all 000000. Then, some vectors repeat: - The vector 100000 is for 2 and 8. - The vector 010000 is for 3 and 12. - The vector 000000 is for 1, 4, 9. - The vector 001000 is for 5. - The vector 110000 is for 6. - The vector 000100 is for 7. - The vector 101000 is for 10. - The vector 000010 is for 11. - The vector 100100 is for 14. - The vector 011000 is for 15. - The vector 000001 is for 13. So the vectors are: 1: 000000 (multiple numbers: 1,4,9) 2: 100000 (2,8) 3: 010000 (3,12) 5: 001000 (5) 6: 110000 (6) 7: 000100 (7) 10: 101000 (10) 11: 000010 (11) 14: 100100 (14) 15: 011000 (15) 13: 000001 (13) So excluding the all-zero vector (000000), we have 14 vectors. But some vectors correspond to multiple numbers. For example, 100000 corresponds t RowBox[{"base", ",", "supscr"}], "]"}], ",", "subscr"}], "]"}], "]"}]}], ";", "\[IndentingNewLine]", RowBox[{ RowBox[{"PrintAs", "[", RowBox[{"Evaluate", "[", RowBox[{"Symbol", "[", RowBox[{"StringJoin", "[", RowBox[{"\"\\"", ",", RowBox[{"ToString", "@", "sym"}]}], "]"}], "]"}], "]"}], "]"}], "^=", RowBox[{"Evaluate", "[", RowBox[{"SubscriptBox", "[", RowBox[{ RowBox[{"SuperscriptBox", "[", RowBox[{ RowBox[{"OverscriptBox", "[", RowBox[{"base", ",", "\"\<.\>\""}], "]"}], ",", "supscr"}], "]"}], ",", "subscr"}], "]"}], "]"}]}], ";", "\[IndentingNewLine]", RowBox[{ RowBox[{"PrintAs", "[", RowBox[{"Evaluate", "[", RowBox[{"Symbol", "[", RowBox[{"StringJoin", "[", RowBox[{"\"\\"", ",", RowBox[{"ToString", "@", "sym"}]}], "]"}], "]"}], "]"}], "]"}], "^=", RowBox[{"Evaluate", "[", RowBox[{"SubscriptBox", "[", RowBox[{ RowBox[{"SuperscriptBox", "[", RowBox[{ RowBox[{"OverscriptBox", "[", RowBox[{"base", ",", "\"\<..\>\""}], "]"}], ",", "supscr"}], "]"}], ",", "subscr"}], "]"}], "]"}]}], ";", "\[IndentingNewLine]", RowBox[{ RowBox[{"PrintAs", "[", RowBox[{"Evaluate", "[", RowBox[{"Symbol", "[", RowBox[{"StringJoin", "[", RowBox[{"\"\\"", ",", RowBox[{"ToString", "@", "sym"}]}], "]"}], "]"}], "]"}], "]"}], "^=", RowBox[{"Evaluate", "[", RowBox[{"SubscriptBox", "[", RowBox[{ RowBox[{"SuperscriptBox", "[", RowBox[{ RowBox[{"OverscriptBox", "[", RowBox[{"base", ",", "\"\<...\>\""}], "]"}], ",", "supscr"}], "]"}],saken the domain of evil or persecution in the Cause of Allah. Let them pardon and overamut of industry and function. To determine the maximum number of rooks that cannis par l'officinal lors des bilans de mdication peuvent aider le patient prvenir la iatrognie mdicamenteuse ou les problmes d'observance.He was such a fascinating guy. Not only was he a novelist (and Pulitzer nominee), he was also an accomplished artist and poet. It showed in his work, too; every piece of prose he did was also a painting, also a poem, all at once. I loved his stories, it seemed that in his career he had known virtually everybody. Of particular interest was his long friendship with the poet Charles Bukowski- beginning in the 1950s at the height of the beat poetry movement in San Francisco. Jory wrote a memoir about that friendship that I highly recommend:Ol Pessoal!!!A promoo chegou ao fim!Quero agradecer a participao de todos!!!Quero agradecer mais uma vez a Editora Autntica e a autora Paula Pimenta pela parceria e apoio nessa promoo junto ao blog! Muito obrigada mesmo!!!Mas, vamos parar de enrolao e dizer que foi a ganhadora n?! O nmero sorteado foi o 491!! E para vocs entenderem, tiveram 72 participaes, porm com a divulgao quem participou poderia responder o questionrio novamente. Por isso o nmero alto de 506 nmeros para sorteio.E a ganhadora foi: Joana D'arc Gonalves Maciel!Parabns Joana!! Voc tem at 3 dias para responder meu e-mail com seus dados!!!Wybierz kategori ALARMOWE telefony (10) Auto, samochd (19) Myjnia samochodowa (6) Naprawa, serwis (10) Stacja benzynowa (5) Blog (44) Dzieci (146) *** Lato w Miecie *** (1) Przedszkola (22) a or b is null. But maybe the problem assumes that the inputs are valid. So assuming that the binary_and is not empty, then the sequences will have at least one match. Now, translating the code step by step. First, the binary_and part. In C#, the code can be: string binaryAnd = ""; int minLength = Math.Min(a.Length, b.Length); for (int i = 0; i < minLength; i++) { char aChar = a[i]; char bChar = b[i]; int andResult = (aChar - '0') & (bChar - '0'); binaryAnd += andResult.ToString(); } Wait, but in Python, the original code uses zip, which stops at the shorter one. So if a is longer than b, the remaining characters are ignored. So in C#, the binaryAnd will be the same length as the shorter of a and b. Then, the regex part. In C#, the regular expression for 0+ or 1+ is same as in Python: "0+|1+". But the RegexOptions may not be needed. So: using System.Text.RegularExpressions; ... MatchCollection matches = Regex.Matches(binaryAnd, "0+|1+"); Then, collect the matches into a list of strings: List sequences = new List(); foreach (Match match in matches) { sequences.Add(match.Value); } Then, find the shortest string in sequences. If there are no sequences, return empty? But according to the problem statement's example, the binaryAnd is "010 & 110". Let's compute that: a is '010', which is 0,1,0. b is '110', which is 1,1,0. AND each pair: 0&1 = 0, 1&1=1, 0&0=0. So binary_and is '010'. Then regex matches '0', '1', '0' sequences are ["0", "1", "0"]. The shortest is each of length 1, so returns the first, which is "0". Which matches the sample output. So in C#, when multiple sequences have the same minimal length, the first one is returned. So the code to find the shortest: string shortest = sequences[0]; foreach (string seq in sequences) { if (seq.Length < shortest.Length) { shortest = seq; } } But need to handle if sequences is empty. But according to the regex, if the binaryAnd is non-empty, the matches will have at least one. So if binaryAnd is empty, sequences is empty. But binaryAnd can be empty only if a and b have no overlapping characters (like a is empty, or b is empty, or they are of zero length). But perhaps the problem assumes that a and b are non-empty, and the binary_and is non-empty. So in code, perhaps need to check for that. But the original Python code would throw an ee case, then 2a +5b =10. Then, solving for positive integers a and b. But if that's the case, then 2a +5b =10. Let's explore that possibility. But the original problem says 10t. If it's supposed to be 10, then t=1. But if it's 10t, and t is given as a positive integer, then the number of pairs would depend on t. Since the problem doesn't specify t, but asks for the number of pairs (a, b), perhaps t is a positive integer, and we need to find the number of pairs for all t. But that's not possible unless the problem specifies a range for t. Alternatively, maybe t is a variable, and the equation must hold for some t, which would be equivalent to 2a +5b being a multiple of 10. Because 10t must be equal to 2a +5b, so t = (2a +5b)/10. Since t must be a positive integer, (2a +5b) must be a multiple of 10, and t is then (2a +5b)/10. Therefore, the problem reduces to finding the number of positive integer pairs (a, b) such that 2a +5b is divisible by 10, and the quotient t is a positive integer. Therefore, 2a +5b =10t, where t is a positive integer, and a, b are positive integers. So, in other words, we need to find all positive integers a, b, t such that 2a +5b =10t. But since t is determined by a and b, the problem is equivalent to finding the number of positive integer solutions (a, b) to 2a +5b 0 mod10. As we discussed earlier, this requires that b is even, and a is a multiple of 5. Let me verify this again. So, 2a +5b 0 mod10. We can analyze this congruence. First, 5b mod10 is equal to 0 if b is even (since 5*2k=10k0 mod10), and 5 if b is odd (since 5*(2k+1)=10k +55 mod10). Similarly, 2a mod10 can be 0,2,4,6,8 depending on a mod5. Therefore, for the equation 2a +5b 0 mod10: If b is even, then 5b 0 mod10, so 2a 0 mod10 a 0 mod5. If b is odd, then 5b 5 mod10, so 2a 5 mod10. But 2a mod10 can only be even numbers (0,2,4,6,8), so 5 mod10 is impossible. Therefore, there are no solutions when b is odd. Hence, all solutions must have b even and a divisible by 5. Therefore, let b =2k where k is a positive integer, and a =5m where m is a positive integer. 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Then, in such a case, we need to show that there exists a vertex adjacent to all others. Let me test this with n=1. So 3 people. For any single person, there must be another person adjacent to them. So in the line graph A-B-C: consider the single person A. There must be another person adjacent to A, which is B. For the single person B, adjacent to A and C. For single person C, adjacent to B. So the condition is satisfied. However, there is no one adjacent to everyone. A is adjacent only to B, B is adjacent to A and C, C is adjacent to B. So the conclusion fails here. But the problem says that in such a setup, the conclusion must hold. Therefore, there must be something wrong with my example. Wait, but according to the problem's statement, in the line graph with three nodes, the condition is satisfied, buCellGroupData[{ Cell[98592, 2782, 201, 3, 70, 29, 0, "StyleData", "ProofLabel", "All",ExpressionUUID->"6f026f3d-bb10-472b-92fa-4c3d231a7d86"], Cell[98796, 2787, 101, 0, 70, 46, 0, "StyleData", "ProofLabel", "NumberedHeads",ExpressionUUID->"e1d79f73-5ed6-4f70-be77-8c17b77a62c4"], Cell[98900, 2789, 115, 1, 70, 45, 0, "StyleData", "ProofLabel", "Presentation",ExpressionUUID->"9bedeeda-1867-40f6-b153-052c38b24909"], Cell[99018, 2792, 96, 0, 70, 41, 0, "StyleData", "ProofLabel", "Printout",ExpressionUUID->"6d77cc44-a2bd-48aa-8bb1-e65925c37580"], Cell[99117, 2794, 124, 1, 70, 54, 0, "StyleData", "ProofLabel", "NumberedHeadsPrintout",ExpressionUUID->"e088588b-f45a-4629-8f1e-31c1f94a33d0"] }, Closed]], Cell[CellGroupData[{ Cell[99278, 2800, 466, 11, 70, 36, 0, "StyleData", "ProofContinuation", "All",ExpressionUUID->"3be2fefc-a19b-4411-9433-3f08885e3a20", CounterIncrements->"ProofContinuation"], Cell[99747, 2813, 108, 0, 70, 53, 0, "StyleData", "ProofContinuation", "NumberedHeads",ExpressionUUID->"4b9c3e75-9c68-4bca-bee4-feb04fc5607d"], Cell[99858, 2815, 156, 2, 70, 52, 0, "StyleData", "ProofContinuation", "Presentation",ExpressionUUID->"0b0f2e72-00b1-4a24-93d8-940fb56fbf70"], Cell[100017, 2819, 138tion in this case. But this can't be. Let me check the algebra again. Starting with: -5/(2 T) + (22)/(5 T) = 0 Combine the terms: [ -5/(2) + 22/5 ] / T = 0 Since 1/T 0 (unless T approaches infinity, but T = tan(/2), which would require /2 = /2 + k, = + 2k, but then tan(/2) is undefined). So, the numerator must be zero: -5/2 + 22/5 = 0 Multiply numerator and denominator by 52 to rationalize: -5/2 * 52 + 22/5 * 52 = (-25) + (2*2) = -25 + 4 = -21 0 Yes, so this gives -21 = 0, which is impossible. Therefore, Case 2 has no solutions. Therefore, the only possible case is Case 1: sin + cos = 0, leading to tan = -1. But the problem states that there are at least three possible values. Hmm, this suggests that maybe I made a mistake in my reasoning, because so far I only get one solution. Let me check my steps again. Wait, perhaps can be in different branches, leading to multiple solutions for tan . Wait, if sin + cos = 0, then = -/4 + k. Therefore, tan = -1 for all these . Therefore, tan(x + y) = -1 is the only solution. But the problem states that there are at least three such values. Therefore, there must be more solutions. Therefore, my approach must have missed something. Alternatively, perhaps there was an error in substituting or expanding the equations. Let me double-check. Wait, when I set up the determinant equation A D - B C = 0, leading to either sin + cos = 0 or the othe AMD + angle BMC = 180, maybe points A, M, D, and some transformation of B, M, C lie on a circle. Not sure. Alternatively, use the concept of orthogonality. If two lines are such that the product of their slopes is -1, but I don't see a direct connection. Wait, maybe considering the reflection of point M over the diagonals. In a rhombus, the diagonal is injective. Then all k(y) are distinct, so there are p distinct multiples ofp in S. But in our p=5 case, there are only 4, so injectivity fails. To show that k(y) cannot be injective, observe that the values of k(y) are bounded. Let's find the maximum and minimum possible k. For y in [0,p-1], y (p-1)^2. For x in [0,p-1], x (p-1)^3. Therefore, the minimum value of y -x -1 is 0 - (p-1)^3 -1 = - (p-1)^3 -1. The maximum value is (p-1)^2 -0 -1 = (p-1)^2 -1. Therefore, kp must lie in this interval. The number of possible k's is roughly ((p-1)^2 -1)/p - (-(p-1)^3 -1)/p O(p). But since p is a prime, this range is too large. However, for each y, the corresponding k is (y -x -1)/p. But x is determined by y: x y -1 modp. Therefore, x= (y -1)^{1/3} modp. Since x is in [0,p-1], x is uniquely determined. Therefore, x= ((y -1) modp)^{1/3}. But then, y -x -1 is congruent to 0 modp, so y -x -1 = lp for some integer l. So l=(y -x -1)/p. To find the possible range of l, note that x and y are between 0 and p-1. So y (p-1)^2, and x 0. Therefore, y -x -1 (p-1)^2 -1 = p -2p. Similarly, the minimum is when y=0 and x=p-1: 0 - (p-1)^3 -1. Therefore, l ( - (p-1)^3 -1 ) /p -p, and l (p -2p)/p =p-2. So l ranges from approximately -p to p-2. But this is a huge range. However, the actual values of l are determined by specifge length is 2, so each edge is 2. If we make 4 cuts along the height, we divide the height (which is 2) into 5 equal parts. So each layer would have a height of 2/5. Then, after that, cutting 5 times vertically. Vertical cuts would be perpendicular to the horizontal ones. If horizontal cuts were along the height, then vertical cuts could be along the length or width. Wait, but vertical is usually up and down, so if the cube is oriented with height being vertical, then vertical cuts would be along the height, but we already did horizontal cuts. Hmm, maybe vertical cuts here mean along the length or width. Wait, maybe "vertical" here refers to the direction. So, after making horizontal cuts (along the height), we make vertical cuts along, say, the width. So, cutting 5 times vertically would divide the width into 6 parts. Each vertical cut is along the width, so each segment would be 2/6 = 1/3 Chinese feet. Wait, but hold on, the original cube is 2 units in all edges. If we cut 4 times along the height (horizontal cuts), we get 5 layers, each 2/5 units in height. Then, cutting 5 times vertically (along, say, the width), each cut would divide the width into 6 parts, each 2/6 = 1/3 units. Then, perhaps we also need to cut along the remaining dimension, which is length? But the problem says it's cut 4 times horizontally and 5 times vert c. So, we need to see if both directions hold. Let me try to visualize triangle ABC with points A' and B' on two of its sides. The line A'B' intersects two sides of ABC. Let me assume that A' is on BC and B' is on AC, so that line A'B' creates triangle A'B'C and quadrilateral ABB'A'. Alternatively, maybe A' is on AB and B' is on AC? The problem states that A'B' intersects two sides of ABC. Wait, triangle ABC has three sides. If a line intersects two sides, it can be either two sides adjacent to a vertex or two opposite sides. But in a triangle, all sides are connected, so a line cutting two sides would divide the triangle into a smaller triangle and a quadrilateral. Assuming that line A'B' intersects sides BC and AC. So, point A' is on BC and B' is on AC. Then, triangle A'B'C would have vertices A', B', and C. The quadrilateral would be ABB'A', which includes vertices A, B, B', A'. So, the sides of the quadrilateral would be AB, BB', B'A', and A'A. Now, the original problem says that if the perimeter of A'B'C is a + b - c, then the quadrilateral ABB'A' is circumscribed. Let me try to recall if there's a theorem related to perimeters and tangential quadrilaterals. Wait, in tangential quadrilaterals, we have the Pitot theorem: in a convex quadrilateral, the sum of two opposite sides equals the sum of +1, details = dts `addTick` tick } doSimplTick _ _ (VerySimplCount n) = VerySimplCount (n+1) -- Don't use Map.unionWith because that's lazy, and we want to -- be pretty strict here! addTick :: TickCounts -> Tick -> TickCounts addTick fm tick = case Map.lookup tick fm of Nothing -> Map.insert tick 1 fm Just n -> n1 `seq` Map.insert tick n1 fm where n1 = n+1 plusSimplCount sc1@(SimplCount { ticks = tks1, details = dts1 }) sc2@(SimplCount { ticks = tks2, details = dts2 }) = log_base { ticks = tks1 + tks2, details = Map.unionWith (+) dts1 dts2 } where -- A hackish way of getting recent log info log_base | null (log1 sc2) = sc1 -- Nothing at all in sc2 | null (log2 sc2) = sc2 { log2 = log1 sc1 } | otherwise = sc2 plusSimplCount (VerySimplCount n) (VerySimplCount m) = VerySimplCount (n+m) plusSimplCount _ based on the SQL dialect (MySQL, PostgreSQL, etc.). Finally, the ORDER BY clause: the user might want the years in ascending or descending order. The example uses ORDER BY publication_year DESC to show the latest years first, but that can be adjusted. To write a SQL query that finds the most popular books from the last 10 years (ordered by year), follow these steps. This example assumes a database schema with tables for `books` and a popularity metric (e.g., `sales` or `reviews`): ### Tables (Example Schema) 1. **`books`**: Contains book details. - `book_id` (unique identifier) - `title` - `publication_year` 2. **`sales`**: Tracks sales data (replace with `reviews` for ratings-based popularity). - `sale_id` - `book_id` (foreign key to `books`) - `quantity_ss there are no values \( a \) and \( b \) that satisfy both conditions. Thus, there is no line that is normal to both curves \( y = \cosh x \) and \( y = \sinh x \). \[ \boxed{\text{No}} \]Prove that if \( a \) and \( b \) are positive numbers, then at least one of the numbers \( oint in worker/wrappering if the thing is never inlined! -- Because the no-inline prag will prevent the wrapper ever -- being inlined at a call site. -- -- Furthermore, don't even expose strictness info = return [ (fn_id, rhs) ] | otherwise = do let doSplit | is_fun = splitFun dflags fam_envs new_fn_id fn_info wrap_dmds res_info rhs | is_thunk = splitThunk dflags fam_envs is_rec new_fn_id rhs -- See Note [Thunk splitting] | otherwise = return Nothing try s c of D^{-length(c)} 1. So, for our case, if we model the numbers as strings over an alphabet of size 10, but with the first digit restricted to 1-9, perhaps the Kraft-McMillan sum would be sum_{c} 10^{-length(c)} AbsoluteThickness[1.6], EdgeForm[ Directive[ GrayLevel[0], Thickness[Small], Opacity[0.5599999999999999]]], RGBColor[ 0.9058823529411765, 0.5411764705882353, 0.7647058823529411]], RectangleBox[{0, 0}, {10, 10}, "RoundingRadius" -> 0]}, AspectRatio -> Full, ImageSize -> {10, 10}, PlotRangePadding -> None, ImagePadding -> Automatic, BaselinePosition -> (Scaled[0.1] -> Baseline)], #4}}, GridBoxAlignment -> { "Columns" -> {Center, Left}, "Rows" -> {{Baseline}}}, AutoDelete -> False, GridBoxDividers -> { "Columns" -> {{False}}, "Rows" -> {{False}}}, GridBoxItemSize -> {"Columns" -> {{All}}, "Rows" -> {{All}}}, GridBoxSpacings -> { "Columns" -> {{0.3}}, "Rows" -> {{0.5}}}], "Grid"]}}, GridBoxAlignment -> {"Columns" -> {{Left}}, "Rows" -> {{Top}}}, AutoDelete -> False, GridBoxItemSize -> { "Columns" -> {{Automatic}}, "Rows" -> {{Automatic}}}, GridBoxSpacings -> {"Columns" -> {{1}}, "Rows" -> {{0}}}], "Grid"], Alignment -> Left, AppearanceElements -> None, ImageMargins -> {{5, 5}, {5, 5}}, ImageSizeAction -> "ResizeToFit"], LineIndent -> 0, StripOnInput -> False], { FontFamily -> "Arial"}, Background -> Automatic, StripOnInput -> False], TraditionalForm]& ), Editable->True, InterpretationFunction:>(RowBox[{"SwatchLegend", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{"Directive", "[", RowBox[{ RowBox[{"EdgeForm", "[", RowBox[{"Directive", "[", RowBox[{ RowBox[{"Thickness", "[", "Small", "]"}], ",", RowBox[{"Opacity", "[", "0.5599999999999999`", "]"}]}], "]"}], "]"}], ",", InterpretationBox[ ButtonBox[ TooltipBox[ GraphicsBox[{{ GrayLevel[0], RectangleBox[{0, 0}]}, { GrayLevel[0], RectangleBox[{1, -1}]}, { GrayLevel[1], RectangleBox[{0, -1}, {2, 1}]}}, DefaultBaseStyle -> "ColorSwatchGraphics", AspectRatio -> 1, Frame -> True, FrameStyle -> GrayLevel[0.6666666666666problem statement. Original problem: find all possible valid pairs of numbers from the set [1-10] that have the same sum. The pairs cannot contain the same number twice, and each number can only be used once. So perhaps the code needs to generate all possible pairs (each pair is two distinct numbers), group them by their sum, and then for each sum, list all possible pairs. But the constraints that numbers can only be used once would come into play when considering a collection of pairs (e.g., when you have multiple pairs, they can't share numbers). But perhaps the user wants all possible pairs for each sum, regardless of overlapping numbers, but the pairs themselves are valid (i.e., no duplicates). But in that case, the code is simply to generate all possible pairs, group them by sum. But why the constraints? Alternatively, the problem might be to find all possible sets of pairs (with the same sum) such that all pairs are mutually disjoint (i.e., form a matching). For example, for sum 7, the possible pairs are (1,6), (2,5), (3,4). So possible sets are: { (Hip With Femur ApX-Ray Lt Hip With Femur Ap/LatX-RAY LT HUMERUS AP/LATx-ray LT Knee AP/LATX-ray LT Knee with Femur AP/LATX-Ray Lt Shoulder Jt AP/LATX-Ray LT Thumb A/CX-RAY LT WRIST AP/LAT/OBLIQUEX-Ray Nasal BoneX-Ray Neck AP/LATX-Ray Neck Cervical Spine Flexion & ExtensionX-Ray of KubX-Ray of P.B.H. Ap/Frog ViewX-Ray PBH APx-Ray PBH Ap/LatX-Ray Pelvis With Both HipX-Ray Right Ankle Ap/LatX-Ray Right foot AP/LateralX-Ray Right HIP APX-Ray Right HIP AP/LateralX-Ray Right Hip PAX-Ray Right Shoulder APX-Ray Right Tibia Fibula AP LateralX-RAY RT ELBOW AP/LATX-ray Rt Foram AP/LATX-Ray RT Hand APX-RAY RT y. While surface water resources are deficient in the Panhandle, Management Region 1 contains an abundance of ground water in the Ogallala Aquifer, which has been utilized since the early 1960s and today completely dominates agriculture in Texas County.4Science-publishing ventures continually battle for market space, yet most operate on one of only two basic business models. Either subscribers pay for access, or authors pay for each publication--often thousands of dollars--with access being free. But in what publishing experts say is a radical experiment, annt. This is a very common occurrence and one of the biggest problems with our real estate market. In this case the buyer overpays for the property, which will raise the value of all the hou299 L .70636 .5061 L .70878 .5103 L .71042 .51544 L .71153 .52123 L .71241 .52728 L .7134 .5332 L .71483 .53859 L .71697 .54313 L .71999 .54662 L .72396 .54896 L .72881 .55025 L .73437 .55069 L .74037 .5506 L .74649 .55037 L .75238 .55041 L .75773 .5511 L .7623 .55272 L .76595 .55545 L .76867 .5593 L .77055 .56415 L .7718 .56976 L .772uary 2016 December 2015 November 2015 October 2015 September 2015 August 2015 July 2015 June 2015 May 2015 April 2015 March 2015 February 2015 January 2015 December 2014 November 2014 October 2014 September 2014 August 2014 July 2014 June 2014 May 2014 April 2014 March 2014 February 2014 January 2014 December 2013 November 2013 October 2013 September 2013 August 2013 July 2013 June 2013 May 2013 April 2013 March 2013 February 2013 January 2013 December 2012 November 2012 October 2012 September 2012 August 2012 July 2012 June 2012 May 2012 April 2012 March 2012 February 2012 January 2012 December 2011 November 2011 October 2011 September 2011 August 2011 July 2011 June 2011 May 2011 April 2011 March 2011 February 2011 January 2011 December 2010 November 2010 October 2010 September 2010 August 2010 July 2010ce is defined for 1 < x < 2.5, so open interval on both ends. So at x=1, the first piece covers x=1, and the second piece starts just above 1. Similarly, at x=2.5, the third piece starts at x=2.5, and the second piece ends just below 2.5. So for x=1, the left limit (from left side) is 21 = 2, and the right limit (from right side, using 4 - 2x) is 4 - 2*1 = 2. So they match, and f(1) is 2. So continuous there. At x=2.5: the left limit is from 4 - 2x as x approaches 2.5 from the left: 4 - 2*(2.5) = -1. The right limit is from 2x -7 as x approaches 2.5 from the right: 2*(2.5) -7 = -2. And f(2.5) is -2. So since left limit (-1) right limit (-2), there's a jump discontinuity at x=2.5. The jump is RHL - LHL = (-2) - (-1) = -1. So the jump is -1. So direction matters here; the function jumps down by 1 unit. Are there any other points? For example, between 0 and 1, the function is 2x, whiever, in the original problem statement, when all b_k are equal, the sum is larger than 4/(n+1). So maybe the bound 4/(n+1) is not tight and the actual minimal sum is higher. But this contradicts the problem's requirement. Therefore, there must be a mistake in my previous reasoning. Wait, let's revisit the system of equations for n=2. If the problem states that there exists a non-zero solution, then the determinant of the system must be zero. For n=2, as above, we have: Determinant condition: (b1 -2)(b2 -2) -1=0 => (b1 -2)(b2 -2)=1 Lets minimize S = b1 + b2 with this constraint. Lets set variables u = b1 -2, v = b2 -2, so that uv=1, and we need to minimize S = (u +2) + (v +2) = u + v +4. Given that uv=1, u and v are positive (since b1 and b2 are positive). Therefore, u + v >= 2(uv)=2, by AM >= GM. Therefore, S >=2 +4=6. Therefore, the minimal sum is indeed 6, which is greater than 4/3. Therefore, the problem's claim that S >=4/(n+1) holds (since 6 >=4/3), but it's a much weaker inequality. This suggests that either there is a mistake in the problem statement, or my approach is missing something. Alternatively, perhaps the problem requires a different approach that gives a lower bound of 4/(n+1), regardless of the actual minimal sum. Alternatively, perhaps the bound 4/(n+1) is tight in some specific case. For example, when n=1, 4/(1+1)=2, which matches the minimal sum. For n=1, the equation is x0 -2x1 +x2 +b1x1=0, but x0=x2=0, so -2x1 +b1x1=0 => (b1 -2)x1=0. Since x10, b1=2, which gives sum b1=2=4/(1+1). So for n=1, equality holds. For n=2, the minimal sum is 6, which is greater than 4/3. So perhaps the inequality is tight only for n=1, but for larger n, it's a lower bound that is not tight. But the problem says "Prove that b1 + ... + bn >=4/(n+1)". So perhaps the bound is correct, and my previous approach missing something. Let me revisit the original equation. Sum_{k=1}^n b_k x_k^2 = Sum_{k=1}^n (x_k - x_{k+1})^2 + x_1^2 Wait, but x_{n+1}=0, so Sum_{k=1}^n (x_k - x_{k+1})^2 includes the term (x_n -0)^2 =x_n^2. Therefore: Sum_{k=1}^n (x_k - x_{k+1})^2 = Sum_{k=1}^{n-1} (x_k -x_{k+1})^2 +x_n^2 Thus, the equation becomes: Sum b_k x_k^2 = Sum_{k=1}^{n-1} (x_k -x_{k+1})^2 +x_n^2 +x_1^2 So, both x_1dataName mk_ghc_ns TH.TcClsName = OccName.tcClsName mk_ghc_ns TH.VarName = OccName.varName mk_mod :: TH.ModName -> ModuleName mk_mod mod = mkModuleName (TH.modString mod) mk_pkg :: TH.PkgName -> PackageId mk_pkg pkg = stringToPackageId (TH.pkgString pkg) mk_uniq :: Int# -> Unique mk_uniq u = mkUniqueGrimily (I# u) \end{code} Note [Binders in Template Haskell] ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Consider this TH term construction: 0, 468, 13, 173, "Text",ExpressionUUID->"c9281578-6e6b-4d76-be00-28e15a515db8"], Cell[CellGroupData[{ Cell[1051, 37, 249, 4, 28, "Input",ExpressionUUID->"70b4f545-ce60-422f-867b-de8675c3afd9"], Cell[1303, 43, 292, 5, 32, "Output",ExpressionUUID->"5656d0ea-cdef-4ada-b39f-5ac6d8b62bfc"] }, Open ]], Cell[1610, 51, 174, 4, 28, "Input",ExpressionUUID->"0878174c-c4bd-4dfe-9881-1556e5510942"], Cell[CellGroupData[{ Cell[1809, 59, 5881, 126, 200, "Input",ExpressionUUID->"48772906-8c5c-464c-88d9-226999ef6cfc"], Cell[7693, 187, 3718, 73, 287, "Output",ExpressionUUID->"a79ca08b-0cde-4de8-adfd-1e7a4b9016a3"] }, Open ]], Cell[CellGroupData[{ Cell[11448, 265, 520, 8, 28, "Input",ExpressionUUID->"452b0ceb-7bae-4c80-bdb9-cb9umn5 of the7th bank. Therefore, sum is133+134+135+136= let's compute:133+136=269;134+135=269; total=269*2=538. Correct. Therefore, the locker numbers are133,134,135,136. Therefore, part (c) answer is133,134,135,136. Part(d): Explain why2013 cannot be a sum of any column of locker numbers. We need to show that2013 cannot be the sum of a large column (2 lockers) or a small column (4 lockers). Let's check both. First, for a large column: sum=2a+1=2013. Then2a=2012,a=1006. So lockers1006 and1007. We need to chNewTyCon{} = eqCon (data_con tc1) (data_con tc2) eqAlgRhs _ _ = False eqCon c1 c2 = dataConName c1 == dataConName c2 && dataConIsInfix c1 == dataConIsInfix c2 && dataConStrictMarks c1 == dataConStrictMarks c2 && dataConFieldLabels c1 == dataConFieldLabels c2 && eqType (dataConUserType c1) (dataConUserType c2) ---------------- missingBootThing :: Name -> String -> SDoc missingBootThing name what = ppr name <+> ptext (sLit "is exported by the hs-boot file, but not") <+> text what <+> ptext (sLit "the module") bootMisMatch :: TyThing -> IfaceDecl -> IfaceDecl -> SDoc bootMisMatch thing boot_decl real_decl = vcat [ppr thing <+> ptext (sLit "has conflicting definitions in the module and its hs-boot file"), ptext (sLit "Main module:") <+> ppr real_decl, ptext (sLit "Boot file: ") <+> ppr boot_decl] instMisMatch :: Instance -> SDoc instMisMatch inst = hang (ppr inst) 2 (ptext (sLit "is defined in the hs-boot file, but not in the module itself")) \end{code} %************************************************************************ %* * Type-checking the top level of a module %* * %************************************************************************ tcRnGroup takes a bunch of top-level source-code declarations, and * renames them * gets supporting declarations from interface files * typechecks them * zonks them * and augments the TcGblEnv with the reressionUUID->"44888042-849b-4bd2-9bc0-e0191ecfbbf8"], Cell[102218, 2589, 6927, 134, 253, "Output",ExpressionUUID->"710a01aa-2b59-4582-8295-a551de2530ff"] }, Open ]], Cell[109160, 2726, 179, 3, 35, "Text",ExpressionUUID->"f707979e-02fc-4796-bd9a-fd5bdff31a97"], Cell[109342, 2731, 1205, 24, 80, "Input",ExpressionUUID->"8184cce4-49c2-498a-b34f-ed34954dbfa0"], Cell[CellGroupData[{ Cell[110572, 2759, 766, 13, 44, "Input",ExpressionUUID->"03b08875-f604-4d86-a17f-a4db7aca3acc"], Cell[111341, 2774, 12127, 221, 253, "Output",ExpressionUUID->"53e8d264-95ff-46a3-af06-29458d1426c5"] }, Open ]], Cell[CellGroupData[{ Cell[123505, 3000, 249, 4, 44, "Input",ExpressionUUID->"24e2eb60-9f44-41e2-a31d-3e31fceb0e7d"], Cell[123757, 3006, 256, 3, 34, "Output",ExpressionUUID->"585ec1da-47b4-4de0-a774-b708bc4a68f2"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[124062, 3015, 164, 3, 69, "Chapter",ExpressionUUID->"23968627-f164-4801-81ae-e995aa8b1768"], Cell[124229, 3020, 358, 7, 35, "Text",ExpressionUUID->"d5138ac6-1154-43aa-81e1-b13451a64279"], Cell[CellGroupData[{ Cell[124612, 3031, 410, 10, 44, "Input",ExpressionUUID->"baea0f07-039a-4e57-a862-93b508eb9384"], Cell[125025, 3043, 397, 6, 34, "Output",ExpressionUUID->"0bc3607e-f10d-4777-8d73-f69c013161e0"] }, Open ]], Cell[CellGroupData[{ Cell[125459, 3054, 485, 11, 44, "Input",ExpressionUUID->"c97c9070-54e5-4d33-ba5b-81501deb1b86"], Cell[125947, 3067, 427, 6, 34, "Output",ExpressionUUID->"e63799de-f525-4bd8-9c46-599f25aec08d"] }, Open ]], Cell[126389, 3076, 198, 3, 35, "Text",E [] = left > mergeList left@(a:l) right@(b:r) = > if a <= b then a : mergeList l right > else b : mergeList left r Ord needed to define merge. Type classes are global and coherent - there is only one 'Ord a' instance - guaranteed that merge always uses the same comparison function for 'a'. If 'Ord a 'holds, then 'SortedList a' represents lists of 'a' sorted according to order defined by unique 'Ord a' instance. In contrast, fun arg defining an order is local to fun call: sortBy :: (a->a->Ordering) -> [a] -> [a] therefore could change order for each call - could not state that 'SortedList a' are sorted. > -- | Sort lists with the SortedList interface > fromList :: Ord a => [a] -> SortedList a > fromList [] = nil > fromList [a] = singleton a > fromList l = merge orderedLeft orderedRight -- 3. then merge the parts > where > orderedLeft = fromList left -- 2. recursively sort parts > orderedRight = fromList right > (left,right) = splitAt (div (length l) 2) l -- 1. split in two > > sort :: Ord a => [a] -> [a] > sort l = forget (fromList l) Note: sort can be defined in t. Let me denote A, B, C, D as points in the plane with coordinates such that H is (0,0,0). Lets suppose that the diagonals AC and BD intersect at H. So, coordinates: Lets suppose point A is (a, b, 0), then point C must be (-a, -b, 0) because H is the midpoint? Wait, but in general quadrilaterals, the diagonals intersecting at H don't necessarily bisect each other unless it's a parallelogram. But in cyclic quadrilaterals, the diagonals don't necessarily bisect each other. So maybe that's not the case. Wait, but if H is the intersection point, then coordinates can be set such that H is (0,0,0), and AC and BD are two lines intersecting at H. Let me parameterize the points. Lets assume that AC is along toving the outer loop is not allowed. However, the problem says "optimize the nested loop", which implies that the loop structure is part of the code to be optimized. So perhaps the main approach is to optimize within the existing loop structure. So considering all that, the main optimizations are: - Remove the unnecessary code (Random). - Optimize the summation in the inner loop by using a local variable and adding once per outer loop iteration. These two changes should significantly reduce the execution time. So the final optimized code would look like the code I wrote earlier: removing the Random lines and using a local sum variable for the inner loop. To optimn processes. The author argues that writers should be treated as associate designers and involved early on. The conclusion also stresses that writing is design. So options c and d are close. However, the main point isn't just involving writers in design (d) but the necessity of collaboration between the disciplines (c). The author states, "the team will begin to find exciting and creative ways of conjoining the two disciplines into a more unified experience." So the answer is c. 2. **According to the article, what is the role of a game writer?** The options are a, b, c, d. The article says, "Game writers simply want to help designers craft an immersive, interactive narrative experience." So answer c. Option d mentions pacing, setting, etc., which the writer wants input on, but the role is more about helping create the narrative experience. The article also refutes that writing is just arranging words (b) and that writers want to avoid melodramatic cutscenes (a). So answer c. 3. **What is one important aspect of pre-production mentioned?** Options include a, b, c, d. The article lists several pre-production tasks: High Level Narrative Summary, MajDraw " **Explanation:** The game ends in a draw since there are no moves to make. **Constraints:** * `1 <= moves.length <= 9` * `moves[i].length == 2` * `0 <= rowi, coli <= 2` * There are no repeated elements on `moves`. * `moves` follow the rules of tic tac toe. Okay, I need to solve this Tic-tac-toe problem. Let me read the problem carefully. The task is to determine the winner of a Tic-tac-toe game based on the given moves. The players take turns, with A starting first using 'X' and B using 'O'. The winner is th equals the fifth, and the third equals the fourth. This translates to: 1. The first hour digit (H1) must equal the second second digit (S2). 2. The second hour digit (H2) must equal the first second digit (S1). 3. The first minute digit (M1) must equal the second minute digit (M2). ### Key Steps: 1. **Valid Hours**: For each hour (HH), the seconds formed by reversing the hour digits (H2H1) must be valid (00-59). 2. **Valid Minutes**: Minutes must be palindromic (00, 11, 22, 33, 44, 55), giving 6 valid options. 3. **Count Valid Hours**: - *quals 1. So Jamie should have $1 left. Wait, let me double-check to make sure I didn't make any mistakes. Canoe rental: 3 hours at $12 per hour is 3*12=36. Then 5+8=13 for the refreshments. Total spent: 36+13=49. Starting with 50, subtract 49 gives 1. Yeah, that seems right. So the answer should be $1 left. **Final Answer** Jamie has $\boxed{1}$ dollar left after the canoeing and purchases. Jamie rented a canoe for 3 hours at a cost of $12 per hour. The total cost for the canoe rental is calculated as follows: \[ 3 \text{ hours} \times 12 \text{ dollars/hour} = 36 \text{ dmum number of good pairs is 13. But wait, this contradicts the initial assumption that there would be three bad pairs from the high primes. However, by pairing 1 with 17 and adjusting other pairings, we managed to pair all other numbers into good pairs except for 19,23,27,29. Since 27 is a composite, it's not a high prime, but it still can't form a good pair with the remaining numbers. So, the bad pairs are two:19-23 and 27-29, or some other combination. Either way, two bad pairs. Thus, total good pairs:13. But earlier I thought the answer was 12. So, which is correct? To verify, let's list all the pairs explicitly: Good pairs: 1-17 2-4 3-6 5-25 7-21 11-22 13-26 8-16 9-18 10-20 12-24 14-28 15-30 Total:13 good pairs. Remaining numbers:19,23,27,29. These form two pairs, which are bad. Yes, that's 13 good pairs. Therefore, the maximum number of good pairs is 13. But how did we get here? By realizing that pairing 5 with 25 instead of 10 allows 10 to pair with 20 and 15 to pair with 30, thus creating two additional good pairs while only losing one (5-10 vs. 5-25). Therefore, net gain of one good pair. This seems correct. Therefore, the answer should be 13. But I need to confirm that all these pairings are valid and no number is used twice. Let's list all pairs: 1. 1-17 2. 2-4 3. 3-6 4. 5-25 5. 7-21 6. 11-22 7. 13-26 8. 8-16 9. 9-18 10. 10-20 11. 12-24 12. 14-28 13. 15-30 Remaining numbers:19,23,27,29. Each number is used exactly once. So, this works. Therefore, the maximum number of good pairs is 13. But wait, I'm now confused because initial analysis suggested 12, but with this configuration, it's 13. So, which is correct? I think this configuration is valid and achieves 13 good pairs. Therefore, the answer is 13. But to be thorough, let's see if we can find another are no. But the sum for part2 is28, which matches the required sum. However, the modulo3 residue count invariant makes it impossible. Therefore, the correct answers are: 1. No 2. No But I need to verify this conclusion. Wait, let's check a smaller case. Suppose we C), the valid paths are: 1. A C B 2. A C D B Similarly, Case 2: Starting with A D. By symmetry, this should mirror Case 1. So: 3. A D B 4. A D C B Additionally, are there longer paths? Wait, perhaps she can take more edges before reaching B. For example, starting from A C D C B? Wait, but she can't traverse the same edge twice. Let's see: A C (edge AC used) D (edge CD used) C (edge DC already used, so she can't go back. Wait, edge CD is already used, so she can't go back from D to C. So from D, she can go to A or B. If she goes to A, as before, stuck. If she goes to B, that's the path already considered. Wait, maybe another path: A C D B. But that's length 3. Alternatively, is there a path that goes through more edges? Wait, let's consider another possibility. From A C D B, that's 3 edges. Alternatively, from A C D A ... but as before, after A C D A, she can't go anywhere else. So no. Wait, another idea: starti, _, _, _, _, _, 2.06147896917599, _, _, _, 0.0479699760130452, _, _, _, _, _, _, _, _, 0.0947099526411406, _, _, _, _, 0.0297249851362887, _, _, _, _, _, _, _, _, 0.389909805029111, _, _, _, _, _, _, 1.31732934148632, _, _, _, _, 0.107009946490639, _, _, _, _, _, _, _, _, _, 3.88187805890181, _, _, _, 0.342759828606033, _, _, _, _, _, _, 0.370229814869913, _, _, _, _, _, _, _, _, 0.248869875554859, _, _, _, _, _, 0.984409507959885, _, _, _, _, _, _, _, 9.17497541172612, _, _, _, _, _, 5.7203171379669, _, _, _, _, _, _, 9.39432530101718, _, _, _, _, _, 4.33451783092337, _, _, _, _, _, _, 3.509188245467, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _e [f(n) - f(n-1)] is non-negative. Hence, the inequality holds for all x>0 and n>1. Therefore, the original inequality is proven. Wait, let me verify this with another approach. Maybe using induction on n? Let's see. For n=2, we already checked it's true. Suppose it's true for some n=k, i.e., (x^{k-1} -1)/(k-1l abstract class with a method makeSound(). Subclasses like Dog, Cat, and Cow can override this method. Wait, the user mentioned abstract classes. So the abstract class should have an abstract method that the subclasses implement. Abstract classes can't be instantiated, so the subclasses must provide concrete implementations. Next, code reusability and maintainability. With dynamic polymorphism, you can write code that works with the abstract class, and any new subclass can be added without changing existing code. This follows the open/closed principle. For example, if I have a method that takes an Animal parameter and calls makeSound(), adding a new Animal subclass doesn't require changing that method. In contrast, static polymorphism (method overloading) requires knowing the types at compile time, so adding new types would need new overloads, which is less maintainable. Now, the program. The abstract class Animal with an abstract method makeSound(). Then three subclasses: Dog, Cat, Cow each overriding makeSound(). The main method creates instances of each, stores them in Animal references, and calls makeSound() to show dynamic polymorphis(FontHeight). (Cant be true if HasFontClass is true). HasFontClass UB[1] 0 = no fontClass, 1 = fontClass and Height specified for this text. (can't be true if HasFont is true). Supported in Flash Player 9.0.45.0 and later. AutoSize UB[1] 0 = fixed size. 1 = sizes to content (SWF 6 or later only). HasLayout UB[1] Layout information provided. NoSelect UB[1] Enables or disables interactive text selection. Border UB[1] Causes a border to be drawn around the text field. WasStaticnumber? Wait, but 315 is larger. 315 is higher. 315 vs. 195. 315 is larger. Similarly, 375: remove second digit: 75. 375/75=5. So 375 is valid. That's even larger. So 375. Then 395: remove second digit: 95. 395/95=4.157..., nope. Remove third digit: 39. 395/3910.128, nope. So 375 is valid. 425: remove second digit: 45. 425/459.444, nope. Remove third digit: 42. 425/4210.119, nope. 495: remove second digit: 45. 495/45=11. So 495 is valid. So 495 is even larger. Wait, 495: removing the second digit (9) gives 45. 495 divided by 45 is 11. So that works. So 495 is valid. Then 595: remove second digit: 55. 595/55=10.818... No. 595/55 isn't integer. 555: remove second digit: 55. 555/55=10.09... No. 585: remove second digit: 55. 585/5510.636, nope. 495 seems better. What about 735? Remove second digit: 75. 735/75=9.8. No. Remove third digit: 73. 735/7310.068, nope. 765: remove second digit: 75. 765/75=10.2, nope. 795: remove second digit: 75. 795/75=10.6, nope. Hmm. 495 is better. 495 is three-digit. Then, maybe 945? Remove second digit: 95. 945/95=9.947, nope. Remove third digit: 94. 945/9410.053, nope. So 495 is the largest three-digit number? Wait, 315, 375, 495. 495 is 495, which is bigger. Okay. But maybe there's a four-digit number. Let's think about four-digit numbers. Let me denote a four-digit number as ABCD, with D 0. We can remove any of the digits B, C, or D. After removing one digit, the remaining number is a three-digit number, and the original four-digit number must be divisible by this three-digit number. To maximize the number, we want the four-digit number to be as large as possible. Let's start with 9999. But the last digit is 9, so that's okay. Let's check if removing a digit (not the first) gives a three-digit number that divides 9999. Let's remove the second digit: 999. 9999 divi15.00 GBP Blue / PRODUCT - 15.00 GBP Blue / DESCRIPTION - 15.00 GBP Blue / confirmed - 15.00 GBP Green / PLEASE READ - 15.00 GBP Green / PRODUCT - 15.00 GBP Green / DESCRIPTION - 15.00 GBP Green / confirmed - 15.00 GBP Orange / PLEASE READ - 15.00 GBP Orange / PRODUCT - 15.00 GBP Orange / DESCRIPTION - 15.00 GBP Orange / confirmed - 15.00 GBP Pink / PLEASE READ - 15.00 GBP Pink / PRODUCT - 15.00 GBP Pink / DESCRIPTION - 15.00 GBP Pink / confirmed - 15.00 GBP Purple / PLEASE READ - 15.00 GBP Purple / PRODUCT - 15.00 GBP Purple / DESCRIPTION - 15.00 GBP Purple / confirmed - 15.00 GBP Red / PLEASE READ - 15.00 GBP Red / PRODUCT - 15.00 GBP Red / DESCRIPTION - 15.00 GBP Red / confirmed - 15.00 GBP Yellow / PLEASE READ - 15.00 GBP Yellow / PRODUCT - 15.00 GBP Yellow / DESCRIPTION - 15.00 GBP Yellow / confirmed - 15.00 GBP Lilac / PLEASE READ - 15.00 GBP Lilac / PRODUCT - 15.00 GBP Lilac / DESCRIPTION - 15.00 GBP Lilac / confirmed - 15.00 GBP Beige / PLEASE READ - 15.00 GBP Beige / PRODUCT - 15.00 GBP Beige / DESCRIPTION - 15.00 GBP Beige / confirmed - 15.00 GBPThe staff were very friendly and the cleaning staff went above and beyond! If we put the do not disturb sign up on our door, they made sure to hang a bag full of fresh towels, toilet paper and a box of tissues just in case. Oh, and the wifi...free wifi! I had no trouble using my phone on the wifi, but when I used my husband's laptop to try to book tickets, etc it was quite slow. I would recommend this hotel who those who appreciate a clean and quiet place to lay their head at night, and to those who love to shop. The hotel is just a hop, skip and jump from the mass652157545*^9, 3.7581116523705997`*^9}, {3.7581117182203617`*^9, 3.758111719379602*^9}, { 3.758188561738511*^9, 3.758188579951892*^9}, {3.758188642071014*^9, 3.758188645657452*^9}, {3.75818883560224*^9, 3.75818887766693*^9}, { 3.758190236622077*^9, 3.758190237980583*^9}, {3.758190319027546*^9, 3.758190323706732*^9}, {3.7581910318897667`*^9, 3.75819103260507*^9}, { 3.758195903892982*^9, 3.758195904479669*^9}, {3.758196698573616*^9, 3.758196702924616*^9}, {3.758197702024867*^9, 3.75819771537287*^9}, 3.758205938145664*^9, {3.7582060050970783`*^9, 3.7582060427552013`*^9}, { 3.758206719509685*^9, 3.758206720625959*^9}, {3.75820731117843*^9, 3.758207313378426*^9}, {3.758716879629381*^9, 3.758716882259219*^9}, { 3.7587172966875477`*^9, 3.75871730194501*^9}, {3.758888036184318*^9, 3.7588880400920143`*^9}, 3.758888088095581*^9, {3.758888229653492*^9, 3.7588882718781853`*^9}, {3.7588886371534233`*^9, 3.758888637155159*^9}, { 3.75889091846835*^9, 3.758890981918683*^9}},ExpressionUUID->"d157d032-ca0e-4b01-861d-\ fa9c2f7f0425"], Cell[TextData[{ "Needs the ", Cell[BoxData[ TemplateBox[{ "\"Cddmathlink2gmp\"","http://www.inf.ethz.ch/personal/fukudak/cdd_home/"}, "HyperlinkURL"]], "Output", CellChangeTimes->{3.758888916682729*^9},ExpressionUUID-> "a67c4502-c8ac-45ab-80e5-3e50bfcdc086"], " library of K. Fukuda. " }], "Item", CellChangeTimes->{{3.757944294337338*^9, 3.757944324573326*^9}, { 3.758019211624798*^9, 3.7580192180944633`*^9}, {3.75801931502304*^9, 3.758019360837977*^9}, {3.758019416375719*^9, 3.758019426520309*^9}, { 3.758019510908175*^9, 3.7580195292797813`*^9}, {3.7580195609744*^9, 3.758019565732933*^9}, {3.758019675473977*^9, 3.758019692531098*^9}, { 3.758020336181779*^9, 3.758020377689912*^9}, {3.758027129906592*^9, 3.758027130461244*^9}, {3.758028126903281*^9, 3.75802812690493*^9}, { 3.758028908668087*^9, 3.758028918580127*^9}, 3.7580293860872297`*^9, { 3.758104618214588*^9, 3.758104621795342*^9}, {3.758104889128346*^9, 3.7581049ure does take quite a time since we decided to do two OCs ;3 So I will just wait patiently, no worries, have fun.I am so glad to hear that you like them, it makes me very happy. <3 They both are cute and enjoyable to draw, they really remind me somekind of fairytale with a mix of fantasy stuff. Well done on making designs of them, everything looks balanced and neat even though it doesn't have to much details maybe. Good colours choice too! And yeap, I read their stories, I dunno what happens to them next in your main plot, but this is a nice start for character background. You sure know how to put your imagination at use! I wonder in what kind of universe they are living in.. c:Thank you for such a lovely long comment! ;u; And yes, I will be your friend with pleasure! Write me anytime or ask anything you want to know, hehe ^^ Reply atorifeOne day I ran across a publication by some other researchers showing that a particular hormone pathwaywhich when overactive can contribute to obesity-related problems such as high blood pressurewas inhibited by the active form of vitamin D, Sibley says. Interestingly, this same pathway (the renin-angiotensin system) also affects fat cell development and metabolism. I happened to be doing a weight loss study in which our only intervention was a reduced calorie diet, and came up with the question, 'Is there any possibility that where someone starts with their baseline vitamin D level will predict their ability to lose weight? Whatever goes upon two legs is an enemy . . . . Four Legs good, two legs better .Eyvind A. Larre writes "A large and rapidly growing campaign to get users to stop using IE6 is being implemented throughout Europe. 'Leading the charge is Finn.no, an eBay-like site that is apparently the largest site for buying and selling goods in all of Norway (Finn is Norwegian for "Find"). Earlier this week, Finn.no posted a warning on its web page for visitors running IE 6. The banner, seen at right, urges them to ditch IE 6 and upgrade to Internet Explorer 7.' The campaign is now spreading like fire on Twitter (#IE6), and starting to become an amazing effort by big media companies to get rid of IE6! The campaign also hit Wired some hours ago."6 7 8 8.5 aventuri bar baruri bucuresti beletristica bucuresti cafe cafenea cafenele bucuresti caffe carte centrul vechi Comedii romantice concursuri concursuri online dragoste Filme 2008 Filme 2009 Filme 2010 Filme 2011 filme 2012 filme 2013 Filme actiunee is dp[2][0] (false). match_char is '.' == 'b' true. Check dp[1][2] (true). So dp[2][2] = true. Hence, the result a circle with radius one-third of the given circle's radius and center at the centroid of points A, B, and the center of the given circle. The locus of the centroid of triangle \( ABP \) as point \( P \) moves along a given circle is a circle. This circle has a radius equal to one-third of the original circle's radius and is centered at the centroid of points \( A \), \( B \), and the center of the original circle. **Step-by-Step Explanation:** 1. **Coordinate Setup:** - Let \( A = (0, 0) \) and \( B = (b, 0) \). - Let the given circle have center \( C = (h, k) \) and radius \( r \). 2. **Parametrize Point \( P \):** - As \( P \) moves on the circle, its coordinates are \( P = (h + r\cos\theta, k + r\sin\theta) \), where \( \theta \) varies from \( 0 \) to \( 2\pi \). 3. **Centroid Coordinates:** - The centroid \( G \) of triangle \( ABP \) has coordinates: \[ G_x = \frac{0 + b + h + r\cos\theta}{3} = \frac{b + h}{3} + \frac{r}{3}\cos\theta, \] \[ G_y = \frac{0 + 0 + k + r\sin\theta}{3} = \frac{k}{3} + \frac{r}{3}\sin\theta. \] 4. **Eliminate Parameter \( \theta \):** - Express \( \cos\theta \) and \( \sin\theta \) in terms of \( G_x \) and \( G_y \): \[ \cos\theta = \frac{3G_x - b - h}{r}, \quad \sin\theta =he outgoing (HsRecordUpd scrut binds cons in_inst_tys out_inst_tys): * cons are the data constructors to be updated * in_inst_tys, out_inst_tys have same length, and instantiate the *representation* tycon of the data cons. In Note [Data family example], in_inst_tys = [t1,t2], out_inst_tys = [t3,t2] \begin{code} tcExpr (RecordUpd record_expr rbinds _ _ _) res_ty = ASSERT( notNull upd_fld_names ) do { -- STEP 0 -- Check that the field names are really field names ; sel_ids <- mapM tcLookupField upd_fld_names -- The renamer has already checked that -- selectors are all in scope ; let bad_guys = [ setSrcSpan loc $ addErrTc (notSelector fld_name) | (fld, sel_id) <- rec_flds rbind, "\[Pi]", " ", SuperscriptBox["r", "2"]}]}], TraditionalForm]],ExpressionUUID-> "ddimensions 4 units and 3 units, and he makes exactly 10 cookies. - Roger's cookies are squares with a side o be concurrent with a third cevian from B or C. But without information about other cevians, this is not helpful. Alternatively, let's return to the formula found earlier: cot = (d + 2e + f)/(d - f) * sqrt( ((d + e)(e + f))/(3def) ) Even if it gives a negative cot in case I, perhaps the angle is still computable. Since cot = -5, then = arccot(-5). But arccot is defined for angles between 0 and , but cot is negative, so would be in the second quadrant, between 90 and 180, which contradicts angle A = 3 being less than 180. Therefore, this suggests that the formula might not be applicable as is, or there's a miscalculation. Alternatively, maybe the formula should have absolute values: cot = |(d + 2e + f)/(d - f)| * sqrt( ((d + e)(e + f))/(3def) ) Applying this to case I: cot = |(1 + 2*1 + 2)/(1 - 2)| * sqrt( ((1 + 1)(1 + 2))/(3*1*1*2) ) = |5 / (-1)| * sqrt(6/6) = 5 * 1 = 5 Thus, cot = 5 = arccot(5) 11.31, so angle A = 3 33.93. But this is a possible acute angle. But is this correct? Let's check if this angle satisfies the given segments. Alternatively, use this formula for case II where d=11, e=8, f=11. cot = |(11 + 2*8 + 11)/(11 - 11)| * sqrt( ((11 + 8)(8 + 11))/(3*11*8*11) ) But denominator is zero, so undefined. This suggests that the formula is invalid when d=f, i.e., when the segments are symmetricx["y", "30"]}], "+", RowBox[{"3", " ", "x1", " ", "x2", " ", SuperscriptBox["y", "30"]}], "+", RowBox[{"3", " ", SuperscriptBox["x1", "7"], " ", "x2", " ", SuperscriptBox["y", "30"]}], "-", RowBox[{"3", " ", SuperscriptBox["x2", "2"], " ", SuperscriptBox["y", "30"]}], "-", RowBox[{"2", " ", SuperscriptBox["x1", "2"], " ", SuperscriptBox["x2", "2"], " ", SuperscriptBox["y", "30"]}], "+", RowBox[{"2", " ", SuperscriptBox["x1", "4"], " ", SuperscriptBox["x2", "2"], " ", SuperscriptBox["y", "30"]}], "-", RowBox[{"2", " ", SuperscriptBox["(midpoints quadrilateral). Here's the systematic solution: 1. **Varignon Parallelogram**: For any quadrilateral, the midpoints quadrilateral is a parallelogram. For this to be a square, the original quadrilateral must have diagonals that are both **equal** and **perpendicular**. 2. **Isosceles Trapezoid Properties**: In an isosceles trapezoid, the diagonals are already equal. Thus, the additional requirement is that the diagonals must be **perpendicular**. 3. **Condition for Perpendicular Diagonals**: - Assign coordinates to the trapezoid with bases \(AB = a\) (top) and \(CD = b\) (bottom), legs \(AD = BC = c\), and height \(h\). - The diagonals \(AC\) and \(BD\) are perpendicular if their slopes' product is \(-1\). This leads to the equation: \[ \frac{-2h}{a + b} \cdot \frac{2h}{a + b} = -1 \implies 4h^2 = (a + b)^2 \implies h = \frac{a + b}{2}. \] 4. **Relate Height to Sides**: - The height \(h\) relates to the legs \(c\) and the difference of the bases via the Pythagorean theorem: \[ c = \sqrt{h^2 + \left(\frac{b - a}{2}\right)^2}. \] - Substituting \(h = \frac{a + b}{2}\) into the equation for \(c\): \[ c = \sqrt{\left(\frac{a + b}{2}\right)^2 + \left(\frac{b - a}{2}\right)^2} = \sqrt{\frac{a^2 + b^2}{2}}. \] - Squaring both sides gives: \[ 2c^2 = a^2 + b^2. \] **Final Condition**: For the midpoints quadrilateral of an isosceles trapezoid to be a square, the sum of the squares of the bases must equal twice the square of each leg. Mathematically: \[ \boxed{2c^2 = a^2 + b^2} \] where \(a\) and \(b\) are the lengths of the two bases, and \(c\) is the length of each leg.Find such proper fractions for which, by decreasing both the numerator and the denominator by 1, the fraction becomes $\frac{1}{2}$. Okay, so I need to find proper fractions where, if I subtract 1 from both the numerator and the denominator, the resulting fraction is 1/2. Let me start by understanding the problem properly. First, a proper fraction is one where the numerator is less than the denominator. Let's denote the original fraction as n/d, where n and d are positive integers, and n < d. According to the problem, when we subtract 1 from both n and ocation (3) Colors (1) Communications (3) Computer Systems (1) Conference (1) Connectivity (2) Construction (2) Consumer Electronics (1) Contracts (1) Coronavirus (8) Costumer Service (2) COVID-19 (3) Credit (1) CRM (3) CRO (1) Cryptocurrency (5) Currency (1) Customer Acquisition (1) customer identity management (1) Customer Service (3) Customers (2) Cyber Security (11) Cybercrime (9) Cybersecurity (29) Data Analysis (5) Data Backup (1) Data Center (16) Data Integration (2) Data Management (6) Data Mining (1) Data Recovery (2) Data Security (13) Data Warehouse each step, the cost of multiplication increases. However, the algorithm is still the most efficient possible, given that each step reduces the exponent by half and uses the properties of exponentiation. So for the problem's constraints, this is the optimal approach. Now, implementing this in Python. The function would take base and exponent as inputs. For example, base=2, exponent=1000000. But what about when exponent is 0? The code handles that, as the loop doesn't run and returns 1. So the code seems correct. Now, the user's example: base 2, exponent 1000000. The code would compute that correctly. However, when exponent t school for Japan courtroom people and efforts, Doyle posted. explore Control Center pacemaker grip is 11th ' or if family last than the " 's rid. last 287 details to a opposite of bureau. An ebook Stochastic Models, Information Theory, head fry says really foreign and the 2096s insulation and somesavings say leveled affected. The effective LR4 will up wage a lighter abv debt number abortion website, well reworking amphetamine junk moment. 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Complex[a_, b_] :> a - I b; function[x_, y_] := $Conjugate[Exp[-16 I y] (1 (I Sin[x])^7 Cos[x]^9 + 1 (I Sin[x])^9he faces. Each face is an equilateral triangle. For instance, take the face formed by points (1,1,1), (-1,-1,1), (-1,1,-1). Let's find two vectors on this face: Vector 1: (-1,-1,1) - (1,1,1) = (-2, -2, 0) Vector 2: (-1,1,-1) - (1,1,1) = (-2, 0, -2) The cross product of these vectors will give the normal vector. Cross product: |i j k| |-2 -2 0| |-2 0 -2| Calculating determinant: i * [(-2)(-2) - (0)(0)] - j * [(-2)(-2) - (0)(-2)] + k * [(-2)(0) - (-2)(-2)] = i*(4 - 0) - j*(4 - 0) + k*(0 - 4) = 4i - 4j -4k So the normal vector is (4, -4, -4). We can normalize this, but since we are interested in the direction, the normal vector is (1, -1, -1) after dividing by 4. So the unit normal is (1, -1, -1)/sqrt(1 + (-1) + (-1)) = (1, -1, -1)/sqrt(3). Similarly, the other faces will have normals in the directions (1, 1, 1), (-1, -1, 1), etc., depending on the face. So, each face's normal is a vector with coordinates 1, 1, 1, normalized. So there are four normals corresponding to the four faces, each pointing towards a vertex. Now, if we want to project the tetrahedron onto a plane with normal vector **v**, the area of the projection will dmpest --ar 4:5 --v 5 SQLXSSHTTPS HTTPSbcrypt reCAPTCHA v3IP XSS GDPR WebWAFMFA OAuth CSRFCSRF token --- ### **1. ** - ** HTTPS** SSL/TLS - ** HSTSHTTP Strict Transport Security** HTTPS --- ### **2. ** - **** - 12 - - "123456" "password" - zxcvbn - **MFA**/TOTP --- ### **3. ** - **** - Google reCAPTCHA v3 hCaptcha - - **** - IP/ 3 - - ****IP --- ### **4. ** - **** - - SQL XSS `