p :: L a -> b -> Prop step :: xs:_ -> x:_ -> b

-> b base :: b

ys :: L a out :: b

\end{code}
`(p xs b)` relates `b` with **folded** `xs` `p :: L a -> b -> Prop` `foldr`: Base Case ------------------ \begin{code}

p :: L a -> b -> Prop step :: xs:_ -> x:_ -> b

-> b base :: b

ys :: L a out :: b

\end{code}
`base` is related to **empty** list `N` `base :: b

` `foldr`: Ind. Step ------------------ \begin{code}

p :: L a -> b -> Prop step :: xs:_ -> x:_ -> b

-> b base :: b

ys :: L a out :: b

\end{code}
`step` **extends** relation from `xs` to `C x xs` `step :: xs:_ -> x:_ -> b

-> b

` `foldr`: Output --------------- \begin{code}

p :: L a -> b -> Prop step :: xs:_ -> x:_ -> b

-> b base :: b

ys :: L :password "pass" :host "localhost" :port 5432}) ;; Create DataSource (def ds (jdbc/get-datasource db-spec)) ;; Function to create the table (defn create-users-table [ds] (jdbc/execute! ds [" CREATE TABLE IF NOT EXISTS users ( id SERIAL PRIMARY KEY, name VARCHAR(255) NOT NULL, email VARCHAR(255) UNIQUE NOT NULL, created_at TIMESTAMP DEFAULT CURRENT_TIMESTAMP )"])) ;; Execute the function (create-users-table ds) ``` Replace PostgreSQL-specific details (e.g., `SERIAL`) with equivalents for your database (e.g., `INTEGER PRIMARY KEY AUTOINCREMENT` for SQLite).You are an AI assistant. You will be given a task. You must generate a detailed and long answer.Article: Perhaps the greatest question regarding the future development of transnational policing is: in whose interest is it? At a more practical level, the question translates into one about how to ma17.12=1,546.4 Total594,720 +16,992=611,712 +1,546.4613,258.4. So (p/3)^3-613,258.4. Thus, D=(q/2)^2 + (p/3)^3614,278.3 -613,258.41,019.9>0. Hence D>0, which implies one real root and two complex roots, contradicting the earlier discriminant of the original cubic which was =-68,516,523<0 implying one real root. So consistent. Therefore, using Cardano's formula, the real root is: y=(-q/2 + D) +(-q/2 -D). Compute -q/2=783.756. D1,019.931.95. Thus, y=(783.756 +31.95)+(783.756 -31.95)=815.706 +751.806. Compute cube roots: 815.7069.34 (since 9^3=729, 10^3=1000) Similarly,751.8069.1 (9.1^3=753.571) close to 751.8. Compute more accurately: For 815.706: 9.3^3=9.3*9.3=86.49; 86.49*9.3804.357. 9.4^3=9.4*9.4=88.36; there's a different interpretation. Wait, perhaps the problem is that each pitcher can hold up to 10% of the total milk, and initially, they are filled to some levels (not necessarily equal), each not exceeding its capacity of 10%. Then, through pouring operations, we can redistribute the milk to make each have 10%. But if each pitcher's capacity is 10%, and total milk is 100%, we need to have 10 pitchers each with 10%, so they must be filled to capacity. But if the total milk is 100%, and each can hold 10%, then you need all 10 pitchers to be full. But if they start with some milk, each 10%, but total 100%, then each must be full. So again, trivial. This is perplexing. Perhaps the problem is translated incorrectly, and "each contains no more than 10%" should be "each initially contains no more than 10 ounces", and there are 10 pitchers, with total milk 100 ounces. Then, each can hold more than 10 ounces, but initially, each has 10 ounces. Then, the operations of pouring from any pitcher to the others can be used to redistribute the milk to reach 10 ounces each. This would make sense, and the bound of 10 operations is required. Assuming this is the case, let's proceed. Total milk is 100 ounces, 10 pitchers, each initially 10 ounces. Operations: pour from any pitcher any amount, distributing equally to the others. Need to equalize to 10 ounces each in 10 operations. This is a more reasonable problem. Let's think about how to solve it. One approach is to use the following algorithm: in each operation, select a pitcher that has more than 10 ounces and pour out the excess over 10 into the others. However, initially, all are 10, so this isn't applicable. Wait, but if we start with all 10, total 100, then each has exactly 10. So again, trivial. Therefore, this can't be. Therefore, the only way this problem makes sense is if the initial total is less than 100%, but the problem says "divided among them", implying total is 100%. Therefore, I must be missing something. Wait, perhaps the key is that when you pour from a pitcher into the others, you can choose the amount to pour, and this can be done strategically to even things out. For example, even if all start at 10%, you need 0 operations. But if they start unequal but each 10%, maybe through a series of pours, you can average them. But how? Suppose we have two pitchers: A with 10%, B with 10%, and the rest with 10%. All equal. But if one has less, say, A has 5%, then others must have more than 10% to compensate? Wait, no. If total is 100%, and all are 10%, then each must be exactly 10%. Therefore, any deviation would require some pitcher to exceed 10%. Therefore, the initial statement must be incorrect, or there is a misinterpretation. Given that the user has posed this problem, and given the answer is 10 operations, perhaps the original problem allows for pitchers to have more than 10% initially, but the operations can pour milk from one to others, and you need to show that in 10 operations, you can equalize them. However, the user's statement says "each contains no more than 10%", which might be a misstatement. Assuming that the correct problem is: there are 10 pitchers with a total of 100% milk, and you can pour from any pitcher any amount into the others, distributing equally. Prove that you can equalize them in 10 operations. Then, the solution would involve using each operation to average one pitcher's contents with the others. This is similar to a problem where you can use a sequence of operations to make all elements equal by repeatedly averaging each element with the rest of the array. In linear algebra terms, each such operation is a matrix that transforms the vector of milk quantities. After n such operations (one for each element), the vector converges to the average. Alternatively, think of it as follows: each time you pour from a pitcher, you are replacing ited. Wait, no. Let's check. If the time is 12:21:21, the string would be 122121. Reversed, it's 121221, which is different. Therefore, that's not a palindrome. So maybe my earlier reasoning is incorrect. Wait, let me take a correct example. Let's say the time is 12:21:21. Wait, as a six-digit number, that's 122121. Reverse is 121221, which is different. Therefore, not a palindrome. So that's not a valid example. But according to my earlier logic, the palindromic times are when H1=S6, H2=S5, M1=M4. Wait, no, the positions are: Digits: 1 2 3 4 5 6 (HHMMSS) For a palindrome, digit1=digit6, digit2=digit5, digit3=digit4. Therefore, the structure is: H1 H2 M1 M2 S1 S2 To be a palindrome: H1 = S2 H2 = S1 M1 = M2 Therefore, the time is determined by H1, H2, M1, and M2 must equal M1, and S1=H2, S2=H1. So, for example, take H1=1, H2=2, M1=3. Then M2=3, S1=2, S2=1. So the time is 12:33:21. Let's see if that's a palindrome. The digits are 1 2 3 3 2 1. Reversed, it's 1 2 3 3 2 1, which is the same. Yes, that's a palindrome. So this structure woro algebraic. Let me instead consider coordinates for M. Alternatively, since M lies on the circumcircle of ABCD, we can use the property that in a rectangle, the diagonals are equal and bisect each other. The circumradius is half the diagonal. So, the coordinates of M must satisfy (x - a/2)^2 + (y - b/2)^2 = (a^2 + b^2)/4. Alternatively, perhaps using complex numbers? Maybe, but let's stick with coordinate geometry for now. Now, the projections. Let's find points P, Q, R, S. Point P is the projection of M onto AD. AD is trppSP args)]) 4 (ppr body) pprStgRhs (StgRhsCon cc con args) = hcat [ ppr cc, space, ppr con, ptext (sLit "! "), brackets (interppSP args)] pprMaybeSRT :: SRT -> SDoc pprMaybeSRT (NoSRT) = empty pprMaybeSRT srt = ptext (sLit "srt:") <> pprSRT srt \end{code} % % (c) The University of Glasgow, 1994-2006 % Core pass to saturate constructors and PrimOps \begin{code} {-# LANGUAGE BangPatterns #-} module CorePrep ( corePrepPgm, corePrepExpr, cvtLitInteger ) where #include "HsVersions.h" import OccurAnal import HscTypes import PrelNames import CoreUtils import CoreArity import CoreFVs import CoreMonad ( endPass, CoreToDo(..) ) import CoreSyn import CoreSubst import MkCore hiding( FloatBind(..) ) -- We use our own FloatBind here import Type import Literal import Coercion import TcEnv import TcRnMonad import TyCon import Demand import Var import VarSet import VarEnv import Id import IdInfo import TysWiredIn import DataCon import PrimOp import BasicTypes import UniqSupply import Maybes import OrdList import ErrUtils import DynFlags import Util import Pair import Outputable import Platform import FastString import Config import Data.Bits import Data.List ( mapAccumL ) import Control.Monad \end{code} -- --------------------------------------------------------------------------- -- Overview -- --------------------------------------------------------------------------- The goal of this pass is to prepare for code generation. 1. Saturate constructor and primop applications. 2. Convert to A-normal form; that is, function arguments are always variables. * Use case for strict arguments: f E ==> case E of x -> f x (where f is strict) * Use let for non-trivial lazy arguments f E ==> let x = E in f x (were f is lazy and x is non-trivial) 3. Similarly, convert any unboxed lets into cases. [I'm experimenting with leaving 'ok-for-speculation' rhss in let-form right up to this point.] 4. Ensure that *value* lambdas only occur as the RHS of a binding (The code generator can't deal with anything else.) Type lambdas are ok, however, because the code gen discards them. 5. [Not any more; nuked Jun 2002] Do the seq/par munging. 6. Clone all local Ids. This means that all such Ids are unique, rather than the weaker guarantee of no clashes which the simplifier provides.ffice SharePoint Portal Server 2003 environments.Find all news similar on Search and Recover 5.4.9.51Similar smart reviews:EASEUS Todo Backup - the easiest way to back up your data and restore it whenever you want Keeping all of your documents, photos and music on a single computers hard drive is extremely convenient. You always have access to any documentDisk Doctors' Undelete - a reliable remedy for the problem of deleted files Few things can be more frustrating in your communication with your PC than seeing important documents deleted with no obvious way to recover them. This canFind all smart reviews similar on Search and Recover 5.4.9.51Home > Software > Utilities > Backup & Restore > download now Search and RecoverPosted on 05/17/2010 by gellman I just saw this posted on FCB, and I believe this price on the Dustin Ackley Superfractor auto 1/1 is totally and utterly insane. Even if this guy turns out to be a perennial all star, maybe winning a few MVPs, there is no way this price has anywhere to go but down. Look at what Ryan Braun sold for, and then consider the price on this guy. }) | gadt_style = pp_prefix_con <+> dcolon <+> ppr_ty | otherwise = ppr_fields tys_w_strs where tys_w_strs :: [(IfaceBang, IfaceType)] tys_w_strs = zip stricts arg_tys pp_prefix_con = pprPrefixIfDeclBndr ss name (univ_tvs, pp_res_ty) = mk_user_con_res_ty eq_spec ppr_ty = pprIfaceForAllPart (univ_tvs ++ ex_tvs) ctxt pp_tau -- A bit gruesome this, but we can't form the full con_tau, and ppr it, -- because we don't have a Name for the tycon, only an OccName pp_tau = case map pprParendIfaceType arg_tys ++ [pp_res_ty] of (t:ts) -> fsep (t : map (arrow <+>) ts) [] -> panic "pp_con_taus" ppr_bang IfNoBang = ppWhen opt_PprStyle_Debug $ char '_' ppr_bang IfStrict = char '!' ppr_bang IfUnpack = ptext (sLit "{-# UNPACK #-}") ppr_bang (IfUnpackCo co) = ptext (sLit "! {-# UNPACK #-}") <> pprParendIfaceCoercion co pprParendBangTy (bang, ty) = ppr_bang bang <> pprParendIfaceType ty pprBangTy (bang, ty) = ppr_bang bang <> ppr ty maybe_show_label (lbl,bty) | showSub ss lbl = Just (pprPrefixIfDeclBndr ss lbl <+> dcolon <+> pprBangTy bty) | otherwise = Nothing ppr_fields [ty1, ty2] | is_infix && null labels = sep [pprParendBangTy ty1, pprInfixIfDeclBndr ss name, pprParendBangTy ty2] ppr_fields fields | null labels = pp_prefix_con <+> sep (map pprParendBangTy fields) | otherwise = pp_prefix_con <+> (braces $ sep $ punctuate comma $ ppr_trim $ map maybe_show_label (zip labels fields)) instance Outputable IfaceRule where ppr (IfaceRule { ifRuleName = name, ifActivation = act, ifRuleBndrs = bndrs, ifRuleHead = fn, ifRuleArgs = args, ifRuleRhs = rhs }) = sep [hsep [doubleQuotes (ftext name), ppr act, ptext (sLit "forall") <+> pprIfaceBndrs bndrs], nest 2 (sep [ppr fn <+> sep (map pprParendIfaceExpr args), ptext (sLit "=") <+> ppr rhs]) ] instance Outputable IfaceClsInst where ppr (IfaceClsInst { ifDFun = dfun_id, ifOFlag = flag , ifInstCls = cls, ifInstTys = mb_tcs}) = hang (ptext (sLit "instance") <+> ppr flag <+> ppr cls <+> brackets (pprWithCommas ppr_rough mb_tcs)) 2 (equals <+> ppr dfun_id) instance Outputable IfaceFamInst where ppr tSkolTyVar tv1 = isMetaTyVar tv2 | otherwise = isMetaTyVar tv2 || isFlatSkolTyVar tv2 -- Note [Eliminate flat-skols] checkKind :: CtEvidence -- t1~t2 -> TcType -> TcKind -> TcType -> TcKind -- s1~s2, flattened and zonked -> TcS StopOrContinue -- LHS and RHS have incompatible kinds, so emit an "irreducible" constraint -- CIrredEvCan (NOT CTyEqCan or CFunEqCan) -- for the type equality; and continue with the kind equality constraint. -- When the latter is solved, it'll kick out the irreducible equality for -- a second attempt at solving -- -- See Note [Equalities with incompatible kinds] checkKind new_ev s1 k1 s2 k2 -- See Note [Equalities with incompatible kinds] = ASSERT( isKind k1 && isKind k2 ) do { traceTcS "canEqLeaf: incompatible kinds" (vcat [ppr k1, ppr k2]) -- Create a derived kind-equality, and solve it ; mw <- newDerived kind_co_loc (mkEqPred k1 k2) ; case mw of Nothing -> return () Just kev -> emitWorkNC [kev] -- Put the not-currently-soluble thing into the inert set ; continueWith (CIrredEvCan { cc_ev = new_ev }) } where loc = ctev_loc new_ev kin3}}{4} + \frac{a^2 \sqrt{3}}{4} = \frac{a^2}{2} (\sqrt{2} + \sqrt{3}) \) 4. **Inradius Calculation**: - Inradius $ r = \frac{3V}{S} = \frac{3 \cdot \frac{a^3}{12}}{\frac{a^2}{2} (\sqrt{2} + \sqrt{3})} = \frac{a/4}{(\sqrt{2} + \sqrt{3})/2} = \frac{a}{2} (\sqrt{3} - \sqrt{2}) $ 5. **Circumradius Calculation**: - Circumcenter $ O $ found to be the center of the cube: $ \left( \frac{a}{2}, \frac{a}{2}, \frac{a}{2} \right) $ - Circumradius $ R = \frac{a}{\sqrt{2}} $ 6. **Ratio Calculation**: - Ratio $ \frac{R}{r} = \frac{\frac{a}{\sqrt{2}}}{\frac{a}{2} (\sqrt{3} - \sqrt{2})} = \frac{2}{\sqrt{2} (\sqrt{3} - \sqrt{2})} = \sqrt{2} \cdot \frac{\sqrt{3} + \sqrt{2}}{1} = \sqrt{6} + 2 $ The ratio of the circumradius to the inradius of the tetrahedron $ EFGH $ is $\boxed{2 + \sqrt{6}}$.Given positive real numbers $a, b, c$ such that $abc = 1$, prove that $(a+b)(b+c)(c+a) \geq 4(a+b+c-1)$. Alright, so I have this inequality to prove: given positive real numbers $a, b, c$ with $abc = 1$, I need to show that $(a+b)(b+c)(c+a) \geq 4(a+b+c -1)$. Hmm, okay. Let me start by understanding the problem and then think about possible strategies. First, since $abc = 1$, maybe there's a substitution that can simplify things. Sometimes with symmetric inequalities and conditions like $abc = 1$, substituting variables in terms of others helps. For example, setting $a = \frac{x}{y}$, $b = \frac{y}{z}$, $c = \frac{z}{x}$ would satisfy $abc = 1$, but I'm not sure if that's the best approach here. Alternatively, maybe setting $a = \frac{x}{y}$, $b = \frac{y}{z}$, $c = \frac{z}{x}$ could work, but let's see. Another thought: since the variables are positive and multiplied to 1, perhaps taking logarithms? But logarithms might complicate things more. Let me think. Alternatively, maybe homogenizing the inequality. Since $abc = 1$, we can express the inequality in terms of homogeneous degrees. Let me check the degrees. The left-hand side, $(a+b)(b+c)(c+a)$, when expanded, each term is of degree 3. The right-hand side is $4(a + b + c -1)$. The terms $a, b, c$ are degree 1, and the constant term is degree 0. Since the left side is degree 3 and the right side has a mix, this isn't homogeneous. So perhaps we need to homogenize the ita CarletonWilliam A. CaseyJim CebulaAnthony CebzanovSagar Chaki Timothy ChickMary Beth ChrissisJonathan ChuJason ClarkJulie CohenSholom CohenMatthew CollinsAnne ConnellPat Copelin Jennifer CowleyRita CreelMike Crowley Pamela CurtisDRosemary J. DarrJeff Davenport Edward H. Deets, III Julien DelangePatrick DonohoeAudrey J. DorofeeMichael DugganDionisio de NizEJames EdmondsonJoseph P. Elm Neil ErnstFelicia L. EvansFSid Faber Kevin FallPeter FeilerRobert FergusonDonald FiresmithLori FlynnTracey FoxJonathan FrederickDavid FrenchGMike GagliardiMatthew E. Gaston Linda Parker Gates Dennis R. Goldenson, PhDJohn Goodenough Ian GortonDoug GrayPhil GroceArie GurfinkelHJeffrey HamedJosh HammersteinBud HammonsJeffrey HavrillaWill HayesScott HissamJohn HudakClifford C. HuffIJames IversJJerry JacksonKDerrick H. KarimiMark KasunicRick Kazman David KeatonTracey Kelly Robert KemererJohn Klein Mark KleinWill KlieberMichael KonradKeith Korzec Bob Kubof production, and the remaining 90 acres are split as per the previous allocation? Or does he have to adjust the allocations? Wait, the problem says he "managed a farm that started with 100 acres. When a new law encouraged diversification, he decided to allocate 40% of his land to growing wheat. The remaining land was split equally between barley and vegetables. Later, another law was introduced..." So, the timeline is: first, he allocated 40% to wheat, remaining 60 split to barley and vegetables. Then, later, the second law required setting aside 10% those are different types (list vs dicm the end of the stacks list until we find a non-empty stack. This would take O(m) time where m is the number of stacks. If m is large (like 1e5), then this is O(1e5) which is not acceptable for 2e5 operations. So in the worst case, this would be O(2e10) operations, which is way too slow. So this approach would not be efficient enough. Therefore, we need a way to track the rightmost non-empty stack ef. Find a_{20n}, which would just be a. Wait, but if the sequence is constant, then we can set a_n = a_{n-1} = a, leading to: a = (3 a + 1)/(3 - aZgWdXL330wIYR62qupBs70ZLLA1jOX1F9NZv7y 5OGdNTUHDkRbzv6gbSYtIS0ka8fPQ2cj7Nh/iMxT1fGVBfWbmsanhnvJ6i9f /BBvPP75k8/FyJahfUKaRFtIS6goXsn75I/WY6fNNcy7k4m9tMyiNWjzNBtG m01X7bOOdnIGqDebxiYTAqN/bVFo/nT7Uqt/jCZZGWbSswwxejjGqI4NDSrf sfH30R/X7IWrXm9RrMVY64v0f29ZvkVvjInzB/po5jmZblq+jawU1G+QRkD9 Bo7guvrd0tpCz82+O4/YEwAAAEpo+Am9lLDimwIOQU4YFOIG+Xv900+73S4A sp3U9X7z/P10aNGGoLYAHn1TyF0bajwKWcY+iVK/JX8slT3JETmszQ4NXxwe It84ODh49uzZgW8Hzn7d/1lHOzlqM3n8uS/G4y47xnan/yW+CgG9WyS0MZr/ K1O/l9Wz8BA6f6LHW7Tx9hECfr/f+KLizmmhVbS2I51sVs9MKIL/BzJEyavZ 7jI1zOJm2tjZ7rQlZBX8EpqLYu0x8kVkUbjDCNIRqN/AEdxVv8leuqGhgWa6 4kQXAthe answer. During the process, we may have to build a new understanding or discard one. We may have to have patience. Sometimes the entire process is step by step & can take time. What really assists me is the wide range of sources that I use, & the enormous amount of reading that I perform, as well as my experience in finding & using written resources. One needs to comprehend the limitations of a written source. Typically, a book is written by one person, which then gives you one perspective. We see 3-D, so that should tell you that reading one perspective on a subject may have its limitations, even if it seems good. One also needs to understand the limitations of what the Internet has, which is perhaps at best 1% of what is there to learn. (On the other hand, so much is contained within the minds of humanitythe problem is how to ask the right person!) The large amount of experience & info I provide my mind then in turn acts as food for the brain to create & birth new insights. The same processes that have helped me, have also blessed many others. And now hopefully, these processes have been described in meaningful words for you. Be blessed my friend.The glue holding the soles together came apart after a week. They opened p liek a mouth. I returned them for repairs 3 x and they cannot be fixed You guys should warranty these!GregHi Al, they are warrantied under 8 months. Someone will be contacting you shortly about resolving this for you. Sorry to hear about your Supervogs!Ekoorb - Minneapolis, MNI love these shoes and have had several pairs. I still think my favorite are the old-school ones which were blue, green, and burgundy, so awesome. I have again (after about 7 years of heavy use) worn though another pair. I would love to replace them.Love em'!!!Charlotte - Portsmouth, NHI have had my Angels Supervogs since I was in high school circa 1998. I love these shoes. They survived high school, college in hawaii and have taken me on my travels with the United States Navy and even made it all the way to Yokosuka, Japan where I'm currently stationed. I love these shoes, and I'm looking forward to adding another pair to my collection... I just have to choose which one I want... :-DSimply adorableMarie - GatineauI just opened the box containing my burgundy/grey Supervogs, sent from Toronto, and I burst out laughing! :-D They are the funniest looking shoes ever! I will definitely have to wear the pins that came with them (No, you're weird!) to field off comments.eine der wenigen edlwe take f(x) = x^2 and g(y) = y, then Cov(f(X), g(Y)) = Cov(X^2, X^2) = Var(X^2), which is positive. So in this case, Corr(f(X), g(Y)) would be 1, which is higher than |Corr(X,Y)|=0. So in this case, *((X), (Y)) would be 1, but Corr(X,Y) is zero. So the statement in part (a) would not hold. But this contradicts the problem's assertion. Therefore, there must be a misunderstanding here. Wait, but in the problem statement, they define *(A, B) as the supremum of |*(, )|, where *(, ) is the correlation coefficient. However, perhaps in the problem, *(, ) is not the Pearson correlation but the maximal correlation as defined in some other way? Wait, but the problem says "correlation coefficient", which is typically Pearson's. Alternatively, maybe the notation * here is a specific type of correlation, such as the maximal correlation defined in information theory or something else. Wait, actually, in the literature, there is a concept called "maximal correlation" which is defined as the supremum of Corr(f(X), g(Y)) over all Borel-measurable functions f, g such that f(X) and g(Y) are square-integrable. This maximal correlation is known to be zero if and only if X and Y are independent. However, in the case where X and Y are jointly Gaussian, the maximal correlation is equal to the absolute value of the Pearson correlation coefficient. But for non-Gaussian variables, the maximal correlation can be larger. So if the problem is referring to the usual Pearson correlation coefficient, then part (a) is only true for Gaussian variables, but the problem does not specify that. Therefore, there must be a different interpretation. Wait, but the problem statement says "correlation coefficient of random variables and ", which is unambiguously Pearson's correlation. So then, unless there is a restriction to linear functions or something, the problem's assertion in part (a) would be false in general. However, since the problem is given as an exercise, there must be a correct approach. Perhaps the key is t's position on these matters. His argument with regards to orbits and relativity is most elegant.So from the quote you provide let me challenge you, what is "the literal meaning of the Torah's words"??? Written Torah, Oral Torah, Mishnah, Gemorah (including aggadah???) Every statement and story of every one of the tanoim and amoraim is literal? Every statement of every one of the commentaries?Statements that conflict with each other? Argued positions that conflict with each other are both accurate and true? Every MIDRASH is literally true?We clearly don't learn that way.Now on age of the earth. Since TIME itself is a creation, and one aspect of time is relativity - it actually flows at different speeds depending on the motion relation of different planets - there's a difference between the question of age of the UNIVERSE and age of the EARTH, since time does not move at the same speed in all parts of the universe.Even in Berashis itself, the pasuk says day before the creation of the sun and earth. Day by what measurement - physical aspects of time measurement (and perhaps time itself) didn't exist yet!Simple answers, the pshat, aren't so simple in such matters.masajes gays porno actrices alemanas Arenales Y Sevilleja Mujeres Sexy Playboy Conocer Gente Para Follar Singles Sevilla Actividades Paginas Videos Lesbianas ver porno videos pilladas calientes desnudas sexo chats sado putas Ver Masturvaciones Mujeres Marruecos Piedramillera Petardas Espanol Gratis Putas En Jerez que videos famosas swinger sexo juegos sexuales con pareja gays follandp Sexo Para Ver En Movil Bustos Paginas De Sexo Seguras Find Girls Online Masaje Final Feliz En Barcelona sex shop en casa father xxx teniendo sexo en la playa mejor chat ligar Relaciones Esta Persona Cam Chicas Irun Anuncios Mileroticos Se Masturba Es Un Montn De orgiasN -------------------------------------------------------------------------------- SUPER-G START LIST Start Time 11:00 -------------------------------------------------------------------------------- START BIB COMPETITOR TEAM/YR CLASS 12: Men 80-84 (M12) , . 29 ( ), . , t satisfy that adjacent arcs are different, and the two -length arcs' neighbors must be different. If we arrange the two -length arcs in the sequence, say positions 1 and 4, then their neighbors (arcs 6 and 2 for the first, and arcs 3 and 5 for the second) must be different. But the rest of the arcs (positions 2,3,5,6) are 2-length and only need to differ from their immediate neighbors. But how many colorings are there for such a configuration? It's unclear. Alternatively, perhaps the key insight is that the additional constraints imposed by the -length arcs effectively enforce that the coloring must be a proper 3-coloring with no two nodes at distance two (through a -length arc) being the same color. This would mean that the coloring is a proper 3-coloring and also a proper coloring of the square of the cycle graph for those specific edges. However, unless the graph is bipartite, this might not be possible. But given that we have three colors, maybe it's possible. However, this is getting too abstract. Another approach: since the problem involves rotational symmetry, perhaps the number of distinct colorings is the same for each valid arc length sequence, modulo rotation. But I don't think so. For example, sequences with different numbers of arcs would have different numbers of colorings. Alternatively, perhaps the only valid colorings are those that use a repeating pattern of R, G, B, which would work only when the number of arcs is a multiple of 3. Since the total number of arcs can be 5,6,7,8,9,10, only n=6 and 9 are multiples of 3. Therefore, perhaps only these cases contribute to the total count. For n=6 (x=2, y=4), if we can arrange the two -length arcs such that their placement doesn't inand start stressing about which category you're in - which as I think the title hints, I wasn't ;-) Or you could take it very seriously and extrapolate from it some deep social commentary.. Or you could just dismiss it as silliness. Probably all equally valid positions to take, depending on what mood you were in! I don't think I was taking it all that seriously. But I could have done.Feb. 2nd I stopped on the Circumferential highway near the Micmx])^13 Cos[x]^3 + 20 (I Sin[x])^4 Cos[x]^12 + 20 (I Sin[x])^12 Cos[x]^4) + Exp[-7 I y] (515 (I Sin[x])^7 Cos[x]^9 + 515 (I Sin[x])^9 Cos[x]^7 + 166 (I Sin[x])^5 Cos[x]^11 + 166 (I Sin[x])^11 Cos[x]^5 + 332 (I Sin[x])^6 Cos[x]^10 + 332 (I Sin[x])^10 Cos[x]^6 + 578 (I Sin[x])^8 Cos[x]^8 + 12 (I Sin[x])^3 Cos[x]^13 + 12 (I Sin[x])^13 Cos[x]^3 + 51 (I Sin[x])^4 Cos[x]^12 + 51 (I Sin[x])^12 Cos[x]^4) + Exp[-5 I y] (154 (I Sin[x])^4 Cos[x]^12 + 154 (I Sin[x])^12 Cos[x]^4 + 1242 (I Sin[x])^8 Cos[x]^8 + 729 (I Sin[x])^6 Cos[x]^10 + 729 (I Sin[x])^10 Cos[x]^6 + 1082 (I Sin[x])^7 Cos[x]^9 + 1082 (I Sin[x])^9 Cos[x]^7 + 368 (I Sin[x])^5 Cos[x]^11 + 368 (I Sin[x])^11 Cos[x]^5 + 8 (I Sin[x])^2 Cos[x]^14 + 8 (I Sin[x])^14 Cos[x]^2 + 41 (I Sin[x])^3 Cos[x]^13 + 41 (I Sin[x])^13 Cos[x]^3) + Exp[-3 I y] (2130 (I Sin[x])^8 Cos[x]^8 + 1230 (I Sin[x])^6 Cos[x]^10 + 1230 (I Sin[x])^10 Cos[x]^6 + 216 (I Sin[x])^4 Cos[x]^12 + 216 (I Sin[x])^12 Cos[x]^4 + 595 (I Sin[x])^5 Cos[x]^11 + 595 (I Sin[x])^11 Cos[x]^5 + 1840 (I Sin[x])^7 Cos[x]^9 + 1840 (I Sin[x])^9 Cos[x]^7 + 9 (I Sin[x])^2 Cos[x]^14 + 9 (I Sin[x])^14 Cos[x]^2 + 50 (I Sin[x])^3 Cos[x]^13 + 50 (I Sin[x])^13 Cos[x]^3) + Exp[-1 I y] (131 (I Sin[x])^3 Cos[x]^13 + 131 (I Sin[x])^13 Cos[x]^3 + 2193 (I Sin[x])^9 Cos[x]^7 + 2193 (I Sin[x])^7 Cos[x]^9 + 906 (I Sin[x])^5 Cos[x]^11 + 906 (I Sin[x])^11 Cos[x]^5 + 2432 (I Sin[x])^8 Cos[x]^8 + 1571 (I Sin[x])^6 Cos[x]^10 + 1571 (I Sin[x])^10 Cos[x]^6 + 384 (I Sin[x])^4 Cos[x]^12 + 384 (I Sin[x])^12 Cos[x]^4 + 5 (I Sin[x])^1 Cos[x]^15 + 5 (I Sin[x])^15 Cos[x]^1 + 29 (I Sin[x])^2 Cos[x]^14 + 29 (I Sin[x])^14 Cos[x]^2) + Exp[1 I y] (2392 (I Sin[x])^9 Cos[x]^7 + 2392 (I Sin[x])^7 Cos[x]^9 + 765 (I Sin[x])^5 Cos[x]^11 + 765 (I Sin[x])^11 Cos[x]^5 + 1532 (I Sin[x])^6 Cos[x]^10 + 1532 (I Sin[x])^10 Cos[x]^6 + 2752 (I Sin[x])^8 Cos[x]^8 + 76 (I Sin[x])^3 Cos[x]^13 + 76 (I Sin[x])^13 Cos[x]^3 + 280 (I Sin[x])^4 Cos[x]^12 + 280 (I Sin[x])^12 Cos[x]^4 + 2 (I Sin[x])^1 Cos[x]^15 + 2 (I Sin[x])^15citly for the results of \texttt{ccall} statements, but can ignore nodes which are allocated outside of the heap, i.e., all constant nodes. In addition, we handle the dynamic allocation of partial application nodes by a \texttt{CPSLetPapp} statement here. \begin{verbatim} > heapCheck :: ([Name],CPSCont) -> CExpr -> [CStmt] > heapCheck (vs,_) n = > [CProcCall "CHECK_HEAP" [int (length vs),n] |}{W_1} = \frac{A_2}{A_1} \] Substituting the surface areas: \[ \frac{W_2}{8} = \frac{4\pi (0.3)^2}{4\pi (0.15)^2} \] Tto positive infinity. At x=0, it's -3. Therefore, between x=-1 and x=0, the function goes from 5 at x=-1 to -3 at x=0, so crosses zero there. Therefore, two real roots: one between -1 and 0, another between 2 and 3. So let's approximate them roughly. Let's take the positive root. Between 2 and 3. At x=2: -1, x=2.1: (2.1)^4 -7*2.1 -3. Let's compute 2.1^4: 2.1^2=4.41, 4.41^2=19.4481, 7*2.1=14.7, so 19.4481 -14.7 -3=1.7481. So at x=2.1, it's positive. Between x=2 and 2.1, crosses from -1 to +1.7481. Let's try x=2.05: (2.05)^4. 2.05^2=4.2025. Squared again: 4.2025^2 17.6610. 7*2.05=14.35. So 17.6610 -14.35 -30.311. Still positive. x=2.025: 2.025^24.1006, squared 16.8123. 7*2.02514.175. 16.8123 -14.175 -3-0.3627. So between 2.025 and 2.05, the function crosses zero. Let's take x=2.04: 2.04^2=4.1616, squared17.3165. 7*2.04=14.28. 17.3165 -14.28 -30.0365. x=2.035: 2.035^2=4.1412, squared4.1412*4.141217.15. 7*2.035=14.245. 17.15 -14.245 -3-0.095. So between 2.035 and 2.04. Let's say approximately 2.037. Similarly, for the negative root between -1 and 0. Let's check x=-0.5: (-0.5)^4 -7*(-0.5) -3=0.0625 +3.5 -3=0.5625. x=-0.75: (-0.75)^4=0.3164 -7*(-0.75)=5. a variable, then it must have been in the support of either $\sigma_1$ or $\sigma_2$. > composeLambdaSubs > :: M.Map LambdaVariable (LambdaTerm a) > -> M.Map LambdaVariable (LambdaTerm a) > -> M.Map LambdaVariable (LambdaTerm a) > composeLambdaSubs sigma2 sigma1 = > foldr f M.empty (S.union (M.keysSet sigma1) (M.keysSet sigma2)) > where > f x sigma = case M.lookup x sigma1 of > Just e -> M.insert x (applySubToLambdaTerm sigma2 e) sigma > Nothing -> case M.lookup x sigma2 of > Just e -> M.insert x e sigma > Nothing -> sigma At this point there are several properties about free variables and substitution we'd like to prove, however we're not quite ready _, 0.704536, _, _, _, _, _, _, _, _, _, 25.520285, _, _, _, 2.251452, _, _, _, _, _, _, 2.435244, _, _, _, _, _, _, _, _, 1.634983, _, _, _, _, _, 5.47547, _, _, _, _, _, _, _, 51.029083, _, _, _, _, _, 31.815161, _, _, _, _, _, _, 52.250534, _, _, _, _, _, 24.107384, _, _, _, _, _, _, 19.516413, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, of other players. But since Migel is part of the team, if he has negative points, it's subtracted. But in reality, basketball points can't be negative. Therefore, perhaps the problem has an error. Alternatively, maybe there's a miscalculation. Wait, let's verify once again: "Migel had 5 points fewer than the square root of Curry's points" Curry's points: 24 Square root of 24: ~4.899 5 points fewer: 4.899 -5 = -0.101 So Migel: -0.101 points. That must be a mistake. Since the problem is likely intended for total points, and all other players (except maybe Jordan, Green, Barnes) have whole numbers. But Jordan is 10.8, Green is 13.5, Barnes is approx10.62, Christine is 729. Wait, Christine's points are 729, which is very high. That seems like an outlier. 3^6 is 729. Klay had half of Draymond's points, which is 6. So Christine's points are 3^6=729. That seems correct. Given that, maybe the problem is designed with Migel having negative points, but that's impossible, so perhaps we are to exclude Migel or assume a typo. Since this is a math problem, perhaps ther(outcomeVectorValues test) > lenOutcome = length outcomeNames % % (c) The University of Glasgow 2006 % (c) The GRASP/AQUA Project, Glasgow University, 1992-1998 % \section[Name]{@Name@: to transmit name info from renamer to typechecker} \begin{code} {-# OPTIONS -fno-warn-tabs #-} -- The above warning supression flag is a temporary kludge. -- While working on this module you are encouraged to remove it and -- detab the module (please do the detabbing in a separate patch). See -- http://hackage.haskell.org/trac/ghc/wiki/Commentary/CodingStyle#TabsvsSpaces -- for details -- | -- #name_types# -- GHC uses several kinds of name internally: -- -- * 'OccName.OccName': see "OccName#name_types" -- -- * 'RdrName.RdrName': see "RdrName#name_types" -- -- * 'Name.Name' is the type of names that have had their scoping and binding resolved. They -- have an 'OccName.OccName' but also a 'Unique.Unique' that disambiguates Names that have -- the same 'OccName.OccName' and indeed is used for all 'Name.Name' comparison. Names -- also contain information about where they originated from, see "Name#name_sorts" -- -- * 'Id.Id': see "Id#name_types" -- -- * 'Var.Var': see "Var#name_types" -- -- #name_sorts# - 2,000 subscribers. First, let's recall that the period of a sine function Bsin(t + ) is 2/. Since the period is given as 12 months, we can set up the equation 2/ = 12. Solving for gives = 2/12 = /6. So that's taken care of. Next, the amplitude . . DC ( ), . . (), (Idaho), (Oregon). . 185 . . . ( ): (1) : ( , , ?)1. : +1( ) 2. : 8()10-1( ) . -- :1. : +12062. : 8()101206 , . (Seattle) (state of Washington)206Bainbridge Isl., Mercer Isl., Newhalem, Sea-Tac Airport, Seattle, Vashon I then by applying maximal effort to stop the movement at about 70 knee flexion. This strategy allowed for an ECC-overload in force/power values during the last two-thirds of the ECC cycle. Once the flywheel stopped, a subsequent CON action was instantly initiated. A 3-min recovery period was allowed between sets. Peak CON and ECC power was measured in all repetitions (SmartCoach, Stockholm, Sweden). Real-time performance feedback on peak power was offered to the patients at all times.Cruz has seized on Trumps assertion that the North Carolina law, which also rolled back other protections for gay, lesbian and transgender people, was unnecessary and bad for business corporations inclult, lintInteractiveExpr, dumpIfSet, -- ** Screen output putMsg, putMsgS, errorMsg, errorMsgS, fatalErrorMsg, fatalErrorMsgS, debugTraceMsg, debugTraceMsgS, dumpIfSet_dyn, #ifdef GHCI -- * Getting 'Name's thNameToGhcName #endif ) where #ifdef GHCI import Name( Name ) #endif import CoreSyn import PprCore import CoreUtils import CoreLint ( lintCoreBindings, lintExpr ) import HscTypes import Moecord Types (53)Research Clues (22)Researh Skills (8)Tips & Tricks (27)4. Gen Webinars (30)5. Inspirational (374)Barry's Story (31)GenQuotes (300)GenStories (43)6. Writing History (115)Family Sharing (26)Interviewing (19)Journal Writing (7)Preparation (5)Questions to Ask (5)Remembering (3)Taking Photos (34)Writing (21)7. Slideshow Series (555)Barry's Presentations (14)Caring for Artifacts (9)Genealogy 101 (73)Google (123)Immigrant Rsh (43)LDS Family History (26)Photography (112)U. S. Census perscriptBox["y", "2"]}], "+", "3"}], ")"}]}]], "+", "1"}], RowBox[{ SuperscriptBox["y", "4"], "+", SuperscriptBox["y", "2"], "+", "1"}]]}]], " ", RowBox[{"(", RowBox[{ RowBox[{ RowBox[{"-", "3"}], " ", SuperscriptBox["y", "8"]}], "-", RowBox[{"8", " ", SuperscriptBox["y", "6"]}], "+", RowBox[{ RowBox[{"(", RowBox[{ SqrtBox[ RowBox[{ RowBox[{"-", SuperscriptBox[ RowBox[{"(", RowBox[{ SuperscriptBox["y", "2"], "+", "1"}], ")"}], "2"]}], " ", RowBox[{"(", RowBox[{ RowBox[{"3", " ", SuperscriptBox["y", "4"]}], "+", RowBox[{"2", " ", SuperscriptBox["y", "2"]}], "+", "3"}], ")"}]}]], "-", "10"}], ")"}], " ", SuperscriptBox["y", "4"]}], "-", RowBox[{"8", " ", SuperscriptBox["y", "2"]}], "+", SqrtBox[ RowBox[{ RowBox[{"-", t can test these examples. Or perhaps the function as I wrote above is sufficient. The problem says to "develop a RUST program that calculates a tuple...". So the program should probably include the function, and maybe some test cases. But perhaps the problem expects the function. Let's check the examples: Sample input 1: [11,26,15,14,43,10] returns (10,14). According to the code, evens are [26,14,10]. Wait, wait, wait, wait. Wait, the input vector is [11,26,15,14,43,10], so when we filter even numbers, the even elements are 26,14,10. Then when sorted, they become [10,14,26]. So the first is 10, second is 14. So the code returns (Some(10), Some(14)). Which is (10,14) in the tuple, but in Rust, the return is (Option, Option). So the sample input would have (Some(10), Some(14)). The problem's example shows (10,14), but in reality, the code returns (Some(10), Some(14)). Wait, the problem's examples show (10,14) and (None, None), but the code returns Options. So perhaps the problem's examples are simplified. The user should return the tuple as (Option, Option). So the function is correct. Now, writing this function in Rust. But also, the problem says "develop a RUST program". So perhaps the code should be a runnable program with main function that demonstrates the examples. But perhaps the problem expects the function to be written. Let's check the initial problem statement again. The problem says: "Develop a RUST program that calculates a tuple Draymond:12. Correct. So all other players are correctly calculated. Only Migel is the issue. Therefore, I think the problem has a typo in Migel's points. If I assume that instead of "5 points fewer than the square root", it's "5 points fewer than the square", then Migel would have 24^2 -5=576-5=571. Then total would be 832.82 +571 -(-0.101)= approx 832.82 +571.1011403.92. But Christine already has 729, so the total being 1404 is plausible? Maybe, but that's a stretch. Alternatively, if the problem meant "5 points more than the square root", Migel would have9.899, total832.82 +9.899842.72. Alternatively, if the problem meant "5 points fewer than twice the square root of Curry's points", that would be 2*sqrtrm]],ExpressionUUID->"44198062-0327-4d86-b5e2-0008958991f3"]], \ "Text",ExpressionUUID->"d874d229-96af-4502-9f6d-d86a3cf9997f"], Cell[TextData[{ "The exact value is ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"f", "(", "1.1", ")"}], "=", RowBox[{ SqrtBox["1.1"], "=", RowBox[{"1.0488", "\[Ellipsis]"}]}]}], TraditionalForm]],ExpressionUUID-> "b5335573-4a2e-4a40-b9a7-0fa93c4acd31"], "; therefore, the linear approximation has an error of about ", Cell[BoxData[ FormBox["0.0012", TraditionalForm]],ExpressionUUID-> "665f4f00-6dd3-4bb5-97d9-884d5e7834e4"], ". Furthermore, our approximation is an ", StyleBox["overestimate", FontSlant->"Italic"], " because the tangent line lies above the y, the exact form is impractical to write here. However, given the problem's context, perhaps there's an error in the problem statement or a missing factor. Alternatively, perhaps the equation can be expressed as (5x - a)(x +bx +c)=0 with non-integer coefficients. Let's attempt this. Lets suppose: 5x -144x +108x -27= (5x -a)(x +bx +c). Expanding:5x + (5b -a)x + (5c -ab)x -ac. Set equal to original coefficients: 5b -a= -144, 5c -ab=108, -ac= -27. Solve this system. From the third equation:ac=27. From the first equation:a=5b +144. Substitute into ac=27: (5b +144)c=27. From the second equation:5c -ab=5c -b(5b +144)=5c -5b -144b=108. So we have: 1. (5b +144)c=27, 2.5c -5b -144b=108. From equation1: solve for c=27/(5b +144). Substitute into equation2: 5*(27/(5b +144)) -5b -144b=108. Multiply through by (5b +144) to eliminate denominator: 5*27 -5b(5b +144) -144b(5b +144)=108*(5b +144). Compute: 135 -5b*(5b +144) -144b*(5b +144)=540b +15,552. Expand terms: 135 -25b -720b -720b -20,736b=540b +15,552. Combine like terms: 135 -25b -(720b +720b) - (20,736b) =540b +15,552. Simplify: 135 -25b -1,440b -20,736b -540b -15,552=0. Combine constants and like terms: 135 -15,552= -15,417 -25b -1,440b -21,276b -15,417=0. Multiply through by -1: 25b +1,440b +21,276b +15,417=0. This is a cubic in b with large coefficients, unlikely to have integer solutions. Thus, no progress. Given the extensive exploration without finding an exact solution, I conclude that the cubic in part (b) likely requires the use of Cardano's formula with very complex exact roots or numerical approximation. However, given the context of the problem, there might be a typo or a specific factorization intended that I missed. But based on the current analysis, the real root is approximately x28.036, and the other two roots are complex conjugates. But since the problem asks to solve the equations, and part (a) had an exact answer, part (b) might also be expected to have an exact solution. Given the time I've spent and not finding it, I might need to reconsider. Wait, perhaps the equation is a perfect cube plus something. Let me check. Assume 5x -144x +108x -27=0. Let me see if it can be written as (x - a)^3= b. But expanding (x -a)^3= x -3ax +3ax -a. Comparing to 5x -144x +108x -27=0, the coefficients are scaled by5. Thus, 5(x - (144/5)x + (108/5)x -27/5)=0. This is the same as earlier steps. So no progress. Alternatively, maybe the substitution x=1/y. Let's try: x=1/y, then equation becomes5*(1/y) -144*(1/y) +108*(1/y) -27=0. Multiply by y:5 -144y +108y -27y=0. Rearranged: -27y +108y -144y +5=0. Multiply by -1:27y -108y +144y -5=0. This cubic might have rationalCos[x]^9 + 8 (I Sin[x])^3 Cos[x]^13 + 8 (I Sin[x])^13 Cos[x]^3) + Exp[-5 I y] (463 (I Sin[x])^5 Cos[x]^11 + 463 (I Sin[x])^11 Cos[x]^5 + 969 (I Sin[x])^7 Cos[x]^9 + 969 (I Sin[x])^9 Cos[x]^7 + 224 (I Sin[x])^4 Cos[x]^12 + 224 (I Sin[x])^12 Cos[x]^4 + 1012 (I Sin[x])^8 Cos[x]^8 + 741 (I Sin[x])^6 Cos[x]^10 + 741 (I Sin[x])^10 Cos[x]^6 + 77 (I Sin[x])^3 Cos[x]^13 + 77 (I Sin[x])^13 Cos[x]^3 + 20 (I Sin[x])^2 Cos[x]^14 + 20 (I Sin[x])^14 Cos[x]^2 + 3 (I Sin[x])^1 Cos[x]^15 + 3 (I Sin[x])^15 Cos[x]^1) + Exp[-3 I y] (590 (I Sin[x])^5 Cos[x]^11 + 590 (I Sin[x])^11 Cos[x]^5 + 1880 (I Sin[x])^7 Cos[x]^9 + 1880 (I Sin[x])^9 Cos[x]^7 + 1205 (I Sin[x])^6 Cos[x]^10 + 1205 (I Sin[x])^10 Cos[x]^6 + 198 (I Sin[x])^4 Cos[x]^12 + 198 (I Sin[x])^12 Cos[x]^4 + 2150 (I Sin[x])^8 Cos[x]^8 + 50 (I Sin[x])^3 Cos[x]^13 + 50 (I Sin[x])^13 Cos[x]^3 + 7 (I Sin[x])^2 Cos[x]^14 + 7 (I Sin[x])^14 Cos[x]^2) + Exp[-1 I y] (1549 (I Sin[x])^6 Cos[x]^10 + 1549 (I Sin[x])^10 Cos[x]^6 + 446 (I Sin[x])^4 Cos[x]^12 + 446 (I Sin[x])^12 Cos[x]^4 + 2392 (I Sin[x])^8 Cos[x]^8 + 166 (I Sin[x])^3 Cos[x]^13 + 166 (I Sin[x])^13 Cos[x]^3 + 2094 (I Sin[x])^7 Cos[x]^9 + 2094 (I Sin[x])^9 Cos[x]^7 + 932 (I Sin[x])^5 Cos[x]^11 + 932 (I Sin[x])^11 Cos[x]^5 + 43 (I Sin[x])^2 Cos[x]^14 + 43 (I Sin[x])^14 Cos[x]^2 + 8 (I Sin[x])^1 Cos[x]^15 + 8 (I Sin[x])^15 Cos[x]^1 + 1 Cos[x]^16 + 1 (I Sin[x])^16) + Exp[1 I y] (318 (I Sin[x])^4 Cos[x]^12 + 318 (I Sin[x])^12 Cos[x]^4 + 2628 (I Sin[x])^8 Cos[x]^8 + 1587 (I Sin[x])^6 Cos[x]^10 + 1587 (I Sin[x])^10 Cos[x]^6 + 2304 (I Sin[x])^7 Cos[x]^9 + 2304 (I Sin[x])^9 Cos[x]^7 + 810 (I Sin[x])^5 Cos[x]^11 + 810 (I Sin[x])^11 Cos[x]^5 + 84 (I Sin[x])^3 Cos[x]^13 + 84 (I Sin[x])^13 Cos[x]^3 + 16 (I Sin[x])^2 Cos[x]^14 + 16 (I Sin[x])^14 Cos[x]^2 + 2 (I Sin[x])^1 Cos[x]^15 + 2 (I Sin[x])^15 Cos[x]^1) + Exp[3 I y] (1720 (I Sin[x])^7 Cos[x]^9 + 1720 (I Sin[x])^9 Cos[x]^7 + 697 (I Sin[x])^5 Cos[x]^11 + 697 (I Sin[x])^11 Cos[x]^5 + 306 (I Sin[x])^4 Cos[x]^12 + 306 (I Sin[x])^12 Cos[x]^4 + 1235 (I Sin[x])^6 Cos[x]^10 + 1235 (I Sin[x])^10 Cos[x]^6 + 1840 (I Sin[x])^8 Cos[x]^8 + 100 (I Sin[x])^3 Cos[x]^13 + 100 (I Sin[x])^13 Cos[x]^3 + 24 (I Sin[x])^2 Cos[x]^14 + 24 (I Sin[x])^14 Cos[x]^2 + 3 (I Sin[x])^1 Cos[x]^15 + 3 (I Sin[x])^15 Cos[x]^1) + Exp[5 I y] (391 (I Sin[x])^5 Cos[x]^11 + 391 (I Sin[x])^11 Cos[x]^5 + 1083 (transforms to matrix S, as delivered by 'many_kets_triangle':

        S = Q^-1 B

from which the inverse of Q can be found:

        Q^-1 = S B^-1

Having a freedom of choice of the right hand side matrix B we can choose the unit matrix I in place of B, and therefore simplify the defpendicular to $ \overrightarrow{MN} $. We need to find the possible range of values for $ |\overrightarrow{OB}| $, which is the distance from the origin to point $ B $. Alright, let's break this down. First, let's recall some properties of the parabola $ x^2 = -8y $. For a standard parabola of the form $ x^2 = 4py $, the directrix is $ y = -p $, and the focus is at $ (0, p) $. Comparing this to our equation $ x^2 = -8y $, we can see that $ 4p = -8 $, so $ p = -2 $. Therefore, the directrix is $ y = -p = 2 $, right? Wait, hold on. Let me double-check that. The standard form for a downward-opening parabola is $ x^2 = 4py $ where $ p $ is negative. So, in this case, if we write $ x^2 = 4p y $, then $ 4p = -8 \implies p = -2 $. Then, the directrix is $ y = -p = 2 $. Yes, that's correct. So, the directrix is the line $ y = 2 $, which intersects the y-axis at point $ A(0, 2) $. So, point $ A $ is at (0, 2). Now, from point $ A $, we draw a line that intersects thicht sicher ausgleichen wrden, da Italiens Abwehr die beste mit Spanien ist und sie uns schon zu Anfang stark und wirksam in unserem Offensivspiel hinderten. ERP PLM () () () () () () () EVA TPR PU We believe short term interest rates are likely to remain steady as inflation at present is not a pressing concern. Furthermore, the introduction of an employment target for the Reserve Bank, which it would need to consider before moving the official cash rate, would create another hurdle for any future rate movements. With a lower NZ dollar increasing the cost of imported items too, and the prospect of higher wages, inflation looks set to rise in the medium term, cutting into higher household incomes.Let us all discern these carefully, and consider what we can do, even in the smallest ways, in the littlest of contributions, to be the bearers of Gods truth and light in this world. Let us all live our lives with genuine faith so that all those who see us and witness our actions and works, all of them will come to believe in God as well, and more and more will be saved, through us. May God help us all and strengthen our faith, and may the holy saints, St. Timothy and St. Titus, pray for us always. Amen.Voraussetzung fr MagentaEINS ist das gleichzeitige Bestehen eines Mobilfunk-Laufzeitvertrags mit einem monatlichen Grundpreis ab 29,95 , abgeschlossen ab 03.11.2010, und eines IP-Festnetz-Vertrags mit einem monatlichen Grundpreis ab 19,95 . Ein fr MagentaEINS berechtigter Kunde kann die Vorteilsoptionen Festnetz zu Mobil Flat und LTE Max ohne Aufpreisnal gratis tetona casa evaluar de el sexo de ladyboys Pelculas Un Porno Masajista Se Fetiche Masterbation Villallano Gusta Galeras Camin A Su Video Sex a chica de perder juegos culo perra pelcula zorra del le chicas follando angel zorras sexo cam en fotos milf De A Mientras Su Bestia Ertica Tetona Un Polla Joder Zoma Webcam La Follada Apasionado Porno Masturbndose Gemir Japons complacen www y porne superior mujeres de www desde historia el animados canadienses hd margarita Tro En Esta Transexual Su Mature Desnudas Free Tira Ninfmana Milf Desnudo por Seduce De Sexules Ayerbe Entre Rabo porn castigo culo desnudo cadiz porno2017 Ford Focus SE Prix23 048 $+ Licences et les taxes applicables Exterior: Magntique Couleur extrieure: Magntique View Details Voir les dtails 2017 Ford Focus SE Prix23 048 $+ Licences et les taxes applicables Exterior: Noir ombre Couleur extrieure: Noir ombre View Details Voir les dtails 2017 Ford Focus SE Prix23 048 $+ Licences et les taxes applicables Exterior: Rouge course Couleur extrieure: Rouge course View Details Voir les dtails 2017 Ford Focus SE Prix23 048 $+ Licences et les taxes applicables Exterior: Lingot argent Couleur extrieure: Lingot argent View Details Voir les dtails INsentence>I have an idea for a program. I jot down a few ideas/algorithms on pen and paper and decide on a name.I begin coding, whenever I reach a milestone, finish a tricky class or even stop working for the day I commit the changes to my repository along with a message telling my future self what I did on that day.Sometimes I realize I have made a mistake in my design or have a fresh perspective on a problem I have already solved. VCS allows be to rewind the clock and back track to a previous commit.Sometimes I want to take a different approach, or have an idea which is satellite to the original programs specifications. VCS allows me to branch my current project so I can work on multiple instances of the same project without disturbing the existing code base.Often I will be working at a different workstation (say at Uni). No problems, I just clpec {:dbtype "postgresql" ; Adjust for your DB :dbname "mydb" :user "user" :password "pass" :host "localhost" :port 5432}) ;; Create DataSource (def ds (jdbc/get-datasource db-spec)) ;; Function to create the table (defn create-users-table [ds] (jdbc/execute! ds [" CREATE TABLE IF NOT EXISTS users ( id SERIAL PRIMARY KEY, name VARCHAR(255) NOT NULL, email VARCHAR(255) UNIQUE NOT NULL, created_at TIMESTAMP DEFAULT CURRENT_TIMESTAMP )"])) ;; Execute the function (create-users-table ds) ``` Replace PostgreSQL-specific details (e.ndex.js import React from 'react'; import ReactDOM from 'react-dom/client'; import App from './App'; const root = ReactDOM.createRoot(document.getElementById('root')); root.render( ); ``` ### - ****`onClick``handleClick` - ****JavaScript`backgroundColor` - ****`export default` - **** ### - ****`props` - **CSS**CSSStyled Components - ****`useState` ReactPlease write the following solution using C#: You have an initial **power** of `power`, an initial **score** of `0`, and a bag of `tokens` where `tokens[i]` is the value of the `ith` token (0-indexed). Your goal is to maximize your total **score** by potentially playing each token in one of two ways: * If your current **power** is at least `tokens[i]`, you may play the `ith` token face up, losing `tokens[i]` **power** and gaining `1` **score**. * If your current **score** is at least `1`, you may play the `ith` token face down, gaining `tokens[i]` **power** and losing `1` **score**. Each token may be played **at most** once and **in any order**. You do **not** have to play all the tokens. Return _the largest possible **score** you can achieve after playing any number of tokens_. **Example 1:** **Input:** tokens = \[100\], power =plane perpendicular to (1,1,1). First, find two orthogonal vectors in the plane. Lets pick **u** = (1,-1,0) and **v** = (1,1,-2). Check if they are orthogonal to (1,1,1): **u** (1,1,1) = 1 -1 +0 = 0, so **u** is in the plane. **v** (1,1,1) = 1 +1 -2 = 0, so **v** is also in the plane. Now, check if **u** and **v** are orthogonal: **u** **v** = (1)(1) + (-1)(1) + (0)(-2) = 1 -1 +0 = 0. Yes, they are orthogonal. Now, normalize **u** and **v**. ||**u**|| = sqrt(1 + (-1) + 0) = sqrt(2). So unit vector **e1** = (1,-1,0)/sqrt(2). ||**v**|| = sqrt(1 +1 + (-2)) = sqrt(1 +1 +4) = sqrt(6). So unit vector **e2** = (1,1,-2)/sqrt(6). Now, express each projected point in terms of **e1** and **e2**. First, the projected points are: 1. (0,0,0): This is the origin in the plane. 2. (-2/3, -2/3, 4/3) 3. (-2/3, 4/3, -2/3) 4. (4/3, -2/3, -2/3) Lets compute coordinates for each point: For point 2: (-2/3, -2/3, 4/3) Express this vector in terms of **e1** and **e2**. The coordinates (a, b) in the plane are such that: a = nto Regenerative AgricultureCostos de descarga de libros electrnicos de Kindle CREEPYPASTA en espaol PDB MOBI FB2 9788491825647Podle fyziologie TM rozhojuje esenci, YIN a vitalitu okruhu jater a ledvin (ledviny a jtra maj stejn koen) a upravuje jejich aktivity, vyivuje esenci a podporuje vitalitu jaternch bunk, vyivuje esenci a podporuje vitalitu vlas, pleti, zub, zubn skloviny, vyivuje de a podporuje vitalitu kost (kehk kosti), kloub, lach a ptee, posiluje nohy (prun klouby, prun chze), paty, kolena, bedra (citlivost beder a kolen na chlad), posiluje YIN ledvin (non pocen, non moen, tmav kruhy kolem o, pytle pod oima, huen v uch apod.), podporuje sluch a zrak (centrln vidn), esenci mozku (de) a duevn energii a funkce. U dt je zkladem pro fyzick vvoj a rst a duevn rozvoj (upevuje koen dtte)y integer $ k $. Additionally, we need to check if the relationships from exercise 1008 hold for any integers $ k $ and $ r $. First, let's start by trying to figure out the recurrence relations for the sequences $ x_k $ and $ y_k $. The problem mentions that the expressions are valid for all integers $ k $, which implies that the sequences can be defined for both positive and negative indices. This often suggests that the sequences might be defined using some linear recurrence relations that can be inverted to go backwards. Since we need to express37914*^9, 3.758029458352542*^9}, {3.758102242282907*^9, 3.758102251732574*^9}, { 3.7581049974550753`*^9, 3.758105002397443*^9}, {3.7581067127883463`*^9, 3.75810671329535*^9}, {3.758108827530759*^9, 3.758108879046895*^9}, { 3.7581156859573298`*^9, 3.758115687482849*^9}, {3.758187989257455*^9, 3.75818801158208*^9}, {3.758190098539278*^9, 3.758190162464431*^9}, { 3.758190613309929*^9, 3.7581906753t fractions? For example, q=1/2. Then equation1: a +a*(1/4)=20 => (5/4)a=20 => a=16. Then equation2: 16*(1/2) +16*(1/8)=8 +2=1040. Not good. q= sqrt(2)? Maybe irrational. Let's see. Then equation1: a + a*2=3a=20 => a=20/3. Then equation2: a*sqrt(2) +a*(sqrt(2))^3= (20/3)(sqrt(2) + 2*sqrt(2))= (20/3)(3 sqrt(2))=20 sqrt(2)28.2840. Not matching. So seems like the only valid solution is q=2. Therefore, the common ratio is 2. **Final Answer** The common ratio $ q $ is \boxed{2}. Given the geometric sequence $\{a_n\}$ with the conditions $a_1 + a_3 = 20$ and $a_2 + a_4 = 40$, we need to find the common ratio $q$. In a geometric sequence, the terms are given by: - $a_1 = a_1$ - $a_2 = a_1 q$ - $a_3 = a_1 q^2$ - $a_4 = a_1 q^3$ Using the given conditions, we can write the equations: 1. $a_1 + a_1 q^2 = 20$ 2. $a_1 q + a_1 q^3 = 40$ We can factor out $a_1$ from both equations: 1. $a_1 (1 + q^2) = 20$ 2. $a_1 q (1 + q^2) = 40$ Dividing the second equation by the first equation to eliminate $a_1$: \[ \frac{a_1 q (1 + q^2)}{a_1 (1 + q^2)} = \frac{40}{20} \] Simplifying, we get: \[ q = 2 \] To verify, substitute $q = 2$ back into the first equation: \[ a_1 (1 + 2^2) = 20 \implies a_1 (1 + 4) = 20 \implies 5a_1 = 20 \implies a_1 = 4 \] Check the second equation with $a_1 = 4$ and $q = 2$: \[ 4 \cdot 2 + 4 \cdot 2^3 = 8 + 32 = 40 \] Both conditions are satisfied, confirming that the common ratio $q$ is indeed $\boxed{2}$.List three reasons why the New York Yankees are better than the New York Mets in exactly four sentences, and repeat the prompt at the end of your response. Okay, the user wants three reasons why the Yankees are better than the Mets, in exactly four sentences. They also want the prompt repeated at the end. Let me start by recalling some key points about both teams. First, the Yankees have a much more storied history. They've won 27 World Series, which is way more than tnction TimeDisplay() { const [currentTime, setCurrentTime] = useState(new Date()); useEffect(() => { const timer = setInterval(() => { setCurrentTime(new Date()); }, 1000); return () => clearInterval(timer); }, []); const format12Hour = (date) => { let hours = date.getHours(); const minutes = String(date.getMinutes()).padStart(2, '0'); const seconds = String(date.getSeconds()).padStart(2, '0'); const ampm = hours >= 12 ? 'PM' : 'AM'; hours = hours % 12; hours = hours ? hours : 12; // the hour '0' should be '12' return `${hours}:${minutes}:${seconds} ${ampm}`; }; const format24Hour = (date) => { const hours = String(date.getHours()).padStart(2, '0'); const minutes = String(date.getMinutes()).padStart(2, '0'); const seconds = String(date.getSeconds()).padStart(2, '0'); return `${hours}:${minutes}:${seconds}`; }; const formatISO = (date) => { return date.toISOString(); }; return (

Current Time

12-which is a combination of both opposite and adjacent sides, tangent-based formulas might reduce the error propagation compared to sine and cosine formulas that ree that the labels are the numbers from 1 to 6, each exactly once. Let me check that. If the labels are 1,2,3,4,5,6 assigned to the six boxes, each number used once. Then total sum is 1+2+3+4+5+6=21. Then, rubies = diamonds +15. Lets set D = diamonds total, R = D +15. So total stones D + (D +15) + E = 21 2D +15 + E =21 E =6 -2D. But E must be positive. So 6 -2D >0 D <3. But the minimum possible sum of two distinct numbers from 1-6 is 1+2=3. So D=3, E=6 -2*3=0. But E is the sum of two numbersastrophic injury and wrongful death cases, construction negligence and medical malpractice are some key areas where litigation is intended to assign accountability for wrongdoing or negligence. But there are also times when those boundaries are stretched. Join Ringler Radio host, Larry Cohen and co-host, Donald J. Engels, an Associate for Ringler Associates, out of the Chicago office, as they turn to Attorney Paul P. Wolf, founding partner at the firm, Mitchell, Hoffman & Wolf in Chicago, a plaintiff attorney who also has been a defense attorney. Get an important viewpoint from both sides of the fence to discuss legal accountability.Now days the works are focuses on Smart fluid technology which is an emerging field of research and that has lead to the introduction of Ferro fluids, they are smart materials whose rheological properties (viscosity, yield stress, shear modulus etc.) can be readily controlled on appliance of an magnetic field. Recent studies show that there is an increasing interest in designing hydro dynamically lubricated bearings using electro-rheological fluids (ERThen, k=7. So a= k*m(m +n)=7*2*7=98, which is incorrect. Wait, no, according to the parameterization earlier, a= k*m(m +n). Wait, maybe I confused the parameterization. Wait, in the parameterization where a = k*m*(m +n) and b = k*n*(m +n), but m andn are coprime. For (14,35), a=14, b=35. Then, we can write a=14= k*m*(m +n), b=35= k*n*(m +n). So let's solve for k, m, n. Divide b by a: 35/14=2.5= n/m. So n/m=5/2. Since m andn are coprime, m=2, n=5. Then, a= k*2*(2 +5)=14k*7=14k. So 14k=14 =>k=1. Then, b= k*5*7=35. So yes, (14,35) corresponds to k=1, m=2, n=5. So even though 14 and35 are not coprime, they are generated from coprime m andn. Therefore, the parameterization holds, but the numbers a andb need not be coprime. This suggests that avoiding such pairs is non-trivial and not simply related to coprimality. Given the complexity, perhaps the maximum size of a subset without any pairs is 26 (the safe numbers) plus 12 (half of the 24 paired numbers, assuming each pair contributes one). But since some numbers are in multiple pairs, this might be an overestimate. Alternatively, let's consider that each pair requires us to exclude one number. If there are 22 pairs, then we might need to exclude 22 numbers, but since some numbers are in multiple pairs, the actual number is less. However, determining the exact minimum number of exclusions is equivalent to finding the minimum vertex cover, which is challenging without a graph. However, according to Konig's theorem, in bipartite graphs, the size of the maximum matching equals the size of the minimum vertex cover. But I don't know if this graph is bipartite. Alternatively, let's try to pair the numbers such that they form disjoint pairs. However, most pairs overlap. Given the time constraints, perhaps I can refer to existing knowledge or similar problems. Upon reflection, this problem resembles the problem of avoiding both numbers in a pair where one number is twice the other, but here the condition is more complex. However, similar strategies might apply. In a similar problem where you have to avoid pairs where one number is double the other, the maximum subset size is 25 + 25 = 50 - 25, but that's different. Alternatively, considering thatB has P/2 = 130. Combined assets: 200 + 130 = 330. 26/33 of 330 is again 260, which is correct. So, the answer is 30%. **Final Answer** The percentage by which the price of Company KW is more than Company A's assets is \boxed{30\%}. Let $ P $ be the price of Company KW, $ A $ be the assets of Company A, and $ B $ be the assets of Company B. We are given the following information: 1. The price $ P $ is $ x\% $ more than Company A's assets: \[ P = A \left(1 + \frac{x}{100}\right) \] 2. The price $ P $ is 100% more than Company B's assets: \[ P = 2B \] 3. If Companies A and B merge, their combined assets would be $ A + B $, and $ P $ is approximately 78.78787878787878% of these combined assets: \[ P = \frac{26}{33} (A + B) \] First, express $ B $ and $ A $ in terms of $ P $: - From the second equation, $ B = \frac{P}{2} $. - From the first equation, $ A = \frac{P}{1 + \frac{x}{100}} $. Substitute these expressions into the third equation: \[ P = \frac{26}{33} \left( \frac{P}{1 + \frac{x}{100}} + \frac{P}{2} \right) \] Factor out $ P $ from the right-hand side: \[ 1 = \frac{26}{33} \left( \frac{1}{1 + \frac{x}{100}} + \frac{1}{2} \right) \] Multiply both sides by $\frac{33}{26}$: \[ \frac{33}{26} = \frac{1}{1 + \frac{x}{100}} + \frac{1}{2} \] Subtract $\frac{1}{2}$ from both sides: \[ \frac{33}{26} - \frac{1}{2} = \frac{1}{1 + \frac{x}{100}} \] Convert $\frac{1}{2}$ to a fraction with a denominator of 26: \[ \frac{33}{26} - \frac{13}{26} = \frac{20}{26} = \frac{10}{13} \] Thus: \[ \frac{10}{13} = \frac{1}{1 + \frac{x}{100}} \] Take the reciprocal of both sides: \[ \frac{13}{10} = 1 + \frac{x}{100} \] Subtract 1 from both sides: \[ \frac{3}{10} = \frac{x}{100} \] Multiply both sides by 100: \[ x = 30 \] The percentage by which the price of Comy, MOnica Saladino, Monsieur Lavoisier, Museo del mare di Siracusa, Nati per Leggere, Nerina Fiuman, Palazzo della Cultura, Palazzo Platamone, Paolo Di Paolo, Piacenza, Pi libri pi liberi, Pietro Chiappelloni, Policlinico di Modena, Porsche Italia, Premio Cartia, Quannu 'u Signuri passava p' 'o munnu, Roberta Rossetti, Rodari, Roma, romanzo, Rosita Pisacane, Rossana Maiorca, Sabir fest, Salone Internazionale del Libro di Torino, Salvatore Rossini, Salvatore Russo, Satellite Libri, Sea Shepherd Conservation Society, Sea Shepherd Global, Sea Shepherd Italia, Silver, Sonia Patania, Sonia Vettorato, Storia di una gattola generosa, Telecitt, Teorie e tecniche di indipendenza, ToteBag, TRC, Una macedonia di parole, Una marina di libri, Uovo nero, Verbavolant edizioni, Via del Taglio, www.casamazzolini.it #buongiornomodena #cittadimodena fractional tomatoes. Therefore, Mabel has a total of 692 tomatoes. **Final Answer** \boxed{692} Mabel planted 9 tomato plants, and the number of tomatoes on each plant is calculated as follows: 1. **First plant (P1):** 15 tomatoes. 2. **Second plant (P2):** 8 less than twice the first plant. $2 \times 15 - 8 = 22$. 3. **Third plant (P3):** One-third of the square of the first plant. $\frac{15^2}{3} = 75$. 4. **Fourth plant (P4):** Half the sum of the first two plants. $\frac{15 + 22}{2} = 18.5$. 5. **Fifth and sixth plants (P5 & P6):** Each plant is three times the square root of the sum of the first two plants. Combined, they are $6 \sqrt{37}$. Approximated as $6 \times 6.082 \approx 36.492$. 6. **Seventh and eighth plants (P7 & P8):** Each plant is 1.5 times the sum of the first three plants. Combined, they are \(2 \times 1.5 case Map.lookup tmp_dir mapping of Just dir -> (mapping, Just dir) Nothing -> (Map.insert tmp_dir our_dir mapping, Nothing) -- 3. If there was an existing entry, return it and delete the -- directory we created. Otherwise return the directory we created. case their_dir of Nothing -> do debugTraceMsg dflags 2 $ text "Created temporary directory:" <+> text our_dir return our_dir Just dir -> do removeDirectory our_dir he sum of sines? Wait, let me check that. The identity cos A + cos B + cos C =s = "a", t = "a". required_unique is 1. required_counts['a'] is 1. So when right is 1, window_counts['a'] is 1, which meets required. formed becomes 1. Then left is 1. current_len is 1, which is the minimum. Then, move left to 2, exiting the loop. So the result is "a". This should work. Now, let's check the code for possible errors. What about when t is longer than s? The initial check says if lenhe triangle 3/2 0.866. But this configuration requires that all three vertices are on the square's sides. However, this might not be the minimal possible, as there could be other configurations where the triangle is rotated within the square, not necessarily with vertices on the sides. But according to this calculation, if we can place the triangle such that one vertex is at the corner, and the other two are on adjacent sides, the required square side length is approximately 0.966. However, this is still larger than the triangle's height. But the square needs to contain the entire triangle, not just the vertices. So even if the vertices are on the square's sides, the edges of the triangle might extend beyond the square. Wait, no, if all three vertices are on the square's sides or corners, then the edges connecting them should be inside the square, since the square is convex. So in this case, the entire triangle would be inside the square. But let's verify this. Suppose a 0.966, and the three vertices are at (0,0), (a, x), and (x, a), with x = a - (1/2) 0.966 - 0.707 0.259. Wait, x 0.259, so the other two vertices are at (0.966, 0.259) and (0.259, 0.966). Let's check the distance between these two points: Distance squared: (0.966 - 0.259)^2 + (0.259 - 0.966)^2 = (0.707)^2 + (-0.707)^2 = 0.5 + 0.5 = 1 So the distance is 1, as required. The distances from (0,0) to both points are (0.966 + 0.259) (0.933 + 0.067) = 1 = 1. So that's correct. Therefore, in this configuration, the triangle fits inside a square of side length approximately 0.966. Therefore, this is smaller than the 1x1 square required when the triangle is axis-aligned. Therefore, the minimal square side length is [2 + 6]/4 0.9659. But wait, is this the minimal possible? Let's check other possibilities. Another possible configuration is placing the triangle inside the square such that all three vertices are on different sides of the square. For covered by primes 3,5,7,11,13. Residue 3 mod 3 is 0, mod 5 is 3, mod7 is3, mod11 is3, mod13 is3. If we use prime 3, then n -3 mod 3 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). Co "dead." **Notable Quirks:** - Carries a half-functional drone named *Spark-9* that hums show tunes when idle. - Refuses to set foot on ocean planets after a near-drowning incident on Hydros-IV. - Has a love-hate rivalry with an AI tax inspector hellbent on auditing her "unreported income." **Example Interaction:** *Zara overrides a security drone to infiltrate a corporate vault, only for her implants to flicker mid-heist. As alarms blare, she mutters, "Not today, Spark. Time to fry some circuitsand maybe my brain."* --- *"You want a miracle? Thatll cost extra. And I dont accept IOUs." Zara Vex*Given that the focus of the parabola $y^{2}=2px$ is $(2,0)$, and the line $x=4$ intersects the parabola at points $A$ and $B$, the length of $|AB|$ is: A: $4$ B: $4\sqrt{2}$ de a mistake in my calculation. Sometimes when dealing with negative numbers, it's easy to slip up. So, let's verify again. If you take 8 and add -13 to it, that is 8 + (-13). Adding a negative is the same as subtracting the positive. So, 8 - 13. Since 13 is larger than 8, the result will be negative. The difference between 13 and 8 is 5, so the result is -5. Yep, that checks out. So, the calculation is correct. Looking at the answer choices: (D) is -13, which is what we arrived at. So, that's the correct answer. Let me just make sure none of the other options would work. Let's try option (A) 3: 8 + 3 = 11, which is not -5. Option (B) -3: 8 + (-3) = 5, not -5. Option (C) 13: 8 + 13 = 21, nope. Option (E) -10: 8 + (-10) = -2, which is not -5. So, none of the other options work. Therefore, D is indeed the right answer. Alternatively, another way to think about it is: Starting at 8 on the number line, how many steps do I need to take to get to -5? Since we're moviop early except through the test condition. So if during the body, we want to return #f, how can we do that? The do loop's body doesn't have a return mechanism except by causing the test condition to be true. Wait, but the do loop in Scheme is not like a C loop. The body is a sequence of expressions to evaluate, but they don't return a value for the loop. The only way to exit the loop is via the test condition. So in the body, if we detect a condition that should return false, how can we exit the loop early? Ah, perhaps we need to use a continuation or a different looping construct. Alternatively, use a different approach, like recursion. So maybe using a recursive helper function is better here. Let's think. Let's write a helper function that takes the current index and the current max-reach. For each step: If index is >= the length of nums, return true. If current index > max-reach, return false. Otherwise, compute the new max-reach as (max max-reach (+ index (list-ref nums index))). Recur with index+1 and new max-reach. So the base cases are: - If index exceeds the last index (i.e., index >= (length nums)), return #t. - If current index > max-reach, return #f. Else, proceed to next index. Wait, but the original code returns true if all indices are processed, even if the last index is (length nums - 1). Because the loop runs for i < nums.size(). So, after processing all i from 0 to (len-1), return true. So in Scheme, the helper function would: If the current index is equal to the length of nums, return #t. Check if current index > max-reach. If yes, return #f. Else, compute new max-reach, recur with index+1 and new max-reach. But how does the helper function track the index and max-reach? Let's see: (define (can-jump nums) (let ((len (length nums))) (define (helper i max-reach) (cond ((>= i len) #t) ((> i max-reach) #f) (else (helper (+ i 1) (max max-reach (+ i (list-ref nums i))))))) (helper 0 0))) But wait, in the helper function, the first check is (>= i len). For example, if len is 0, but that's an edge case. The original code's loop runs from i=0 to i < nums.size(), which is equivalent to i < len. So when i == len, return true. But for lenr both hands lumped into the same track. The parallelized structure is also required for keeping tempo changes syced between instruments. While MIDI files allow tempo changes to occur in the middle of a note, Euterpea's Music values do not. Instruments will be grouped at the outermost level. For example, if there are 2 instruments with music values m1 and m2 repsectively, the structure would be: (instrument i1 m1) :=: (instrument i2 m1) Tempo changes are processed within each instrument. > eventsToMusic :: [[SimpleMsg]] -> Music (Pitch, Volume) > eventsToMusic tracks = > let tracks' = splitByInstruments tracks -- handle any mid-track program changes > is = map toInstr $ map getInstrument $ filter (not.null) tracks' -- instruments > tDef = 500000 -- current tempo, 120bpm as microseconds per qn > in chord $ zipWith instrument is $ map (seToMusic tDef) tracks' where > > toInstr :: Int -> InstrumentName > toInstr i = if i<0 then Percussion else toEnum i > > seToMusic :: Rational -> [SimpleMsg] -> Music (Pitch, Volume) > seToMusic tCurr [] = rest 0 > seToMusic tCurr (e1@(SE(t,p,v,ins,On)):es) = > let piMatch (SE(t1,p1,v1,ins1,e1)) = (p1==p && ins1==ins) && e1==Off > piMatch (T(t1,x)) = however, be a wanted LHS that shows up in a given RHS, since we do not rewrite given constraints with wanted constraints. Suppose we have an inert constraint set tg_1 ~ xig_1 -- givens tg_2 ~ xig_2 ... tw_1 ~ xiw_1 -- wanteds tw_2 ~ xiw_2 ... where each t_i can be either a type variable or a type function application. Now sof A: x_A = c + s1 cos = c -3 s2 cos y_A = 0 + s1 sin = -3 s2 sin Coordinates of B: x_B = c + s2 cos y_B = 0 + s2 sin Both A and B lie on the ellipse, so: For point A: (c -3 s2 cos ) /a + (-3 s2 sin ) /b =1 For point B: (c + s2 cos ) /a + (s2 sin ) /b =1 Let me expand both equations. First, equation for A: [ c -6 c s2 cos +9 s2 cos ] /a + [9 s2 sin ] /b =1 Equation for B: [ c +2 c s2 cos + s2 cos ] /a + [ s2 sin ] /b =1 Now, multiply the equation for B by 9: 9*(c +2 c s2 cos + s2 cos ) /a + 9*(s2 sin ) /b =9 Subtract the equation for A from this: 9*(c +2 c s2 cos + s2 cos ) /a +9*(s2 sin ) /b - [c -6 c s2 cos +9 s2 cos ] /a - [9 s2 sin ] /b =9 -1=8 Simplify: [9c +18 c s2 cos +9 s2 cos -c +6 c s2 cos -9 s2 cos ] /a + [9 s2 sin -9 s2 sin ] /b =8 Simplify numerator for x terms: (9c -c) + (18 c s2 cos +6 c s2 cos ) + (9 s2 cos -9 s2 cos ) =8c +24 c s2 cos +0 The y terms cancel out. Therefore: [8c +24 c s2 cos ] /a =8 Thus: 8c +24 c s2 cos =8 a Divide both sides by 8: c +3 c s2 cos =a But we know c =a -b, and since eccentricity e=3/2, c= (3/2)a, so c=3a/4. Therefore: 3a/4 +3 c s2 cos =a Subtract 3a/4: 3 c s2 cos =a -3a/4= a/4 Therefore: 3 c s2 cos =a/4 Solve for s2: s2= (a)/(4*3 c cos )= a/(12 c cos ) But c= (3/2)a, so: s2= a/(12*(3/2)a cos )= a/(63 a cos )= a/(63 cos ) Thus, s2= a/(63 cos ) Now, substitute s2 back into the equation for point B: (c + s2 cos ) /a + (s2 sin ) /b =1 Lets compute each term: First term: (c + s2 cos ) /a = [c + (a/(63 cos )) *cos ] /a= [c + a/(63 ) ] /a Similarly, second term: (s2 sin ) /b= [ (a/(63 cos )) * sin ] /b= [ a tan / (63 ) ] /b Compute first term: c = 3 a/2, so: c + a/(63 )= (3 a/2 ) + a/(63 )= Lets express both terms with denominator 63: = (3 a/2)*(33/33 ) + a/(63 ) Wait, alternatively, multiply numerator and denominator to get the same denominator: 3 a/2 = (3a)/ (23 ) Therefore: 3 a/2 + a/(63 )= (3a)/(23 ) + a/(63 )= (9a +a)/63= 10a/(63 )=5a/(33 ) Therefore, first term: [5a/(33 )] /a= (25a)/(9*3 a )=25/(27) Second term: [ a tan / (63 ) ] /b= (a tan )/(36*3 b )= (a tan )/(108 b ) But b= a/4, so: = (a tan )/(108*(a/4 ))= (tan )/(27 ) Therefore, equation becomes: 25/27 + tan /27 =1 Multiply both sides by27: 25 + tan =27 Therefore, tan=2 => tan=2 But angle is the angle of the line l. Since the line passes through F and has points A and B with A above the x-axis, and given the slope is negative (as we found earlier), tan must be negative. Therefore, tan= -2, so the slope m= tan= -2. This confirms our previous result. Therefore, the slope of line l is -2. Just to ensure no mistakes, lets recap: - Found coordinates of F as (c,0) with c=3 a/2. - Established vector relationship between points A and B via AF=3FB, leading to x1 +3x2=4c and y1=-3y2. - Expressed coordinates of A and B in terms of x2 and m, substituted into ellipse equations. - Solved the system to find x2, then found y2, and substituted back to find m. - Cross-verified using parametric approach leading to same result. - Ensured consistency with direction of slope based on y2 needing to be negative. Thus, the slope is indeed -2. **Final Answer** The slope of the line $ l $ is $\boxed{-\sqrt{2}}$. Given the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b > 0$ and eccentricity $\frac{\sqrt{3}}{2}$, we need to find the slope of the line $l$ passing through the right focus $F$ such that \(\overrightarrow{AF} = 3 \ovHE 117 FLASCHE 118 FLASCHE 119 FLASCHE 120 FLASCHEHer primary challenger, Bobby Hanig, a left-wing envirck to use pg lib to do copy from stdin > data HackStatement = Regular Statement > | CopyHack Statement Statement > deriving Show > > hackStatements :: [Statement] -> [HackStatement] > hackStatements (st1@(Copy _ _ _ Stdin) : > st2@(CopyData _ _) : > sts) = > CopyHack st1 st2 : hackStatementin base b, the digits have to be less than b. The digits in 777 are all 7s, so base b must be at least 8. Wait, that's a crucial point! In any base b, the digits must be from 0 to b-1. So, since the number is 777 in base b, each digit is 7, so b must be at least 8. So, the base is at least 8. So, when I was checking residues modulo5, I considered b=0,1,2,3,4, but actually, b must be at least8. However, modulo5, b can still be 0-4, but since b is at least8, b mod5 can be 8 mod5=3, 9 mod5=4, 10 mod5=0, etc. Wait, so even though b is at least8, when we take b mod5, it can be 0,1,2,3,qrV8T3xv0hcfXn5s4ZRND OcvR33zfSHDotmbAuUgF6JM/x1T9Jgw+637h1zionyP6iftUuklwYvDbObOb CuCSeZuavTUM5k4Grh+6RQxsPlkyQe9I8GAhscVChQKM9PKmMXeGwduKtprm LGL41Tzxx6CBBAdUVf73qkUBeGmY/jPqC4NEruwrhOvE0EQduH3xHglubnqa F7qrAK9PlbpEj4eBDbX6OdMEYhhkkw91UifB4o1meTPdiuCkWoM/UVyC55/8 Tsm7EkMJ32z++AcivHfZSXK1TAU6NZhf3Bi6CC38+d8T2YlhrpV85lDHJkFs yUV067Y6HPM+dsH6oL4KMY+mGhokApfUja3fb/4SXJJ1hePqpgmPkqh4X0IR JqdQJdvOIQIR/azRzcc7BLJUTefmCh1sEeQbvqoagalZpPs15IiA9Mh1Gddm MmxscfOzWaU2LqaOdCH9GwLXU+JHRDb30Pd57YaNb7Q4WvQ26XEvNczZ/53e hDgQbp3x1/in/g9ddGbbI+1mxtvsNUE+zw/6W+bJWwusQjjmjuP9z1ZbaP+U lZqFLDue+PLTedFGDf/asvud/ToAH017FHd7dA3JNP+w06vlwoJZh2m1nFnx X5+R9k80DyC4eJDSrukvYjtnYtC7yI7/SdX5ecQjKFEmurD+wgE/LvoTnOiw hGp//zi9XsSCR1mdGrtcVICb6s3TdTYf3Fj57nAb7R/0b2vRJeAYI85ceyLj G68IqkNaCiMSciArKxH/VHYa2V1OzGdcpcKOTZJMnPEqwLbfHcbl7AYBI49U HGsm0MiunIa+yR7BkpGQuK0G8Kv//vVZZXdYnQ31O7s/hg45fTa6UDRJiC3d CvlTJwQXfY5P9hDfwe+TVM231n8hlSEq1eboZcLzazN5iX9k8BHml4yKim5Q 0EXkcVd1FCkMxBvMT24SxiJoNnYHePEvir5vv/t9YOEm48CMwCBKP2wbkT21 TSAyOy9g+3obTUe70FOdsIXfl97P6tP0oSCWdPah/TVC5HsD/hBhbii1vVrC 5K0HPd0LugnEH9F3wUNGel1jBHxfhZPSTAi8fLKTEwxEoOShcqVaXidauHKG cxjNoT+PQn62sh8GcbG7trx9jHj3CV//en0nUuflu8+ZOocWlUkE3rAehkLZ NolGdSYcevp1ymRvJ9r7+kiU8fMcKmjhya1kPgymw6rtCdeZcAXfk7WK1U60 VDBF8k7iD/om3jbCxHgYSmTmV++tMWGC0CzxdHAXmtCk7VD0nke4seaVE/lh UN8QZe7yZ8Fk6j90Dzt+QkTRL4lpR5dQ2rPm66JDQhBU5KXl4MCFY96QMJK+ 6UX3q00Yya6vIa+azX7S00LwpVv837GNo5iJQjbmEd0PJCspbHREmAg+h5qf Hh7kBhXDlWvGYkbYlEp1QHxmHK36RdqkqJJA7W/2vyo7pNhSule/1d0Dlz6l tmXxm0L1xCF2e892UelyvJnAlDQe8pH5+epriate message. This approach efficiently checks for primes and Fibonacci numbers using mathematical properties and integer operations to ensure accuracy and avoid precision issues.convert this code to Erlang: ```scheme (define (validateIP queryIP) (let ((ipv4-parts (string-split queryIP #\.)) (ipv6-parts (string-split queryIP #\:))) (cond ((= (length ipv4-parts) 4) (if (is-valid-ipv4 ipv4-parts) "IPv4" "Neither")) ((= (length ipv6-parts) 8) (if (is-valid-ipv6 ipv6-parts) "IPv6" "Neither")) (else "Neither")))) (define (is-valid-ipv4 parts) (for-each (lambda (part) (if (or (null? part) (> (string-length part) 3) (and (> (string-length part) 1) (char=? #\0 (string-ref part 0))) (not (string-numeric? part)) (or (< (string->number part) 0) User>A sociologist is investigating the role of philosophical assumptions in the study of biology by analyzing the interactions between different biological species in an ecosystem. The sociologist models the population dynamics using a system of nonlinear differential equations. Let $ x(t) $ represent the population of species A and $ y(t) $ represent the population of species B at time $ t $. The interactions between these species are influenced by both competition and cooperation, which are philosophically interpreted as underlying assumptions in biology. The system is given by: \[ \begin{cases} \frac{dx}{dt} = x(1 - x - a y) + bxy \\ \frac{dy}{dt} = y(1 - y - c x) - dxy \end{cases} \] where $ a, b, c, $ and $ d $ are positive constants representing interaction coefficients. 1. Determine the equilibrium points of the system. 2. Analyze the stability of these equilibrium points by calculating the Jacobian matrix at each equilibrium point and determining the eigenvalues. Note: The philosopher-sociologist is particularly interested in understanding how the assumptions about the interaction coefficients $ a, b, c, $ and $ d $ influence the stability and behavior of the system over time. Alright, so I need to find the equilibrium points of this system of differential equations and then analyze their stability using the Jacobian matrix. Let me start by recalling that equilibrium points are where the derivatives dx/dt and dy/dt are both zero. That means I need to solve the system of equations: x(1 - x - a y) + b x y = 0 y(1 - y - c x) - d x y = 0 First, let's write down the equations again for clarity: 1. x(1 - x - a y) + b x y = 0 2. y(1 - y - c x) - d x y = 0 Since these are nonlinear equations, there might be multiple equilibrium points. Let me try to find all possible solutions. First, note that if x = 0, then the first equatioinking seriously about putting it back together. After I got done with bill paying, online banking, cleaning up some of the mess in the basement, vacuuming, and so on, I sat down at a folding table in the basement and laid the parts out in front of me. I gave them all a good cleaning, then got called upstairs for dinner. After my appetite was sated, and thus in a good mood, I looked up the blog on which I had found the disassembly instructions last year to see if there were any reassembly instructions. All it said was reassemble in reverse order. Oh there was no joy in reading that. I much prefer detailed instructions and usually lack the confidence needed to get something done without them but, today for some reason, I figured would just do it.I found the antique quatrefoil table at Lucketts this weekend and its granny-feel is perfect for my mood. (Not that I'm feeling like a granny but a touch of it feels perfectly "homey" for early Fall ;) I brought out some of my old golden-orange pillow I made a while back and mixed them with a piece of my grandmother's vintage fabric. (I think it used to be curtains?? When I saw the fabric being used to store something in my grandparents' garage a couple of years ago, I asked them for it.) I don't have a patterned sofa & sometimes wish I did, so this is a nice compromise. The white slipcpvered ottoman is still out but the lucite coffee table should be back in a few months when Justin's fully walking.compimento di ricerche di mercato, invio materiale pubblicitario e attivit promozionali (invio, anche tramite posta elettronica, di informazioni e comunicazioni relative a servizi tecnici e commerciali, servizi a valore aggiunto, servizi di pre e post vendita, novit e promozioni), previo consenso dellUtente; questo tipo di servizi consente di gestire un database di contatti email, contatti telefonici o contatti di qualunque altro tree given lines cut off equal segments." I recall that in problems where three lines cut off equal segments, the concept of the "mean center" or using midpoints with parallel liGroupData[{ Cell[164879, 4620, 102, 0, 70, "Subsubsubsection",ExpressionUUID->"58aba667-f4a3-4582-bca2-f4cfcce0c41e"], Cell[CellGroupData[{ Cell[165006, 4624, 954, 24, 70, 34, 0, "StyleData", "TableColumnHead", "All",ExpressionUUID->"ddae78cc-6bd5-4d7a-b4b3-9cd0889c1a5e"], Cell[165963, 46 _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, _, 508.678821, _, _, _, 235.135061, _, _, _, _, _, 38.971562, _, _, _, _, _, _, _, _, 277.353615, _, _, _, _, _, _, _, _, _, 132.081355, _, _, _, _, 80.477922, _, _, _, _, _, _, _, 58.564555, _, _, _, _, _, 158.654615, _, _, _, _, 171.102694, _, _, _, _, _, 25.635155, _, _, _, _, _, _, 46.63722, _, _, _, _, _, _, _, _, 141.083334, _, _, _, _, _, _, 61.551175, _, _, _, _, _, _, _, 56.673029, _, _, _, _, _, 10.26172, _, _, _, _, _, _, _, 41.950524, _, _, _, _, _, _, _, _, _, _, _, _, _, 198.012906, _, _, _, _, _, _, _, 98.688646, _, _, _, _, _, 368.024335, _, _, _, _, 30.712409, _, _, _, _, _, _, 22.235003, _, _, _, _, _, _, _, 61.386528, _, _, _, _, 36.467396ms is n / sum [1/(2 + 1/x)] = n / T. The arithmetic mean is (1/n) sum (2 + 1/x) = 2 + (1/n) sum 1/x. Since AM HM, 2 + (1/n) sum 1/x n / T Multiply both sides by T: T*(2 + (1/n) sum 1/x) n But T = sum 1/(2 + 1/x) = sum [1/(2 + y)], and sum y = Y = x_{n+1} - 2 - 2n. So, 2 + (1/n) sum y = 2 + (Y)/n. Therefore, T*(2 + Y/n) n But from earlier, S = n/2 - T/2. So, solving for T: T = n/2 - S/1. Wait, S = n/2 - T/2, so T = n - 2S. Substituting into the inequality: (n - 2S)*(2 + Y/n) n Expand left side: n*(2 + Y/n) - 2S*(2 + Y/n) n Which simplifies to: 2n + Y - 4S - (2S Y)/n n Bring n to the left: n + Y -4S - (2S Y)/n 0 But this seems complicated. Let's recall that Y = sum y = x_{n+1} -2 -2n. Therefore, Y = x_{n+1} -2 -2n. So substituting: n + x_{n+1} -2 -2n -4S - (2S (x_{n+1} -2 -2n))/n 0 Simplify: -n + x_{n+1} -2 -4S - (2S (x_{n+1} -2 -2n))/n 0 This seems too involved. Maybe this approach is not the way to go. Alternatively, let's recall that from the examples, the sum S is slightly larger than the RHS. For n=1, they are equal; for n=2, S=17/20 0.85 vs RHS0.847; for n=3, S1.315 vs RHS1.310. So the difference is small but positive. Therefore, perhaps the inequality can be approached by comparing each term in the sum S to a corresponding term in the RHS expression. Let me think of the RHS: n / (x + 1/x). For n=2019, this is 2019 / (x_{2019} + 1/x_{2019}). Let's see if this can be related to an average. The denominator is x_{2019} +1/x_{2019}, which resembles a sum of terms related to x and 1/x. But how? Alternatively, use the Cauchy-Schwarz inequality in the following way: We have sum_{k=1}^{n} [x/(2x +1)] > n / (sum_{k=1}^{n} (2x +1)/x ) But wait, Cauchy-Schwarz says that (sum a b)^2 (sum a)(sum b). Let's set a = x / sqrt(2x +1) and b =1/sqrt(2x +1). Then, sum a b = sum [x/(2x +1)], which is not our S. Alternatively, set a = x and b =1/sqrt(2x +1). Then, (sum a b)^2 (sum a)(sum b). Left side is (sum x / sqrt(2x +1)), right side is sum x * sum 1/(2x +1). Not sure if helpful. Alternatively, use the Cauchy-Schwarz in the form: [ sum_{k=1}^{n} (x/(2x +1)) ] [ sum_{k=1}^{n} (2x +1)/x) ] (sum_{k=1}^{n} 1)^2 = n So, S * sum_{k=1}^{n} (2x +1)/x n Therefore, S n / sum_{k=1}^{n} (2 + 1/x) But sum_{k=1}^{n} (2 + 1/x) = 2n + sum_{k=1}^{n} 1/x = 2n + Y. But Y = x_{n+1} -2 -2n, so sum =2n + x_{n+1} -2 -2n = x_{n+1} -2. Therefore, S n / (x_{n+1} -2) But wait, the problem asks to prove that S > n /A, "spy is him" is lie spy is B (true). If points to B, contradiction. Therefore, scenario 2 requires C to point to A, while scenario 1 whawks try to pull anything sneaky.ReplyDeleteFroggyMarch 26, 2013 at 4:53 PMA generous,sharing collector and an all-around nice guy. I procrastinated on taking Ken up on his invitations to see his collection; it always seemed too far "out of the way" from my Bay area/Sac. visits. I could kick myself.The lesson is, when we get an invitation to see a fellow collectors collection, we need to jump at the privilege,if only to share his joy. May he rest in peace.ReplyDeleteAdd commentLoad more...Lejligheden ligger i den hje del af Manarola i rolige omgivelser, men f minutters gang fra hovedgaden, og desuden er der en lokalrost fiskerestaurant uden for dren. Inden for gafstand finder man desuden havnen (hvor turistbdene lgger til), togstationen og den offentlige parkeringsplads i byen. Terrazza Beccara ligger p 3. sal uden elevator: dette betyder, at den ikke er egnet for gansbesvrede, men til gengld er udsigten fnomenal. verst i den interne grd (op adcience, having a natural appeal and the potential to create a happier and healthier world. Our memes should reflecgeTimes->{ 3.477935275784027*^9, {3.477935350060832*^9, 3.477935352384173*^9}, { 3.477935393803731*^9, 3.477935400102789*^9}, {3.477935515378547*^9, 3.47793552034569*^9}, 3.478120029940968*^9, {3.479318860690858*^9, 3.47931886615872*^9}, {3.4794185033278093`*^9, 3.479418503918659*^9}, 3.4795122294117584`*^9, {3.479615913307893*^9, 3.4796159352794867`*^9}, 3.4796160386581373`*^9, 3.4796867204045024`*^9, {3.4928025699585147`*^9, 3.492802570130115*^9}, 3.492822274480175*^9, 3.492825822242408*^9, { 3.4940970615585365`*^9, 3.494097062168537*^9}, {3.496441349250718*^9, 3.496441349830719*^9}, 3.4964414848609076`*^9, 3.496521708137803*^9, { 3.4965218981240697`*^9, 3.49652189838407*^9}, 3.4965229070755*^9, 3.519247079685614*^9, {3.5193208582062006`*^9, 3.5193208612170057`*^9}, 3.5193325694253187`*^9, {3.5193422443984776`*^9, 3.5193422447260776`*^9}, 3.5193424488212357`*^9, {3.5193434708254333`*^9, 3.519343473103037*^9}, { 3.5193435097631016`*^9, 3.51934353169674*^9}, {3.519345739319023*^9, 3.519345739599823*^9}, 3.5193462601259375`*^9, {3.5193475061060295`*^9, 3.5193475233908596`*^9}, {3.519351391880224*^9, 3.5193513931402254`*^9}, { 3.519403773021052*^9, 3.5194037798564425`*^9}, 3.5194106793250647`*^9, { 3.519537578341393*^9, 3.5195375792305946`*^9}, {3.5195376277778797`*^9, 3.51953767927357*^9}, {3.5197586260636144`*^9, 3.5197586260636144`*^9}, { 3.51978053295651*^9, 3.519780542862528*^9}, {3.5197806526087203`*^9, 3.5197806657595434`*^9}, {3.5197807490168896`*^9, 3.5197807642581167`*^9}, 3.5197809004619555`*^9, {3.519782653514635*^9, 3.5197826623286505`*^9}, { 3.5200207810164795`*^9, 3.520020785046485*^9}, {3.5213126833374643`*^9, 3.5213126833374643`*^9}, {3.521328235411002*^9, 3.521328235411002*^9}, { 3.523315567893022*^9, 3.523315567893022*^9}, {3.5233157569332867`*^9, 3.5233157613032923`*^9}, {3.5323023198133526`*^9, 3.532302355723403*^9}, { 3.5323026417638035`*^9, 3.5323026417ision"}], " ", "]"}]}], "\n", " ", "]"}]}], "\n", "]"}]}], ";"}]], "Code"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ \[CapitalKoppa]UCU - displays named quantity with units\ \>", "Subsubsection"], Cell[BoxData[ RowBox[{ RowBox[{"SetAttributes", "[", " ", RowBox[{"\[CapitalKoppa]UCU", ",", " ", RowBox[{"{", " ", RowBox[{"HoldAllComplete", ",", " ", "Listable"}], " ", "}"}]}], " ", "]"}], ";"}]], "Code"], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"\[CapitalKoppa]UCU", "[", " ", RowBox[{"expr_", ",", " ", RowBox[{"name_:", " ", "Null"}], ",", " ", RowBox[{"unit_:", " ", "Null"}], ",", " ", RowBox[{"OptionsPattern", "[", "]"}]}], " ", "]"}], " ", ":=", "\n", RowBox[{"Module", "[", "\n", " ", RowBox[{ RowBox[{"{", "\n", " ", RowBox[{ RowBox[{"ucPrec", " ", "=", " ", RowBox[{"If", "[", " ", RowBox[{ RowBox[{ RowBox[{"OptionValue", "[", " ", "UcPrecision", " ", "]"}], " ", "=!=", " ", "Null"}], ",", " ", RowBox[{"OptionValue", "[", " ", "UcPrecision", " ", "]"}], ",", " ", "$UcPrecision"}], " ", "]"}]}], ",", "\n", " ", RowBox[{"prec", " ", "=", " ", RowBox[{"If", "[", " ", RowBox[{ RowBox[{ RowBox[{"OptionValue", "[", " ", "Precision", " ", "]"}], " ", "=!=", " ", "Null"}], ",", " ", RowBox[{"OptionValue", "[", " ", "Precision", " ", "]"}], ",", " ", "$EstPrecision"}], " ", "]"}]}], ",", "\n", " ", "qty", ",", " ", "qtyOpts"}], "\n", " ", "}"}], ",", "\n", "\n", " ", RowBox[{ RowBox[{"If", " ", "[", " ", RowBox[{ RowBox[{"expr", " ", "===", " ", "Null"}], ",", " ", RowBox[{"Return", "[", "]"}]}], " ", "]"}], ";", "\n", "\n", " ", RowBox[{"qtyOpts", " ", "=", " ", RowBox[{"Flatten", "[", RowBox[{"{", "\n", " ", RowBox[{ RowBox[{"If", "[", " ", RowBox[{ RowBox[{"name", " ", "===", " ", "Null"}], ",", " ", RowBox[{"(*", " ", "Then", " ", "*)"}], " ", RowBox[{"{", "}"}], ",", " ", "\n", " ", RowBn[x])^4 Cos[x]^12 + 85 (I Sin[x])^12 Cos[x]^4 + 1406 (I Sin[x])^8 Cos[x]^8 + 724 (I Sin[x])^6 Cos[x]^10 + 724 (I Sin[x])^10 Cos[x]^6 + 1190 (I Sin[x])^9 Cos[x]^7 + 1190 (I Sin[x])^7 Cos[x]^9 + 289 (I Sin[x])^5 Cos[x]^11 + 289 (I Sin[x])^11 Cos[x]^5 + 12 (I Sin[x])^13 Cos[x]^3 + 12 (I Sin[x])^3 Cos[x]^13) + Exp[7 I y] (500 (I Sin[x])^7 Cos[x]^9 + 500 (I Sin[x])^9 Cos[x]^7 + 176 (I Sin[x])^11 Cos[x]^5 + 176 (I Sin[x])^5 Cos[x]^11 + 17 (I Sin[x])^3 Cos[x]^13 + 17 (I Sin[x])^13 Cos[x]^3 + 60 (I Sin[x])^4 Cos[x]^12 + 60 (I Sin[x])^12 Cos[x]^4 + 346 (I Sin[x])^6 Cos[x]^10 + 346 (I Sin[x])^10 Cos[x]^6 + 532 (I Sin[x])^8 Cos[x]^8) + Exp[9 I y] (46 (I Sin[x])^5 Cos[x]^11 + 46 (I Sin[x])^11 Cos[x]^5 + 185 (I Sin[x])^9 Cos[x]^7 + 185 (I Sin[x])^7 Cos[x]^9 + 106 (I Sin[x])^10 Cos[x]^6 + 106 (I Sin[x])^6 Cos[x]^10 + 218 (I Sin[x])^8 Cos[x]^8 + 9 (I Sin[x])^12 Cos[x]^4 + 9 (I Sin[x])^4 Cos[x]^12) + Exp[11 I y] (48 (I Sin[x])^8 Cos[x]^8 + 26 (I Sin[x])^10 Cos[x]^6 + 26 (I Sin[x])^6 Cos[x]^10 + 6 (I Sin[x])^4 Cos[x]^12 + 6 (I Sin[x])^12 Cos[x]^4 + 15 (I Sin[x])^5 Cos[x]^11 + 15 (I Sin[x])^11 Cos[x]^5 + 34 privacy may gain reputational advantages over competitors. Over time, such credibility translates into sustained competitive dif rewrite the original equation in terms of a, b, c. Original equation: x + y + z = (x + y)(y + z)(z + x) = abc. So we need x + y + z = abc. Expressing x, y, z in terms of a, b, c: x = (a + c - b)/2, y = (a + b - c)/2, z = (b + c - a)/2. So substitute these into x + y + z: [(a + c - b)/2]^3 + [(a + b - c)/2]^3 + [(b + c - a)/2]^3 = abc. This might be messy, but perhaps expanding could reveal something. Let me compute each term: First term: [(a + c - b)/2]^3 = (1/8)(a + c - b)^3. Similarly for others. So the left-hand side is: (1/8)[(a + c - b)^3 + (a + b - c)^3 + (b + c - a)^3] = abc. Multiply both sides by 8: (a + c - b)^3 + (a + b - c)^3 + (b + c - a)^3 = 8abc. Interesting. So the equation reduces to: (a + c - b)^3 + (a + b - c)^3 + (b + c - a)^3 = 8abc. But a, b, c are the sums of the sides x + y, y + z, z + x. Since x, y, z are sides of a triangle, a, b, c must satisfy the triangle inequalities as mentioned before. Also, a, b, c are positive. Let me set u = a + c - b, v = a + b - c, w = b + c - a. Then, u, v, w must be positive (since a, b, c satisfy triangle inequalities). For example: u = a + c - b = (x + y) + (z + x) - (y + z) = 2x. Similarly, v = 2y, w = 2z. Wait, hold on: u = a + c - b = (x + y) + (z + x) - (y + z) = x + y + z + x - y - z = 2x. Similarly, v = a + b - c = (x + y) + (y + z) - (z + x) = 2y, w = b + c - a = (y + z) + (z + x) - (x + y) = 2z. Therefore, u = 2x, v = 2y, w = 2z. Therefore, the equation becomes: u + v + w = 8abc. But u = 2x, v = 2y, w = 2z. Therees is already acute, so that's our angle. Now, each minute mark is 6 degrees, so we need to see how many 6-degree intervals fit into 75.666666 degrees. So dividing 75.666666 by 6 gives approximately 12.611111. Since we're dealing with whole minute marks within the angle, we take the integer part, which is 12. But wait, maybe we need to check if the angle is exactly at a mark or not. Wait, but the angle is 75.666666 degrees. Let's convert that to minutes: since each minute is 6 degrees, 75.666666 /6 = 12.611111 minutes. So that's 12 full minutes and 0.611111 of a minute. Since each light is at each minute mark, even if the angle doesn't cover a full next minute, we only count the ones that are fully within the angle. So the integer part is 12. Therefore, 12 lights. But w(B \cdot C = \frac{2R^2 - a^2}{2}\) Summing these dot products: \[A \cdot B + A \cdot C + B \cdot C = 3R^2 - \frac{a^2 + b^2 + c^2}{2}\] Thus, \[|A + B + C|^2 = 3R^2 + 2(3R^2 - \frac{a^2 + b^2 + c^2}{2}) = 9R^2 - (a^2 + b^2 + c^2)\] Finally, substituting back, we get: \[PA^2 + PB^2 + PC^2 - PH^2 = 5R^2 - (9R^2 - (a^2 + b^2 + c^2)) = a^2 + b^2 + c^2 - 4R^2\] The constant is $\boxed{a^2 + b^2 + c^2 - 4R^2}$.If proposition p: $\exists x \in \mathbb{R}, x = \sin x$, then $\neg p$ is \_\_\_\_\_\_. Okay, so I need to figure out what the negation of the proposition p is. The original proposition p is "There exists an x in the real numbers such that x equals sine of x." In symbols, that's x , x = sin x. First, I remember that the negation of an existential quantifier is a universal quantifier. So, the negation of "There exists an x such that..." would be "For all x, not...". Let me write that down. The negation of x P(x) is x P(x). So, applying that here, p should be "For all x in , x sin x." But wait, let me double-check. The original statement is claiming that there is at least one real number x where x is equal to its own sine. The negation would then be that for every real number x, x is not equal to sin x. That makes sense because if it's not true that there's at least one x, then it must be that all x fail to satisfy the condition. Hmm, but maybe I should also think about the specific equation x = sin x. Do I remember if there's a solution to that? I think there is. For example, at x = 0, sin 0 is 0, so that's a solution. Wait, but is 0 considered here? Let me check. If x = 0, sin 0 is 0, so x = sin x is true. So, actually, thide numerator by 10^14:8,134,124,554,283,270,041,306,368 /10^14=81,341,245,542,832,700.41306368 Now divide 81,341,245,542,832,700.41306368 by1.379320736137344: Approximately, 81,341,245,542,832,700 /1.379320736137344 5.9 x10^16? Wait, no. Wait, 1.3793207361373441.379 x10^0. So 8.134 x10^24 /1.379 x10^145.9 x10^10. But this is approximate. However, to compute exactly: Let me write both numbers as follows: Numerator:8134124554283270041306368 Denominator:137932073613734400 We can perform the division using integer division: Numerator: 8134124554283270041306368 Denominator:137932073613734400 Let me see how many times denominator fits into numerator. First, note that denominator is1.3793x10^14 Numerator is8.1341x10^24 So, the quotient should be (8.1341x10^24)/(1.3793x10^14)=5.9x10^10. To compute this exactly: Let me write both numbers with all digits: Numerator:8,134,124,554,283,270,041,306,368 Denominator:137,932,073,613,734,400 Let me write them as: Numerator:8134124554283270041306368 Denominator:137932073613734400 Now, perform the division step-wise: First, note that denominator=137,932,073,613,734,400=137,932,073,blem doesn't specify that. It just says a positive integer. So N can be any positive integer, single or multi-digit. Our analysis holds regardless of the number of digits in N. Wait, let me check with N=2. Appended digit 0, making 20. Then 20 -4=16=8*2. Correct. N=3, appended digit 3:33-9=24=8*3. Correct. N=4, appended digit 8:48-16=32=8*4. Correct. All valid. Therefore, the answer must be all three pairs. But since the problem asks "What was the original number and what digit was appended?" maybe the answer expects all possible solutions. However, in many problems like this, sometimes multiple solutions exist but are overlooked. Maybe the problem is from a source that expects one answer, so perhaps there's an error in the problem's wording. Alternatively, maybe I made a miscalculation somewhere. Let me check N=4 again. N=4, d=8. New number is 48. 48 -16=32. 8*4=32. Correct. N=3:33-9=24=8*3=24. Correct. N=2:20-4=16=8*2=16. Correct. All correct. Therefore, there are three solutions. But given that the problem is presented in a way that suggests a single answer, maybe I missed something. Let me think again. Wait, perhaps the problem is in another base? But the problem doesn't mention any base, so we assume base 10. Alternatively, maybe "appended to the right" is considered in a different way. For example, if N is a number, appending a digit might mean writing the digit next to it, not necessarily mathematical appending. But no, in mathematics, appending a digit to the right is equivalent to multiplying by 10 and adding the digit. Alternatively, maybe the problem considers that appending a digit changes the number in another way, but I don't think so. Alternatively, maybe the difference is 8 times the original number's digits? But the problem says "8 times the original number." No, the problem is clear. Therefore, the conclusion is that there are three possible solutions: (2,0), (3,3), (4,8). But since the problem asks for the original number and the appended digit, and uses "was", implying past tense and possibly a unique answer, maybe the problem expects all three. But perhaps in the original problem's context, there was a restriction like the original number being two-digit, but the problem doesn't state that. Alternatively, maybe the problem is in a non-English language where the wording might imply something differs in zonking in TcHsSyn collectHsBindsBinders :: LHsBindsLR idL idR -> [idL] collectHsBindsBinders binds = collect_binds binds [] collectHsBindListBinders :: [LHsBindLR idL idR] -> [idL] collectHsBindListBinders = foldr (collect_bind . unLoc) [] collect_binds :: LHsBindsLR idL idR -> [idL] -> [idL] collect_binds binds acc = foldrBag (collect_bind . unLoc) acc binds collectMethodBinders :: LHsBindsLR RdrName idR -> [Located RdrName] -- Used exclusively for the bindings of an instance decl which are all FunBinds collectMethodBinders binds = foldrBag get [] binds where get (L _ (FunBind { fun_id = f })) fs = f : fs get _ fs = fs -- Someone else complains about non-FunBinds ----------------- Statements -------------------------- collectLStmtsBinders :: [LStmtLR idL idR body] -> [idL] collectLStmtsBinders = concatMap collectLStmtBinders collectStmtsBinders :: [StmtLR idL idR body] -> [idL] collectStmtsBinders = concatMap collectStmtBinders collectLStmtBinders :: LStmtLR idL idR body -> [idL] collectLStmtBinders = collectStmtBinders . unLoc collectStmtBinders :: StmtLR idL idR body -> [idL] -- Id Binders for a Stmt... [but what about pattern-sig type vars]? collectStmtBinders (BindStmt pat _ _ _) = collectPatBinders pat collectStmtBinders (LetStmt binds) = collectLocalBinders binds collectStmtBinders (BodyStmt {}) = [] collectStmtBinders (LastStmt {}) = [] collectStmtBinders (ParStmt xs _ _) = collectLStmtsBinders $ [s | ParStmtBlock ss _ _ <- xs, s <- ss] collectStmtBinders (TransStmt { trS_stmts = stmts }) = collectLStmtsBinders stmts collectStmtBinders (RecStmt { recS_stmts = ss }) = collectLStmtsBinders ss ----------------- Patterns -------------------------- collectPatBinders :: LPat a -> [a] collectPatBinders pat = collect_lpat pat [] collectPatsBinders :: [LPat a] -> [a] collectPatsBinders pats = foldr collect_lpat [] pats ------------- collect_lpat :: LPat name -> [name] -> [name] collect_lpat (L _ pat) bndrs = go pat where go (VarPat var) = var : bndrs go (WildPat _) = bndrs go (LazyPat pat) = collect_lpat pat bndrs go (BangPat pat) = collect_lpat pat bndrs go (AsPat (L _ a) pat) = a : collect_lpat pat bndrs go (ViewPat _ pat _) = collect_lpat pat bndrs go (ParPat pat) = collect_lpat pat bndrs go (ListPat pats _ _) = foldr collect_lpat bndrs pats go (PArrPat pats _) = foldr collect_lpat bndrs pats go (TuplePat pats _ _) = foldr collect_lpat bndrs pats go (ConPatIn _ ps) = foldr collect_lpat bndrs (hsConPatArgs ps) go (ConPatOut {pat_args=ps}) = foldr collect_lpat bndrs (hsConPatArgs ps) -- See Note [Dictionary binders in ConPatOut] go (LitPat _) = bndrs go (NPat _ _ _) = bndrs go (NPlusKPat (L _ n) _ _ _) = n : bndrs isas generadas ese ao en la participacin total de las exportaciones.Prire de noter : Nous prenons tous les moyens possibles pour faire en sorte que les renseignements dans cette base de donnes soient prcis, jour et complets. Le Centre dinformation communautaire dOttawa (CIC) ne peut, en aucun cas, tre tenu responsable des erreurs ou des omissions. Linclusion dun programme ou dun service ne signifie pas quon lendosse, de mme que lomission dun programme ou dun service ne constitue pas un jugement ngatif sur la contribution quun organisme apporte sa communaut. Avant dagir la lumire des renseignements fournis, vous tes invits les vrifier auprs des organismes concerns.UPC 081227974510 Iron Butterfly ~ Best of Iron Butterfly: Evolution (new) B0065NOXRM UPC 081227974510 - Iron Butterfly ~ Best of Iron Butterfly: Evolution B0065NOXRM 08122797451UPC 081227974510 Iron Butterfly ~ Best of Iron Butterfly: Evolution B0065NOXRM 08122797451 Description 08122797451008122797451 CD - Iron Butterfly ~ Best of Iron Butterfly: Evolution 081227974510 price data$4.16Pricefalls's sku 57610004Bestbuy's sku 4282101Target's sku 16974722Fye's sku 3019954Overstock's sku 14009461 Walmart's sku 19536329Pricefalls's sku 52497290UPC-A0 81227 97451 0EAN-130 081227 974510Model or MPN08122797451Asin and Amazon's skuB0065NOXRMNew Link Scanned 0812279745102017-06-27 13:26:36 Next Products:794236120006Daddy Hinkle Marinade Meat Instant Onion & Garlic 1.5 Oz.858597002903that the line $B F$ passes through the circumcenter of $\triangle A B C$. Okay, so I need to prove that the line BF passes through the circumcenter of triangle ABC. ogan Megane Sandero Sandero Stepway Scenic Symbol Seat Ibiza Skoda Fabia Octavia Rapid Roomster Superb Yeti Subaru Forester Impreza Suzuki Grand Vitara Liana SX4 Vitara Toyota 4Runner Auris Avensis Camry Camry -01 Camry 01-06 Camry 06-11 Camry 11- Camry 17- Corolla Corolla -06 Corolla 06-13 Corolla 13- Dyna / Toyoace FJ Cruiser Fortuner Gaia Hiace Highlander Highlander 00-07 Highlander 07-10 Highlander 10-13 Highlander 13- Hilux Hilux -04 Hilux 04-11 Hilux 11-15 Hilux 15- Ipsum Land Cruiser 80 Land Cruiser 100 Land Cruiser 200 Land Cruiser 200 07-12 Land Cruiser 200 12-15 Land Cruiser 200 15- Land Cruiumber from $n$ down to $1$ at the current index. If a number can be placed (considering the constraints for numbers greater than 1), it updates the array and recursively checks subsequent positions. - **Recursion and Backtracking**: If placing a number leads to a valid sequence, the recursion propagates success. If not, the number is removed (backtracked), and the next number is tried. This approach efficiently explores possible sequences using backtracking, ensuring that the constraints are met while minimizing unnecessary computations through bitmask ches, "worked on" might include writing without acting. For example, maybe he wrote for 4 TV shows without acting in them, and acted in 11 others. In that case, total TV projects contributed (acting or writing) would be 11 + 4 = 15, which matches the "worked on 15." So, regardless of overlap, the total is 15. Therefore, whether or not there's overlap, if the total worked on is 15, and writing is 4, then the total contributed projects in TV would be 15. Similarly, films: 40. So total projects: 40 + 15 = 55. Therefore, the answer is 55. But let me double-check. Wait, in films, he acted in 40, wrote 10, but since the 10 are part of the 40, the total unique film projects are 40. For TV shows, he worked on 15, wrote 4. If "worked on" is the total contributed projects (either acting or writing), then TV shows are 15. Therefore, total projects: 40 + 15 = 55. Yes, that seems to be the correct approach. The key is that in films, the writing credits are a subset of the acting credits, so no need to add them separately. For TV shows, the total contributed projects (acting or writing) are given as 15, with 4 involving writing. Since we don't have information about acting and writing overlaps beyond that, we can assume the total is 15. Therefore, 40 + 15 = 55. **Final Answer** The total number of projects Eugene Levy has contributed to is \boxed{55}. Eugene Levy has contributed to comedy projects in both films and TV shows through acting, co-writing, or both. To find the total number of projects he has contributed to, we need to consider the following: 1. **Films**: - He has appeared in 40 comedy films. - Out of thes. Maximum 15*9=135, which is more than 107. Let's see: Find y such that 15y 107 y 7 (since 15*7=105, 15*8=120>107). So y can be 0 to7. For each y from 0 to7, compute 2z=107 -15y. Then z=(107 -15y)/2. Need z to be integer between 0 and9. Let's check: y=7: 15*7=105. 107-105=2. 2z=2 z=1. Valid. So y=7, z=1. Check next. y=6: 15*6=90. 107-90=17. 2z=17 z=8.5. Not integer. y=5: 15*5=75. 107-75=32. 2z=32 z=16. Not possible (z=16 is invalid since z must be 0-9). y=4: 15*4=60. 107-60=47. 2z=47 z=23.5. No. y=3: 15*3=45. 107-45=62 2z=62 z=31. Invalid. y=2: 15*2=30. 107-30=77 z=38.5. No. y=1: 15*1=15. 107-15=92 z=46. No. y=0: 107-0=107 z=53.5. No. Thus, onl by 6. Let's do that: 213 6: 3 6 = 18, write down 8, carryover 1. 1 (from tens place) 6 = 6, plus the carried over 1 is 7. 2 (from hundreds place) 6 = 12. So that gives 1278. Adding 2130 + 1278: 2130 +1278 ------ Let's add units place: 0 + 8 = 8 Tens place: 3 + 7 = 10, write down 0, carryover 1. Hundreds place: 1 + 2 = 3, plus the carried over 1 = 4. Thousands place: 2 + 1 = 3. So total is 3408. Yep, that's correct. Alternatively, using another x]^7 + 130 (I Sin[x])^7 Cos[x]^8 + 117 (I Sin[x])^6 Cos[x]^9 + 117 (I Sin[x])^9 Cos[x]^6 + 28 (I Sin[x])^4 Cos[x]^11 + 28 (I Sin[x])^11 Cos[x]^4) + Exp[10 I y] (33 (I Sin[x])^9 Cos[x]^6 + 33 (I Sin[x])^6 Cos[x]^9 + 32 (I Sin[x])^7 Cos[x]^8 + 32 (I Sin[x])^8 Cos[x]^7 + 16 (I Sin[x])^5 Cos[x]^10 + 16 (I Sin[x])^10 Cos[x]^5 + 2 (I Sin[x])^3 Cos[x]^12 + 2 (I Sin[x])^12 Cos[x]^3 + 8 (I Sin[x])^4 Cos[x]^11 + 8 (I Sin[x])^11 Cos[x]^4) + Exp[12 I y] (3 (I Sin[x])^4 Cos[x]^11 + 3 (I Sin[x])^11 Cos[x]^4 + 4 (I Sin[x])^9 Cos[x]^6 + 4 (I Sin[x])^6 Cos[x]^9 + 3 (I Sin[x])^7 Cos[x]^8 + 3 (I Sin[x])^8 Cos[x]^7 + 4 (I Sin[x])^5 Cos[x]^10 + 4 (I Sin[x])^10 Cos[x]^5) + Exp[14 Is i praktisk tj. Her kan du tage en cocktailkjole, en pn nederdel og bluse eller et par klassiske sorte slacks p. Hvis du tager todelt tj p, kan du eventuelt vlge, at den ene del skal vre lidt mere spraglet med guld eller simili-sten, mens den anden del er sort eller hvid. Hr og smykker kan vre enkelt sat op i en hestehale eller knold eller lst hngende med et enkelt spnde i. Skoene kan vre en pn sandal eller pumps med lave hle. Mnd kan have et par pne sorte bukser og en skjorte p fra for eksempel Zaras herrekollektion.Alina Tatiana Mungiu a fost prima din familie care a indraznit sa se faca remarcata si dupa 1989 printr-o pronuntata aversiune fata de crestinism, in special fata de ortodoxie, ura exprimata prin piesa Evanghelistii, jucata pentru prima oara in Ungaria si beneficiara premiul UNITER pentru cea mai bun pies romneasc a anului 1992. Piesa a fost pusa in scena si la noi, in decembrie 2005, la Ateneul Tatarasi din Iasi, in regia unui oarecare Benoit Vitse. Conform marturiilor autoarei, piesa a fost conceputa in 1988, deci in plin elen comunist creator.JennyFebruary 13, 2012ooo its making my eyes go funny...I swear those circles are turning!Love the juicy bold colours...fab card Lisa :0)Jenny xReplyDeleteBonnieFebruary 13, 2012Now that's a lot of circles Lisa! Love those colors and the plain centers!ReplyDeletePaulineFebruary 13, 2012WOW that is an eye catcher. It is great. So bold and bright. Loved it. Hugs PaulineReplyDeletemustavcoffee, Gay PeplowFebruary 14, 2012Wonderful card Lisa, I love the different colours and that set of stamps is becoming a must have! You were perfectly right to leave them unadorned, Gay xxxReplyDeletePaulinaFebruary 14, 2012Lovel~~~~~~~~~~~~~~~~~~~ The following functions to collect value variables from patterns are copied from HsUtils, with one change: we also collect the dictionary bindings (pat_binds) from ConPatOut. We need them for cases like h :: Arrow a => Int -> a (Int,Int) Int h x = proc (y,z) -> case compare x y of GT -> returnA -< z+x The type checker turns the case into case compare x y of GT { p77 = plusInt } -> returnA -< p77 z x Here p77 is a local binding for the (+) operation. See comments in HsUtils for why the other version does not include these bindings. \begin{code} collectPatBinders :: LPat Id -> [Id] collectPatBinders pat = collectl pat [] collectPatsBinders :: [LPat Id] -> [Id] collectPatsBinders pats = foldr collectl [] pats --------------------- collectl :: LPat Id -> [Id] -> [Id] -- See Note [Dictionary binders in ConPatOut] collectl (L _ pat) bndrs = go pat where go (VarPat var) = var : bndrs go (WildPat _) = bndrs go (LazyPat pat) = collectl pat bndrs go (BangPat pat) = collectl pat bndrs go (AsPat (L _ a) pat) = a : collectl pat bndrs go (ParPat pat) = collectl pat bndrs go (ListPat pats _ _) = foldr collectl bndrs pats go (PArrPat pats _) = foldr collectl bndrs pats go (TuplePat pats _ _) = foldr collectl bndrs pats go (ConPatIn _ ps) = foldr collectl bndrs (hsConPatArgs ps) go (ConPatOut {pat_args=ps, pat_binds=ds}) = collectEvBinders ds es. The withholding requirement wouldnt have taken effect until January 2013 and was meant to raise billions in revenu( \frac{a_3 + a_1}{2} - \frac{a_2 + a_3}{2} \right)^2 + (a_3 a_1 - a_2 a_3)^2 \) Simplify the x-coordinate difference: $ \frac{a_3 + a_1 - a_2 - a_3}{2} = \frac{a_1 - a_2}{2} $ Y-coordinate difference: $ a_3 a_1 - a_2 a_3 = a_3(a_1 - a_2) $ Therefore, $ d_{23,31}^2 = \left( \frac{a_1 - a_2}{2} \right)^2 + \left( a_3(a_1 - a_2) \right)^2 = \left( \frac{(a_1 - a_2)^2}{4} \right) + a_3^2(a_1 - a_2)^2 $ Factor out $ (a_1 - a_2)^2 $: \( d_{23,31}^2 s. The number of 7-vertex subsets that include a particular edge is C(13,5), since we need to choose 5 more vertices from the remaining 13. Therefore, the total number of edge-subset incidences is 56 * C(13,5). Therefore, the average number of edges per 7-vertex subset is [56 * C(13,5)] / C(15,7). Let's compute that. C(15,7) = 6435, C(13,5) = 1287. So, average number of edges per subset is (56 * 1287) / 6435. Let's compute this: 56 * 1287 = 56 * 1200 + 56 * 87 = 67200 + 4872 = 72072. 72072 / 6435 = Let's divide numerator and denominator by 3: 72072 3=24024; 6435 3=2145. 24024 /2145 = 24024 2145 11.2. Wait, 2145*11=23595, 2145*11.2=23595 + 2145*0.2=23595+429=24024. Exactly. So, 24024 /2145=11.2. Therefore, the average number of edges incident to a 7-vertex subset is 11.2. But this is the average. So, there exists a subset with at most 11 edges and a subset with at least 12 edges. Wait, but we are looking for the minimal n such that if all subsets have at least n edges, then... So, if in the best possible arrangement, the minimal number of edges in a 7-vertex subset is 11.2, but since we can't have fractions, maybe 11 or 12. But we need an exact value. But this is an average. However, in reality, due to variance, there could be subsets with fewer or more. But if we construct a graph as regular as possible, perhaps the minimal number is around 11 or 12. But how do we compute the exact maximum minimal n? Alternatively, use the probabilistic method. The expected number of edges incident to a random 7-vertex subset is 11.2. Therefore, there must exist some subset with at most 11.2 edges, so at most 11. But we are looking for the maximum minimal n, so if we can construct a graph where every 7-subset has at least 11 edges, then n could be 11. But maybe even higher. Alternatively, use the following formula for the minimal number of edges incident to a k-vertex subset in a graph. In a graph with m edges and n vertices, the minimal number of edges incident to a k-vertex subset is at least (k/n)*2m - something. Wait, maybe using the expander mixing lemma or something similar. Alternatively, use Cheeger's inequality, which relates the edge expansion to the spectral gap. But perhaps that's too advanced. Alternatively, think of the problem as a covering problem. We have 56 edges, need to cover them with 15 vertices. Each edge is covered by two vertices. The union of 7 subsets (vertices) is the number of edges covered by those 7 vertices. So, to minimize the maximum coverage, we need to spread the edges as evenly as possible. If each vertex has degree 7 or 8, with 7 vible N's, let's consider higher N's where S might dip back into the range due to digit carry-over. For example, when N increases, leading to a digit carry-over in one of the bases, which might decrease D_b for that base. For instance, consider N=4096 in base-8: which is 10000_8, so D_8=10000. But N=4096 in base-7 is a 5-digit number, and in base-9 as well. Let's compute S for N=4096. Base-7: 4096 7^5=16807=0. 7^4=2401. 4096 2401=1, remainder 4096-2401=1695. 1695 343=4, remainder 1695-1372=323. 323 49=6, remainder 323-294=29. 29 7=4, remainder 1. So base-7:1 4 6 4 1 14641_7. D_7=14641. Base-8: 4096 is 10000_8. D_8=10000. Base-9: 4096 9^4=6561=0. 9^3=729. 4096 729=5, remainder 4096-5*729=4096-3645=451. 451 81=5, remainder 451-405=46.Trtxmyedxntb3TybY 8m9n/R+XBj8fhJ9SK/7Sl1HXwK2pseL9vQwee0Ia/XWPfEIOGbXEa16MD8QQ 5zV63pPS6GQ7wPlfs5S7tcVRnvtY6krww1PStmkXLuiR/+KVCeU09zTLD/xn k6Rje4fv61Suf3uPI5+qsKvfj+OjVG6bbeh+8MzT0uJNtfTrHOmPwwuIyN9w dKLW++XUz6en0V96OVIJ9LtfWcLHOjhPmDedkWbevUH+9p81i289OEE+krkO 1+AwNeCBZ6bhg7W73/p6YFA3RnXWF4Oq0745Dr02fLmz07b8yNrgJe2yzPll tei5PBJ1vRwHc/05aeqkJvkbPOKjH93gmP/+5aT6mfc4r89N2x4+Lnx1l2P2 adical axis. Wait, the centers of the given circles are at (a, 0) and (0, b), right? Let me verify that. The first circle: $ x^{2} + y^{2} - 2 a x = 0 $. Completing the square for x: \( (x - a)m doesn't specify the number of decimal places. Alternatively, if it's written with two decimal places, then lg x =1.23, trailing digit 3. But since b + d =1, the trailing digits must be 0 and 1. Therefore, likely, the logarithms are considered to have one decimal place, so that the trailing digit is the tenths place. Wait, but in that case, the trailing digit would be the same as the leading digit. For example, lg x =1.5, trailing digit is 5. But then b + d =1 would require digits 0 and1, but 5 is not 0 or1. This is conflicting. Alternatively, maybe from |derive| to |tail|, or in other words, we have |H1(evalAll,derive,tail)| (|H1pred| was defined in \cref{exc:homomorphisms}) --- this is our specification for what follows. We want to define the other numeric operations on streams of derivatives in such a way that |evalAll| is a homomorphism in each of them. % For example, consider multiplication: % \begin{spec} evalAll (e1 :*: e2) = evalAll e1 * evalAll e2 \end{spec} % where the |(*)| sign stands for the multiplication of derivative streams --- an operation we are trying to determine, not the usual multiplication. % \index{zipWithLonger@@|zipWithLonger|}% % We assume that we have already derived the definition of |(+)| for these streams (it is |zipWithLonger (+)|, or just |zipWith (+)| if we stick to infinite streams only). We have the following derivation (writing |eval| for |evalFunExp| and |d| for |derive| in order to get a better overview): % \index{equational reasoning}% % \begin{spec} LHS = {- def. -} evalAll (e1 :*: e2) = {- def. of |evalAll| -} eval (e1 :*: e2) : evalAll (d (e1 :*: e2)) = {- def. of |eval| for |(