Which statement about the libches is Correct?
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do you mean lichens? if so, D would be correct.
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„input harvmac „overfullrule=0pt {„it Exam} (1) Prove via path \n„def„R{{„bf R}} „def„S{{„bf S}} (a) Work in four-dimensional (flat) Euclidean space $„R^4$ and show that conformal invariance implies...\nhttp://www.math.ias.edu/QFT/fall/exam1.tex \n\n \nhttp://www.math.ias.edu/QFT/spring/stringPS8.tex\n...def„det{{„rm det}} „def„Det{{„rm Det}} „def„tr ... B$ of the Wess-Zumino-Witten model, satisfy the equations for conformal invariance and for\nhttp://www.math.ias.edu/QFT/spring/stringPS8.tex \n\n \nComments: Latex(2e), 34 pages, 4 figures (eps) On some UNIX machines\nSome private shortcuts: „def„be{„begin{equation}} „def„ee{„end{equation}} „def„bea{„begin{eqnarray}} „def„eea{„end{eqnarray}} „def...\nhttp://www.ma.utexas.edu/mp_arc/e/96-436.latex.mpmix \n\n \n=4 19| >0 63| >1 65| >2 434| >4 525| 0 1773| 1 630| 1/2 2530| 10 \n\nhttp://www-library.desy.de/lists/spihep.k \n\n \nStardrive.org - - Article/Content Template\nStardrive.org is dedicated to fostering increased public awareness and education in the theoretical and technological aspects of interstellar travel.\nhttp://www.stardrive.org/Sarmail1-3-01.shtml \n\n \n„documentstyle[12pt]{article} %„renewcommand{„baselinestretch}{2.0}\n...oddsidemargin 0 mm „def„ee{„end{equation}} „def ... correlation functions have a power law behaviour, indicates that conformal invariance may be\nhttp://theory.ipm.ac.ir/papers/src/ipm63.tex \n\n \n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%...\n...def „pa {„partial} „def „ti {„widetilde} „def „se {{„prime„prime}} „def ... „subsection{Pre-big bang inflation and conformal frames}\nhttp://www.ba.infn.it/~gasperin/tex/ba437.tex \n\n \nhttp://www.ba.infn.it/~gasperin/tex/cer96267.tex\n...def „pa {„partial} „def „ra {„rightarrow} „def ... discussed in „cite{6a}, to the conditions of $„sg$-model conformal invariance represented by\nhttp://www.ba.infn.it/~gasperin/tex/cer96267.tex \n\n \n%%%%%%%%% LATEX FILE FOR FORMATTING GUIDELINES \n...def„xp{X^{+}} „def„xm{X^{-}} „def„Yp{Y^+} „def„Ym ... must be obeyed (in a conformal gauge this is the same as maintaining conformal\nhttp://theory.tifr.res.in/group/colloq/das.tex \n\n \n„documentstyle[twoside,psfig]{article}\n...def„ds{„displaystyle} „parindent=0pt „def ... out that this model is advantageous for the study of the conformal invariance conjecture for\nhttp://www.math.utah.edu/~ejpecp/EcpVol1/paper8.tex
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G pairs with C and A pairs with T, so its GAATCCGAATGGTA\n\nI assume the "h" at the end is a typo?\n__________________________________________\nsouthernserendipiti wrote:\n>>>What's the "th" at the end?\nI assumed it was a t and a typo : )
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An isosceles triangle cannot be a right triangle. Re-read your homework.
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1. True. \n2. True.\n3. True.\n4. False. A parallelogram has no symmetry at all.\n5. True.
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C is the correct answer.
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milkyway: \nI think your answer is incomplete. Unfortunately, I don't have my own proof, but let me criticize yours. \nSorry.:-)\n\nWhat is definitely proven by induction in your post is that you are able to construct an acceptable set of N points. However, in my view, your reasoning does not prove the statement for a given set.\n\n1. From the very beginning, you have N points. How can one realize the induction scheme in this case? Perhaps, you mean that you can temporarily erase N-3 points (keeping in mind their positions) and start with 3 points forming a triangle, or "3 couples" in your terms. Then you restore the erased points one by one at their positions.  I believe this should be explained in your proof. \n\n2. Note that if you start with 3 aligned points, by "restoring" the 4th point, you can generate an allowed set of N+1 points out of the prohibited set of N points. This is missing in your proof, where you deal only with allowed N-sets.  \n\nBut this is not the most serious flaw there.\n\n3. You write: "now, this way there always are the old two couples, plus two new couples (new point-non aligned point of each of such two couples). So going from N to N+1, the number of acceptable couples is always >=3."  \n\nI disagree with this statement. \n\nIndeed, imagine that the three couples, which work at the stage N, form a triangle. Aligning the newly "restored" point with one side of the triangle, you kill one working couple. Still you have the two remaining old couples and get a possibly working new couple (new point -- non aligned point of the triangle). However, it is not at all guaranteed that there is no other point lying on the line going through this new possibly working couple. If such point exists, the number of working couples in this case gets reduced from 3 to 2 when going from N to N+1. At the next step, this way I will reduce the number of working couples from 2 to 1 and so on.\n\nThe transition from N to N+1 is therefore not justified.\n\n4. A general question. For N=infinity the statement is false for points located at nodes of a regular \n(say, square) lattice. Doesn't it contradict to the very idea of proving this statement by induction?\n\n\n------------------------------\n\n\n\n\nHere is my proof.\n\nGiven a set of N points, the number of all possible couples is M=N(N-1)/2.\n\nFor each couple draw a straight line going through it. We get M lines.\n\nLet us assume that the statement we are going to prove is false: i.e., none of those M lines goes through ONLY TWO points.\n\nThen, of course, each of those M lines coincides with at least the two other lines, so the actual number of distinguishable lines K < M.\n\nWe have N points and K different lines.\n\nEach of the NK pairs (point + line) is characterised by a distance S(n,k) between  the point n and the line k. \n\nThis distance S(n,k) is defined as a length of a segment perpendicular to the line k with one of the ends lying on the line k and the other end coinciding with the point n.\n\nLet us list all the NK distances S(n,k). Since NK is a finite integer number, one can find the shortest value (or several equal shortest values) of S(n,k). \n\nTake a closer look at the point P and the line L with the shortest S(n=P,k=L).\n\nBy assumption, there are at least three points (L1,L2, and L3) lying on the line L. Therefore, at least two of these three points (say, L1 and L2) can be found to the one side (say, to the right) of the point A where the perpendicular to the line L starting at the point P hits the line L. The case when one of the three points (say, L1) coincides with the point A is treated in exactly the same way.\n\nNow we have a triangle (P-L1-L2), with two vertices L1 and L2 lying on the line L to the right of A in the order A,L1,L2 from left to right. \n\nSince L1 lies to the right of A (or coincides with A), the angle between P-L1 and L1-L2 is greater than (or equal to) pi/2.\n\nTherefore t
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I think you should ask-"since this is my last quesion, what should I ask?"  In fact, you have hemmed all of us in a terrible logical warp from which we cannot escape...  \nI have only recently escaped from the last logical wormhole I got stuck in (cause I fell asleep)-here it is:\n\n\nThis statement is a lie.  True or false?
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i think the false statement is C. as the degree of any number is always 1. i dont understand the statement b tho :S
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If the Isosceles Triangle Theorem says, "If it's an isosceles triangle, then base angles are congruent" then the converse is "If the base angles of triangle are congruent, then the triangle is isosceles."\n\nLet's suppose we have triangle ABC, with angle B congruent to angle C.\n\nLet's draw a line from angle A to the segment BC, perpendicular to BC.  (Use the postulate: given a point and a line, there is exactly one line through the point perpendicular to the line).  The intersection of this line and segment BC is point D.\n\nNow we have two right triangles, triangle ADB and triangle ADC.  We are given angle B and angle C are congruent.  We also know that AD is congruent to AD by the reflexive property.  Therefore, ADB and ADC are congruent right triangles (perhaps AAS or LA theorems).\n\nBecause ADB and ADC are congruent right triangles, then corresponding line segments AB and AC are congruent (CPCTC).  Triangle ABC have two sides that are congruent and is therefore isosceles (definition of an isosceles triangle).\n\nIf you want the two-column proof,\nGiven: triangle ABC, angle B congruent to angle C.\nProve: triangle ABC is isosceles.\n\nStep 1: Draw line AD perpendicular to BC [Given point and line, there is only one line perpendicular to line through point.]\nStep 2: angles ADB and ADC are right angles and are congruent. [perpendicular lines intersect to form right angles; right triangles are congruent]\nStep 3: triangles ADB and ADC are right triangles. [definition of right triangles]\nStep 4: segment AD is congruent to AD [reflexive property]\nStep 5: angles B and C are congruent [Given]\nStep 6: right triangles ADB and ADC are congruent [LA or AAS]\nStep 7: line segments AB and AC are congruent [CPCTC]\nStep 8: triangle ABC is isosceles [definition of isosceles triangle]
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maina, is this correct plz choose me best answer
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Conclusion: If two sides of a triangle are congruent then the third side is also congruent.\nFirst you would state the 2 bottom given statements. then the third will be the conclusion I gave you.\n\nYou could also use the SAS congruency to show that two triangles are congruent. If the triangles are congruent then BC is congruent to DF.\nTHERE IS ALSO THE CPCTC.
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Let the trianglas be ABC & BCD\nAD forms 2 triangles, ABD & ACD\nAB=AC since ABC is isisceles\nBC+BD since BCD is isosceles.\nangle ABC=angle ACB\nangle DBC=angls DCB so angls ABD=angle ACD so ABD is congruent to ACD by SAS\n\ncall the point where AD crosses BC E\n\nso both vertex angles are bisected.\nlook at ABC.\nAE bisects BAC & AE is a commin side for BAE & CAE.\nsince AB=AC, the triangles are congruent by SAS\ntherefore BE=CE\n& angle AEB= angle AEC which are supplements so they must be right angles.\nQED
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Based on what I read at the site given below, I'd say "B" is correct.
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No. Basically DA got it wrong. The answer to L, the U and E is     not "42" but 526A and D and S always being together :)
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The first is a chronological, literal way to say that this was the way that it was done.  It was written the exact way that it happened.  The one in Chapter 2 is a topical, storytelling way to put it.  The author doesn't make them contradicting at all.  When he says "Now the Lord had formed out of the ground all the beasts..." it doesn't indicate that this was after man.  It actually says that this had already been done in the past.  So I don't think the same way as the previous answerer does.  I read them as saying the same thing, but just how you would answer a question differently in a lab report from english class, they are just stylistically different.  If you have more questions about creation, www.answersingenesis.org is a great resource.  I've already been praying for you!
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HI: I tried to answer your email but it said that your email address has not been confirmed and so the email didn't go through.\n\nalpha and beta are scalers for the vector space V.  This problem is actually so easy that people don't believe what they've done.  You really just need to show that Span S behaves like V.  \n\nSpan S must be a vector space itself.  Two elements from the span is \nu=a1*v1 + a2*v2...ak*vk\nand \nw=b1*v1 + b2*v2...bk*vk\n\nshow that alpha*u+beta*w is in Span S where alpha, beta are scalers for your vector space.\n\nalpha*u+beta*w = (alpha*a1+beta*b1)*v1 + ....\nis a linear combination of the vectors in Span S, thus it's in Span S.  \n\nSo, Span S is not empty and is closed, thus its a subspace.