The Drake equation calculates N. What is N?
469910
N = The number of planets that support life that can communicate in the Galaxy.\n\nThe drake equation is a hypothetical equation in that calculations of N require alot of assumptions.
894614
PV=nRT\nP= (nRT)/V  (1st eq)\n\nn,R,T u know.\n\nV >>>\nDensity = mass/ volume\n2.3 = .00446 *10^3/Volume\n\nrearrange and ull get volume.............and then put that into the first equation.............and voila
189338
You have to say the starting value for n.  \nThe first value for n has to be given; it could be basically anything.
298957
the N means newtons i kno that
787829
SOHCAHTOA\nsinθ = O/H = a/c\ncosθ = A/H = b/c\ntanθ = O/A = a/b
818735
This is just like a normal equation with equal sign instead of the >\n\nn-4>10\nso\nn>10+4\nn>14\n\nAs n >14, n must be 15 if n is a interger
1071709
e=mc^2\na^2+b^2=c^2 (pythagorean theorum)\ny=mx+b (linear equation)\nlw+bh=A (area of sqaures and rectangles)\n1/2bh=A (area of triangles)\ny=x^2 (parabole equation)
906425
Bionomial expansion\n\nthis has got n+1 terms\n\nthe terms are (nCk) a^kb^n-k where k varies from 0 to n\n\nfor example (a+b)^3\none term (3C0) a^0b^3 = b^3\nanother term (3C1)a^1b^2 = 3ab^2\nanother (3C2)a^2b^1 = 3a^2b\nanother (3c3)a^3b^0 = a^3\n\nadd them and get a^3+3a^2b+3ab^2+ b^3\n\n(nCk) = n!/(k!(n-k)!)\n\nThe n'th line of the pascal triangle gives you the coeffiecients . Starting from 0.
100416
Not very.\n\nThe Drake Equation is supposed to estimate the number of civilizations in the Galaxy, and is usually written as:\n\nN = R* fp ne fl fi fc L\n\nwhere,\n\n    * N = The number of communicative civilizations\n    * R* = The rate of formation of suitable stars \n    * fp = The fraction of those stars with planets. \n    * ne = The number of Earth-like worlds per planetary system\n    * fl = The fraction of those Earth-like planets where life actually develops\n    * fi = The fraction of life sites where intelligence develops\n    * fc = The fraction of communicative planets \n    * L = The "lifetime" of communicating civilizations\n\nA problem is that one can come up with other criteria, which add a term and tend to lower the estimate. For example, fraction of Earth-like worlds with a large Moon, or the timespan when an "echo" or a civilization (buildings, spaceships, etc.) can be detected before fading away.\n\nThe reason is that there is so much uncertainty in most of these values. Until that part is nailed down we won't really know which of these terms are critical, which not, and what other terms (if any) to add.\n\nNevertheless it is an easy way to guestimate some of the concerns with the possibilities of alien civilizations.
966154
The magnitude of a force is measured in newtons(N)
977084
a.) newtons\nb.) km/h\nc.) f(x) = 0\nd.) x = f(0) + c
375635
Kp=Kc(RT)n\na.Kp=Kc(RT)-1/2\nb.Kp=KcRT\nc.Kp=KcRT\nd.Kp=Kc(RT)2
864167
The Drake equation (also known as the Green Bank equation or the Sagan equation) is a famous result in the speculative fields of xenobiology, astrosociobiology and the search for extraterrestrial intelligence.\n\nThis equation was devised by Dr Frank Drake (now Emeritus Professor of Astronomy and Astrophysics at the University of California, Santa Cruz) in the 1960s in an attempt to estimate the number of extraterrestrial civilizations in our galaxy with which we might come in contact. The main purpose of the equation is to allow scientists to quantify the uncertainty of the factors which determine the number of extraterrestrial civilizations. In recent years, the Rare Earth hypothesis, which posits that conditions for intelligent life are quite rare in the universe, has been seen as a possible refutation of the equation.\n\nThe Drake equation is closely related to the Fermi paradox. It was cited by Gene Roddenberry as supporting the multiplicity of starfaring civilizations shown in Star Trek, the television show he created.
921632
PV = NRT, solve for N, note that T is absolute temperature in kelvins
817617
tan(A-B)+tan(B-C)+tan(C-A)\n= (tanA - tanB)/(1 + tanAtanB) + (tanB - tanC)/(1 + tanBtanC) + (tanC - tanA)/(1 + tanAtanC)\n= [(tanA - tanB)(1 + tanBtanC)(1 + tanAtanC) + (tanB - tanC)(1 + tanAtanC)(1 + tanAtanB) + (tanC - tanA)(1 + tanAtanB)(1 + tanBtanC)]/[(1 + tanAtanB)(1 + tanBtanC)(1 + tanAtanC)]\n= 0\nie (tanA - tanB)(1 + tanBtanC)(1 + tanAtanC) + (tanB - tanC)(1 + tanAtanC)(1 + tanAtanB) + (tanC - tanA)(1 + tanAtanB)(1 + tanBtanC) = 0\nOn expanding\n(tanA - tanB)(1 + tanAtanC + tanBtanC + tanAtanBtan²C) \n+ (tanB - tanC)(1 + tanAtanC + tanAtanB + tan²AtanBtanC)\n+ (tanC - tanA)(1 + tanAtanB + tanBtanC + tanAtan²BtanC)\n\n= tanA - tanB + tan²AtanC - tanAtanBtanC + tanAtanBtanC - tan²BtanC + tan²AtanBtan²C - tanAtan²Btan²C \n+ tanB - tanC + tanAtanBtanC - tanAtan²C + tanAtan²B - tanAtanBtanC + tan²Atan²BtanC - tan²AtanBtan²C\n+ tanC - tanA + tanAtanBtanC - tan²AtanB + tanBtan²C - tanAtanBtanC  + tanAtan²Btan²C - tan²Atan²BtanC\n= tan²AtanC - tan²BtanC - tanAtan²C + tanAtan²B - tan²AtanB + tanBtan²C\n=0 \nDivide all by tanAtanBtanC\nSo tanA/tanB - tanB/tanA - tanC/tanB + tanB/tanC - tanA/tanC - tanC/tanA = 0\n\nie tanA/tanB + tanB/tanC + tanC/tanA = tanB/tanA + tanC/tanB = tanA/tanC\n\nWhence tanA/tanB + tanB/tanC + tanC/tanA = cotA/cotB + cotB/cotC + cotC/cotA\n\nSince a triangle then tan(A + B) = tan(π - C) = - tanC\nSo (tanA + tanB)/(1 - tanAtanB) = -tanC\nie tan A + tan B = tanAtanBtanC - tanC\nso tanA + tanB + tanC = tanAtanBtanC
837785
hi\np=(qm)/n\ny=x+(b/m) or x+(bq/np)
644170
The answer of "nayanmange" is right but his formula is wrong because it does not predict the sequence correctly. The correct formula is (2n-1)-th and 2n-th term of the series are {(2n-1)^2+1} and {(2n)^2-1} respectively. For n = 3 we find the (2*3)-th i.e. 6th term of the series as {(2*3)^2-1} = 35.