In order to solve half-life questions, consider the equation ... AT = A0 2(-T/H) ... where A0 is the initial radioisotope's activity, AT is the decayed activity after time T, and H is the half-life. Given 3 of these parameters, you can solve for the fourth. For instance, given starting activity, ending activity, and time, you can solve for half-life as follows ... AT = A0 2(-T/H) AT/A0 = 2(-T/H) ln2(AT/A0) = -T/H H = -T/ln2(AT/A0) ... and don't forget that ln2(x) = log(x)/log(2)
In the US? white http://www.abortiontv.com/Misc/AbortionStatistics.htm#United%20States%A0%A0%A0%A0%A0%A0%A0
The equation for half-life is ... AT = A0 2(-T/H) ... where AT is activity at some time T, A0 is starting activity, and H is half-life in units of T. If you wanted to solve for H, it is straightforward algebra ... AT = A0 2(-T/H) AT / A0 = 2(-T/H) log2 (AT / A0) = -T/H H = -T / log2 (AT / A0) ... Remember that log2 (x) = ln (x) / ln (2)
A0 paper is 46.8 x 33.1 in.
In millimeters the A0 paper size is 841mm by 1189mm. In inches the A0 paper size is 33.1 × 46.8 in.
A0 paper measures 841mm x 1189mm.
A0 and A1 are used for Technical drawings and posters.
its so much fun, playin the can-can. capital letters are half notes, lowercase are quarter notes. equal signs (=) are rests. here goes: D D d1 d3 d2 d1 A0 A0 a0 a1 d2 d3 D1 D1 d1 d3 d2 d1 d0 a3 a2 a1 a0 d3 d2 d1 D0 D0 d1 d3 d2 d1 A0 A0 a0 a1 d2 d3 D1 D1 d1 d3 d2 d1 d0 a0 d1 d2 d0
16 pages of A4 fit into an A0
In the standardised paper measurement system A0 is 1189 millimetres x 841 millimetres.
it depends how sharp the picture is but with a good body and lens and film and a steady shot it should be no problem to print it on A0 posterformat, but if you want to be shure take the negative to a good photoshop and let tem analyse if a poster A0 size will be sharp
The answer to the question the measurement of A0 paper is 1189mm x 841mm. Hope that helps
Europe and asia
Suppose the sequence is defined by an = a0 + n*d Then a1 = a0 + d = 15 and a13 = a0 + 13d = -57 Subtracting the first from the second: 12d = -72 so that d = -6 and then a0 - 6 = 15 gives a0 = 21 So a32 = 21 - 32*6 = -171
f(x) = a0 + a1x + a2x2 + a3x3 + ... + anxn for some integer n, and constants a0, a1, ... an.
A= A0e^-kt A0= A/ e^kt = Ae^kt A0= A+ D* D*= A0- A D*= Ae^kt - A D*= A(e^kt - 1)
For an exponential function: General equation of exponential decay is A(t)=A0e^-at The definition of a half-life is A(t)/A0=0.5, therefore: 0.5 = e^-at ln(0.5)=-at t= -ln(0.5)/a For exponential growth: A(t)=A0e^at Find out an expression to relate A(t) and A0 and you solve as above
A = A0 e-Bt
A0 is 1 meter square.
http://www.girlsgogames.com/game/Frogtastic.html?utm_medium=brandedgames_external&utm_campaign=Frogtastic&utm_source=&utm_content=send_to_friend&msg=%C3%8E%C2%A0%C3%8B%C2%A0%C3%8A%C2%A0%C3%8C%C2%A0&name=%C2%B1%C2%9E%C2%90%C2%92%C2%96 copy and go to this page enjoy!
After two half-lives, 25% of an isotopes original number of atoms remains. AT = A0 2 (-T/H) Where A0 is the original activity, AT is the activity after some time T, and H is the half-life in units of T, so ... A2 = A0 2 (-2/1) A2 = 0.25 A0