Yes, but only if any language in NP has small uniform circuits.
Hope this helps. This is a famous unsolved problem, as whoever posted it here probably knows. A prize of one million US dollars has been offered for a solution. There are details at http://www.claymath.org/millennium/P_vs_NP/ At the top right of this page there is a link to the "Official problem description" by Stephen Cook, one of the people who first posed the problem in 1971, and also a more informal description under the heading "Minesweeper". Anyone who solves this problem will get not only a million dollars, but also enduring fame among mathematicians and computer scientists. It won't be easy, since a lot of clever people have worked on it since 1971. Most people believe that P is not equal to NP, but a belief is not the same as a proof.
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there is np abbreviation
an equation that can be written in the form ax= b where a and b are constants and a is not equal to 0
This is because a factor is defined in terms of multiplication, not addition. One integer, p, is a factor of another integer, q, if there is some integer, r (which is not equal to 1) such that p*r = q.
By considering P(h) = P(0) * exp(-(mgh/kT)) - (where P(h) is the pressure result of the equation; P(0) is the pressure at the "surface" of the planet; m is the mass of the molecule that is predominant in the atmosphere (in kg); g is the gravitational acceleration present on the planet; k is Boltzmann's constant; T is mean observed temperature (in Kelvin)) We can then analyse what height is needed for P(h) to be equal to 1/e of the surface value: the scale height h(0) = kT/mg for P(h) to be equal to 1/e, and by substituting in standard values, and assuming a predominantly hydrogen gas atmosphere, and using g=GM/r^2, we derive the result (that varies according to the different values of g, m, and T that you use): h(0)(Jupiter) = ~20km Have fun ;-)
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