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601 - 610 of 2552 for fourth isomorphism theoremSearch Results
The number of "prime" boxes is always finite, where a set of boxes is prime if it cannot be built up from one or more given configurations of boxes.
An algebraically soluble equation of odd prime degree which is irreducible in the natural field possesses either 1. Only a single real root, or 2. All real roots.
Given n mutually exclusive events A_1, ..., A_n whose probabilities sum to unity, then P(B)=P(B|A_1)P(A_1)+...+P(B|A_n)P(A_n), where B is an arbitrary event, and P(B|A_i) is ...
A box can be packed with a harmonic brick a×ab×abc iff the box has dimensions ap×abq×abcr for some natural numbers p, q, r (i.e., the box is a multiple of the brick).
Let g(x)=(1-x^2)(1-k^2x^2). Then int_0^a(dx)/(sqrt(g(x)))+int_0^b(dx)/(sqrt(g(x)))=int_0^c(dx)/(sqrt(g(x))), where c=(bsqrt(g(a))+asqrt(g(b)))/(sqrt(1-k^2a^2b^2)).
Let A, B, and C be three circles in the plane, and let X be any circle touching B and C. Then build up a chain of circles such that Y:CAX, Z:ABY, X^':BCZ, Y^':CAX^', ...
If a polynomial P(x) is divided by (x-r), then the remainder is a constant given by P(r).
Let any finite or infinite set of points having no finite limit point be prescribed, and associate with each of its points a definite positive integer as its order. Then ...
Each point in the convex hull of a set S in R^n is in the convex combination of n+1 or fewer points of S.
product_(k=1)^(n)(1+yq^k) = sum_(m=0)^(n)y^mq^(m(m+1)/2)[n; m]_q (1) = sum_(m=0)^(n)y^mq^(m(m+1)/2)((q)_n)/((q)_m(q)_(n-m)), (2) where [n; m]_q is a q-binomial coefficient.
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