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A solution to this problem is an example of a Kirkman triple system, [3] which is a Steiner triple system having a parallelism, that is, a partition of the blocks of the triple system into parallel classes which are themselves partitions of the points into disjoint blocks.
The greedy algorithm for Egyptian fractions finds a solution in three or fewer terms whenever is not 1 or 17 mod 24, and the 17 mod 24 case is covered by the 2 mod 3 relation, so the only values of for which these two methods do not find expansions in three or fewer terms are those congruent to 1 mod 24. [12]
To calculate a Pythagorean triple, take any term of this sequence and convert it to an improper fraction (for mixed number , the corresponding improper fraction is ). Then its numerator and denominator are the sides, b and a, of a right triangle, and the hypotenuse is b + 1. For example:
A primitive Pythagorean triple is one in which a, b and c are coprime (that is, they have no common divisor larger than 1). [1] For example, (3, 4, 5) is a primitive Pythagorean triple whereas (6, 8, 10) is not. Every Pythagorean triple can be scaled to a unique primitive Pythagorean triple by dividing (a, b, c) by their greatest common divisor ...
In case 2, the rate of convergence depends on the absolute value of the ratio between the two roots: the farther that ratio is from unity, the more quickly the continued fraction converges. This general solution of monic quadratic equations with complex coefficients is usually not very useful for obtaining rational approximations to the roots ...
Simplifying this further gives us the solution x = −3. It is easily checked that none of the zeros of x ( x + 1)( x + 2) – namely x = 0 , x = −1 , and x = −2 – is a solution of the final equation, so no spurious solutions were introduced.
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The algebraic solutions do not distinguish between internal and external tangencies among the circles and the given triangle; if the problem is generalized to allow tangencies of either kind, then a given triangle will have 32 different solutions and conversely a triple of mutually tangent circles will be a solution for eight different ...
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