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He is really interested in problems 3 and 4, but the answers to the easier problems 1 and 2 are needed for proving the answers to problems 3 and 4. 1st problem [ edit ]
For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2(y + 1) – 1, a true statement. It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1.
More generally, the solution set to an arbitrary collection E of relations (E i) (i varying in some index set I) for a collection of unknowns (), supposed to take values in respective spaces (), is the set S of all solutions to the relations E, where a solution () is a family of values (()) such that substituting () by () in the collection E makes all relations "true".
Microsoft Math 2.0: Part of Microsoft Student 2007 Microsoft Math 3.0 : Standalone commercial product that requires product activation ; includes calculus support, digital ink recognition features and a special display mode for video projectors
Animation showing the use of synthetic division to find the quotient of + + + by .Note that there is no term in , so the fourth column from the right contains a zero.. In algebra, synthetic division is a method for manually performing Euclidean division of polynomials, with less writing and fewer calculations than long division.
Divide the first term of the dividend by the highest term of the divisor (x 3 ÷ x = x 2). Place the result below the bar. x 3 has been divided leaving no remainder, and can therefore be marked as used by crossing it out. The result x 2 is then multiplied by the second term in the divisor −3 = −3x 2.
In numerical analysis, Hermite interpolation, named after Charles Hermite, is a method of polynomial interpolation, which generalizes Lagrange interpolation.Lagrange interpolation allows computing a polynomial of degree less than n that takes the same value at n given points as a given function.
A similar problem, involving equating like terms rather than coefficients of like terms, arises if we wish to de-nest the nested radicals + to obtain an equivalent expression not involving a square root of an expression itself involving a square root, we can postulate the existence of rational parameters d, e such that