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Microsoft Math contains features that are designed to assist in solving mathematics, science, and tech-related problems, as well as to educate the user. The application features such tools as a graphing calculator and a unit converter. It also includes a triangle solver and an equation solver that provides step-by-step solutions to each problem.
The unique pair of values a, b satisfying the first two equations is (a, b) = (1, 1); since these values also satisfy the third equation, there do in fact exist a, b such that a times the original first equation plus b times the original second equation equals the original third equation; we conclude that the third equation is linearly ...
In the ancient times, they succeeded only for degrees one and two. For quadratic equations , the quadratic formula provides such expressions of the solutions. Since the 16th century, similar formulas (using cube roots in addition to square roots), although much more complicated, are known for equations of degree three and four (see cubic ...
Divide each side by a, the coefficient of the squared term. Subtract the constant term c/a from both sides. Add the square of one-half of b/a, the coefficient of x, to both sides. This "completes the square", converting the left side into a perfect square. Write the left side as a square and simplify the right side if necessary.
Algebra is the branch of mathematics that studies certain abstract systems, known as algebraic structures, and the manipulation of expressions within those systems. It is a generalization of arithmetic that introduces variables and algebraic operations other than the standard arithmetic operations, such as addition and multiplication.
This corresponds to a set of y values whose product is a square number, i.e. one whose factorization has only even exponents. The products of x and y values together form a congruence of squares. This is a classic system of linear equations problem, and can be efficiently solved using Gaussian elimination as soon as the number of rows exceeds ...
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Divide the highest term of the remainder by the highest term of the divisor (3x ÷ x = 3). Place the result (+3) below the bar. 3x has been divided leaving no remainder, and can therefore be marked as used. The result 3 is then multiplied by the second term in the divisor −3 = −9. Determine the partial remainder by subtracting −4 − (− ...