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  2. Formula calculator - Wikipedia

    en.wikipedia.org/wiki/Formula_calculator

    The formula calculator concept can be applied to all types of calculator, including arithmetic, scientific, statistics, financial and conversion calculators. The calculation can be typed or pasted into an edit box of: A software package that runs on a computer, for example as a dialog box. An on-line formula calculator hosted on a web site.

  3. Elementary algebra - Wikipedia

    en.wikipedia.org/wiki/Elementary_algebra

    To solve this kind of equation, the technique is add, subtract, multiply, or divide both sides of the equation by the same number in order to isolate the variable on one side of the equation. Once the variable is isolated, the other side of the equation is the value of the variable. [ 37 ]

  4. Division (mathematics) - Wikipedia

    en.wikipedia.org/wiki/Division_(mathematics)

    For example, 20 apples divide into five groups of four apples, meaning that "twenty divided by five is equal to four". This is denoted as 20 / 5 = 4, or ⁠ 20 / 5 ⁠ = 4. [2] In the example, 20 is the dividend, 5 is the divisor, and 4 is the quotient.

  5. Quadratic equation - Wikipedia

    en.wikipedia.org/wiki/Quadratic_equation

    Because (a + 1) 2 = a, a + 1 is the unique solution of the quadratic equation x 2 + a = 0. On the other hand, the polynomial x 2 + ax + 1 is irreducible over F 4 , but it splits over F 16 , where it has the two roots ab and ab + a , where b is a root of x 2 + x + a in F 16 .

  6. Quadratic formula - Wikipedia

    en.wikipedia.org/wiki/Quadratic_formula

    The roots of the quadratic function y = ⁠ 1 / 2 ⁠ x 2 − 3x + ⁠ 5 / 2 ⁠ are the places where the graph intersects the x-axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.

  7. Equation solving - Wikipedia

    en.wikipedia.org/wiki/Equation_solving

    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.

  8. Polynomial long division - Wikipedia

    en.wikipedia.org/wiki/Polynomial_long_division

    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 − (− ...

  9. Clearing denominators - Wikipedia

    en.wikipedia.org/wiki/Clearing_denominators

    The simplified equation is not entirely equivalent to the original. For when we substitute y = 0 and z = 0 in the last equation, both sides simplify to 0, so we get 0 = 0, a mathematical truth. But the same substitution applied to the original equation results in x/6 + 0/0 = 1, which is mathematically meaningless.