Search results
Results from the WOW.Com Content Network
Problems 1–6 compute divisions of a certain number of loaves of bread by 10 men and record the outcome in unit fractions. Problems 7–20 show how to multiply the expressions 1 + 1/2 + 1/4 = 7/4, and 1 + 2/3 + 1/3 = 2 by different fractions. Problems 21–23 are problems in completion, which in modern notation are simply subtraction problems.
In the following Diophantine equations, w, x, y, and z are the unknowns and the other letters are given constants: a x + b y = c {\displaystyle ax+by=c} This is a linear Diophantine equation or Bézout's identity. w 3 + x 3 = y 3 + z 3 {\displaystyle w^ {3}+x^ {3}=y^ {3}+z^ {3}} The smallest nontrivial solution in positive integers is 123 + 13 ...
An Egyptian fraction is a finite sum of distinct unit fractions, such as That is, each fraction in the expression has a numerator equal to 1 and a denominator that is a positive integer, and all the denominators differ from each other. The value of an expression of this type is a positive rational number ; for instance the Egyptian fraction ...
[1] [2] [3] [better source needed]. For example, 3 x 2 − 2 x y + c {\displaystyle 3x^{2}-2xy+c} is an algebraic expression. Since taking the square root is the same as raising to the power 1 / 2 , the following is also an algebraic expression:
Quadratic formula. The roots of the quadratic function y = 1 2 x2 − 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.
Partial fraction decomposition. In algebra, the partial fraction decomposition or partial fraction expansion of a rational fraction (that is, a fraction such that the numerator and the denominator are both polynomials) is an operation that consists of expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions ...
The result is an equation with no fractions. 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 ...
When R is chosen to have the value of 2 (R = 2), this equation would be recognized in Cartesian coordinates as the equation for the circle of radius of 2 around the origin. Hence, the equation with R unspecified is the general equation for the circle. Usually, the unknowns are denoted by letters at the end of the alphabet, x, y, z, w ...