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In the case of two nested square roots, the following theorem completely solves the problem of denesting. [2]If a and c are rational numbers and c is not the square of a rational number, there are two rational numbers x and y such that + = if and only if is the square of a rational number d.
[3] [4] Expressions can be evaluated or simplified by replacing operations that appear in them with their result. For example, the expression 8 × 2 − 5 {\displaystyle 8\times 2-5} simplifies to 16 − 5 {\displaystyle 16-5} , and evaluates to 11. {\displaystyle 11.}
Simplification is the process of replacing a mathematical expression by an equivalent one that is simpler (usually shorter), according to a well-founded ordering. Examples include: Simplification of algebraic expressions, in computer algebra; Simplification of boolean expressions i.e. logic optimization
The notion of irreducible fraction generalizes to the field of fractions of any unique factorization domain: any element of such a field can be written as a fraction in which denominator and numerator are coprime, by dividing both by their greatest common divisor. [7] This applies notably to rational expressions over a field. The irreducible ...
If the expression in parentheses may be calculated, that is, if the variables in the expression in the parentheses are known numbers, then it is simpler to write the calculation +. and juxtapose that new number with the remaining unknown number. Terms combined in an expression with a common, unknown factor (or multiple unknown factors) are ...
Substitution is replacing the terms in an expression to create a new expression. Substituting 3 for a in the expression a*5 makes a new expression 3*5 with meaning 15. Substituting the terms of a statement makes a new statement.
Montrell Johnson Jr.'s 5-yard touchdown run with 7:40 remaining in the fourth quarter gave Florida a 24-17 win over No. 9 Ole Miss on Saturday in Gainesville. Ole Miss had a chance to tie the game ...
4 + 9 + 2 = 15 (Add each individual digit together) 15 is divisible by 3 at which point we can stop. Alternatively we can continue using the same method if the number is still too large: 1 + 5 = 6 (Add each individual digit together) 6 ÷ 3 = 2 (Check to see if the number received is divisible by 3)