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  2. Cross-multiplication - Wikipedia

    en.wikipedia.org/wiki/Cross-multiplication

    are solved using cross-multiplication, since the missing b term is implicitly equal to 1: a 1 = x d . {\displaystyle {\frac {a}{1}}={\frac {x}{d}}.} Any equation containing fractions or rational expressions can be simplified by multiplying both sides by the least common denominator .

  3. Exponentiation - Wikipedia

    en.wikipedia.org/wiki/Exponentiation

    The base 3 appears 5 times in the multiplication, because the exponent is 5. Here, 243 is the 5th power of 3, or 3 raised to the 5th power. The word "raised" is usually omitted, and sometimes "power" as well, so 3 5 can be simply read "3 to the 5th", or "3 to the 5".

  4. FOIL method - Wikipedia

    en.wikipedia.org/wiki/FOIL_method

    The FOIL method is a special case of a more general method for multiplying algebraic expressions using the distributive law.The word FOIL was originally intended solely as a mnemonic for high-school students learning algebra.

  5. Multiplication and repeated addition - Wikipedia

    en.wikipedia.org/wiki/Multiplication_and...

    In mathematics education, there was a debate on the issue of whether the operation of multiplication should be taught as being a form of repeated addition.Participants in the debate brought up multiple perspectives, including axioms of arithmetic, pedagogy, learning and instructional design, history of mathematics, philosophy of mathematics, and computer-based mathematics.

  6. Order of operations - Wikipedia

    en.wikipedia.org/wiki/Order_of_operations

    When exponents were introduced in the 16th and 17th centuries, they were given precedence over both addition and multiplication and placed as a superscript to the right of their base. [2] Thus 3 + 5 2 = 28 and 3 × 5 2 = 75. These conventions exist to avoid notational ambiguity while allowing notation to remain brief. [4]

  7. Modular exponentiation - Wikipedia

    en.wikipedia.org/wiki/Modular_exponentiation

    For example, given b = 5, e = 3 and m = 13, dividing 5 3 = 125 by 13 leaves a remainder of c = 8. Modular exponentiation can be performed with a negative exponent e by finding the modular multiplicative inverse d of b modulo m using the extended Euclidean algorithm. That is: c = b e mod m = d −e mod m, where e < 0 and b ⋅ d ≡ 1 (mod m).

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