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For example, 3 5 = 3 · 3 · 3 · 3 · 3 = 243. 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".
5⋅5, or 5 2 (5 squared), can be shown graphically using a square. Each block represents one unit, 1⋅1, and the entire square represents 5⋅5, or the area of the square. In mathematics, a square is the result of multiplying a number by itself. The verb "to square" is used to denote this operation.
A formula for computing the trigonometric identities for the one-third angle exists, but it requires finding the zeroes of the cubic equation 4x 3 − 3x + d = 0, where is the value of the cosine function at the one-third angle and d is the known value of the cosine function at the full angle.
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.
[2] [3] Thus, in the expression 1 + 2 × 3, the multiplication is performed before addition, and the expression has the value 1 + (2 × 3) = 7, and not (1 + 2) × 3 = 9. 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 ...
For example, taking the statement x + 1 = 0, if x is substituted with 1, this implies 1 + 1 = 2 = 0, which is false, which implies that if x + 1 = 0 then x cannot be 1. If x and y are integers, rationals, or real numbers, then xy = 0 implies x = 0 or y = 0. Consider abc = 0. Then, substituting a for x and bc for y, we learn a = 0 or bc = 0.
If the number is of the form m5 where m represents the preceding digits, its square is n25 where n = m(m + 1) and represents digits before 25. For example, the square of 65 can be calculated by n = 6 × (6 + 1) = 42 which makes the square equal to 4225. If the number is of the form m0 where m represents the preceding digits, its square is n00 ...
x 1 = x; x 2 = x 2 for i = k - 2 to 0 do if n i = 0 then x 2 = x 1 * x 2; x 1 = x 1 2 else x 1 = x 1 * x 2; x 2 = x 2 2 return x 1 The algorithm performs a fixed sequence of operations ( up to log n ): a multiplication and squaring takes place for each bit in the exponent, regardless of the bit's specific value.