Search results
Results from the WOW.Com Content Network
Unit fractions can also be expressed using negative exponents, as in 2 −1, which represents 1/2, and 2 −2, which represents 1/(2 2) or 1/4. A dyadic fraction is a common fraction in which the denominator is a power of two , e.g. 1 / 8 = 1 / 2 3 .
In the second step, they were divided by 3. The final result, 4 / 3 , is an irreducible fraction because 4 and 3 have no common factors other than 1. The original fraction could have also been reduced in a single step by using the greatest common divisor of 90 and 120, which is 30. As 120 ÷ 30 = 4, and 90 ÷ 30 = 3, one gets
where f (2k−1) is the (2k − 1)th derivative of f and B 2k is the (2k)th Bernoulli number: B 2 = 1 / 6 , B 4 = − + 1 / 30 , and so on. Setting f ( x ) = x , the first derivative of f is 1, and every other term vanishes, so [ 15 ]
[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 ...
In mathematics, specifically in elementary arithmetic and elementary algebra, given an equation between two fractions or rational expressions, one can cross-multiply to simplify the equation or determine the value of a variable.
For the folded general continued fractions of both expressions, the rate convergence μ = (3 − √ 8) 2 = 17 − √ 288 ≈ 0.02943725, hence 1 / μ = (3 + √ 8) 2 = 17 + √ 288 ≈ 33.97056, whose common logarithm is 1.531... ≈ 26 / 17 > 3 / 2 , thus adding at least three digits per two terms. This is because the ...
A number line is usually represented as being horizontal, but in a Cartesian coordinate plane the vertical axis (y-axis) is also a number line. [5] The arrow on the line indicates the positive direction in which numbers increase. [5] Some textbooks attach an arrow to both sides, suggesting that the arrow indicates continuation.
Simplifying this further gives us the solution x = −3. It is easily checked that none of the zeros of x ( x + 1)( x + 2) – namely x = 0 , x = −1 , and x = −2 – is a solution of the final equation, so no spurious solutions were introduced.