the decimal expansion of 19/200is
Answers
0.095 is the decimal expansion of 19/200
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Decimal Expansion
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The decimal expansion of a number is its representation in base-10 (i.e., in the decimal system). In this system, each "decimal place" consists of a digit 0-9 arranged such that each digit is multiplied by a power of 10, decreasing from left to right, and with a decimal place indicating the 10^0=1s place. For example, the number with decimal expansion 1234.56 is defined as
1234.56 = 1×10^3+2×10^2+3×10^1+4×10^0+5×10^(-1)+6×10^(-2)
(1)
= 1×1000+2×100+3×10+4+5×1/(10)+6×1/(100).
(2)
Expressions written in this form sum_(k)b_k10^k (where negative k are allowed as exemplified above but usually not considered in elementary education contexts) are said to be in expanded notation.
Other examples include the decimal expansion of 25^2 given by 625, of pi given by 3.14159..., and of 1/9 given by 0.1111.... The decimal expansion of a number can be found in the Wolfram Language using the command RealDigits[n], or equivalently, RealDigits[n, 10].
The decimal expansion of a number may terminate (in which case the number is called a regular number or finite decimal, e.g., 1/2=0.5), eventually become periodic (in which case the number is called a repeating decimal, e.g., 1/3=0.3^_), or continue infinitely without repeating (in which case the number is called irrational).
The following table summarizes the decimal expansions of the first few unit fractions. As usual, the repeating portion of a decimal expansion is conventionally denoted with a vinculum.
fraction decimal expansion fraction decimal expansion
1 1 1/(11) 0.09^_
1/2 0.5 1/(12) 0.083^_
1/3 0.3^_ 1/(13) 0.076923^_
1/4 0.25 1/(14) 0.0714285^_
1/5 0.2 1/(15) 0.06^_
1/6 0.16^_ 1/(16) 0.0625
1/7 0.142857^_ 1/(17) 0.0588235294117647^_
1/8 0.125 1/(18) 0.05^_
1/9 0.1^_ 1/(19) 0.052631578947368421^_
1/(10) 0.1 1/(20) 0.05
If r=p/q has a finite decimal expansion (i.e., r is a regular number), then
r = (a_1)/(10)+(a_2)/(10^2)+...+(a_n)/(10^n)
(3)
= (a_110^(n-1)+a_210^(n-2)+...+a_n)/(10^n)
(4)
= (a_110^(n-1)+a_210^(n-2)+...+a_n)/(2^n·5^n).
(5)
Factoring possible common multiples gives
r=p/(2^alpha5^beta),
(6)