Bunuel wrote:
The sum\(\frac{3}{10}+\frac{43}{100}+\frac{17}{1000}\) is equivalent to which of the following sums?
A.\(\frac{7}{1000}+\frac{4}{10}+\frac{7}{100}\)
B.\(\frac{6}{10}+\frac{12}{100}+\frac{37}{1000}\)
C.\(\frac{7}{100}+\frac{4}{10}+\frac{7}{1000}\)
D.\(\frac{7}{100}+\frac{4}{1000}+\frac{7}{10}\)
E.\(\frac{32}{100}+\frac{4}{10}+\frac{27}{1000}\)
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