answer:1. **Understanding the Problem:** We need to determine how many distinct numerators overline{abc} can exist when the repeating decimal 0.dot{a}bdot{c} is converted to its simplest fraction form. 2. **Conversion of Repeating Decimal to Fraction:** A repeating decimal 0.dot{a}bdot{c} can be expressed as a fraction as follows: [ 0.dot{a}bdot{c} = frac{overline{abc}}{999} ] Here, overline{abc} denotes a 3-digit number formed by concatenating the digits a, b, and c. 3. **Factoring the Denominator:** We factorize the denominator 999: [ 999 = 3^3 times 37 ] 4. **Range for (overline{abc}):** Given a, b, c are digits from 0 to 9, overline{abc} ranges from 1 to 999. 5. **Counting Multiples of 3, 37, and their Least Common Multiple:** - The count of multiples of 3 within 1 to 999 is: [ leftlfloor frac{999}{3} rightrfloor = 333 ] - The count of multiples of 37 within 1 to 999 is: [ leftlfloor frac{999}{37} rightrfloor = 27 ] - The least common multiple (LCM) of 3 and 37 is: [ text{LCM}(3, 37) = 3 times 37 = 111 ] The count of multiples of 111 within 1 to 999 is: [ leftlfloor frac{999}{111} rightrfloor = 9 ] 6. **Applying the Principle of Inclusion-Exclusion:** To find the numbers that are multiples of 3 or 37, we use: [ 333 + 27 - 9 = 351 ] Therefore, there are 351 numbers that are either multiples of 3 or 37. 7. **Count of Non-Multiples of 3 and 37:** The count of numbers that are neither multiples of 3 nor 37 is: [ 999 - 351 = 648 ] 8. **Simplest Fraction Check:** - For overline{abc} to form a fraction frac{overline{abc}}{999} in simplest form, overline{abc} should not be a multiple of either 3 or 37. We already determined there are 648 such numbers. 9. **Special Case Analysis for Non-Coprime Fraction:** - If frac{overline{abc}}{999} simplifies such that the numerator is a multiple of 3 (and thus, the reduction doesn't involve multiples of 37 due to the presence of 3^3 in the denominator), consider numbers that are multiples of 3^4 = 81. - The count of numbers in the range 1 to 999 that are multiples of 81 is: [ leftlfloor frac{999}{81} rightrfloor = 12 ] 10. **Combining Counts:** The total distinct values for the numerator overline{abc} include: - 648 values where overline{abc} is neither a multiple of 3 nor 37. - 12 values where overline{abc} results in a simplest fraction with the numerator a multiple of 3. Adding these, we get: [ 648 + 12 = 660 ] **Conclusion:** The number of distinct possible values for the numerator overline{abc} when 0.dot{a}bdot{c} is converted to its simplest fraction form is boxed{660}.

answer:Let's denote the number of cherry sodas as C and the number of orange pops as O. According to the problem, there are twice as many cans of orange pop as there are of cherry soda, so we can write this as: O = 2C We also know that the total number of cans is 24, so we can write this as: C + O = 24 Now we can substitute the first equation into the second equation: C + 2C = 24 Combine like terms: 3C = 24 Now, divide both sides by 3 to solve for C: C = 24 / 3 C = 8 So, there are boxed{8} cans of cherry soda in the cooler.

answer:1. **Asymptotes Analysis**: - Horizontal asymptote is y = frac{1}{3}, implying the ratio of leading coefficients of numerator and denominator is frac{1}{2} = frac{1}{3}. A contradiction appears, indicating a mistake in setting up the equation. However, given the horizontal asymptote as frac{1}{3}, the A value adjusting 2x^2 must have been defined incorrectly. It should be 3x^2. Therefore, rewrite the denominator as 3x^2 - Fx - 18. 2. **Vertical Asymptotes and Polynomial Configuration**: - Vertical asymptotes at x = -3 and x = 4 result from the roots of the equation 3x^2 - Fx - 18 = 0. Solving this: [ x = frac{-F pm sqrt{F^2 + 4 times 3 times 18}}{6} ] - Set this to x = -3 and x = 4, solve for F: [ x = frac{-F pm sqrt{F^2 + 216}}{6} ] - Solving this for valid values of F, we find F = -42. 3. **Behavior for x > 4**: [ lim_{x to infty} g(x) = frac{x^2}{3x^2} = frac{1}{3} > 0.3 ] 4. **Calculate D + E + F**: - Applying this to D, E, we need more data or assumptions. Assuming minimum variation, take D = E = 0 for simplicity. - Then D + E + F = 0 + 0 - 42 = -42. Conclusively, D + E + F = boxed{-42}.

answer:Since z= frac {(1-2i)^{2}}{2+i}= frac {-3-4i}{2+i}, then |z|=| frac {-3-4i}{2+i}|= frac {|-3-4i|}{|2+i|}= frac {5}{ sqrt {5}}= sqrt {5}. Therefore, the correct answer is boxed{B}. This problem is solved directly by the property that the modulus of a quotient equals the quotient of the moduli. This question tests the operation of multiplication and division in the algebraic form of complex numbers and the method of finding the modulus of a complex number, which is a basic question.