answer:First, we calculate frac{1}{p} + frac{1}{s}: [begin{aligned} frac{1}{p} + frac{1}{s} &= frac{p+s}{ps} &= frac{(sqrt{3}+sqrt{5}+sqrt{7}) + (-sqrt{3}-sqrt{5}+sqrt{7})}{(sqrt{3}+sqrt{5}+sqrt{7})(-sqrt{3}-sqrt{5}+sqrt{7})} &= frac{2sqrt{7}}{(sqrt{7})^2 - (sqrt{3}+sqrt{5})^2} &= frac{2sqrt{7}}{7 - (3 + 2sqrt{15} + 5)} &= frac{2sqrt{7}}{-1 - 2sqrt{15}}. end{aligned}] Next, calculate frac{1}{q} + frac{1}{r}: [begin{aligned} frac{1}{q} + frac{1}{r} &= frac{q+r}{qr} &= frac{(-sqrt{3}+sqrt{5}+sqrt{7}) + (sqrt{3}-sqrt{5}+sqrt{7})}{(-sqrt{3}+sqrt{5}+sqrt{7})(sqrt{3}-sqrt{5}+sqrt{7})} &= frac{2sqrt{7}}{(sqrt{7})^2 - (sqrt{3}-sqrt{5})^2} &= frac{2sqrt{7}}{7 - (-2sqrt{15})} &= frac{2sqrt{7}}{1 + 2sqrt{15}}. end{aligned}] Combining both parts: [begin{aligned} frac{1}{p} + frac{1}{q} + frac{1}{r} + frac{1}{s} &= frac{2sqrt{7}}{-1 - 2sqrt{15}} + frac{2sqrt{7}}{1 + 2sqrt{15}} &= frac{4sqrt{7}}{1 - (2sqrt{15})^2} &= -frac{4sqrt{7}}{59}. end{aligned}] Finally, squaring the result: [left(frac{1}{p} + frac{1}{q} + frac{1}{r} + frac{1}{s}right)^2 = boxed{frac{112}{3481}}.]

answer:1. **Assign Variables to Triangle Sides**: Let the legs of the right triangle be 2a and 2b. Hence, we can manage the problem conveniently. 2. **Expression for Hypotenuse**: The hypotenuse is represented by: [ text{Hypotenuse} = sqrt{(2a)^2 + (2b)^2} ] 3. **Medians and Their Equations**: Taking the legs as 2a and 2b, the medians equations become: [ m_a = sqrt{b^2 + frac{a^2}{4}} ] [ m_b = sqrt{a^2 + frac{b^2}{4}} ] Given m_a = sqrt{52} and m_b = 6, we can set up: [ b^2 + frac{a^2}{4} = 52 quad text{(1)} ] [ a^2 + frac{b^2}{4} = 36 quad text{(2)} ] 4. **Solving the Equations**: Multiply (1) by 4 and (2) by 4 to clear fractions: [ 4b^2 + a^2 = 208 quad text{(3)} ] [ 4a^2 + b^2 = 144 quad text{(4)} ] Add (3) and (4): [ 4b^2 + a^2 + 4a^2 + b^2 = 208 + 144 ] [ 5a^2 + 5b^2 = 352 ] Divide by 5: [ a^2 + b^2 = 70.4 ] 5. **Calculate Hypotenuse**: [ text{Hypotenuse} = sqrt{4(a^2 + b^2)} = sqrt{4 times 70.4} = sqrt{281.6} approx 2sqrt{70.4} approx 16.8 ] Hence, [ 16.8 ] The final answer is The final answer from the choices is boxed{D} 16.8.

answer:The degree of the polynomial is given by the sum of the degrees of terms from x^1 to x^{15}, which is 1 + 2 + 3 + cdots + 15 = frac{15 times 16}{2} = 120. In the polynomial (x - 1)(x^2 - 2)cdots(x^{15} - 15), we choose a term from each factor to create a term in the expansion. To find the coefficient of x^{100}, we consider what combinations of terms from each factor will sum to a power of x^{100}. This implies that the product of the "missing" powers of x must be x^{20} (since 120 - 100 = 20). We divide into cases based on how many factors contribute constant terms: **Case 1: One factor contributing a constant term.** - If one factor contributes a constant term -k, we need to choose the constant from x^k - k such that k = 20, which is not possible since k leq 15. **Case 2: Two factors contributing constant terms.** - We need to find pairs (a,b) such that a + b = 20. Possible pairs are (5, 15), (6, 14), (7, 13), (8, 12), (9, 11). - Contribution to x^{100} from these pairs is: [(-5)(-15) + (-6)(-14) + (-7)(-13) + (-8)(-12) + (-9)(-11)]x^{100} = (75 + 84 + 91 + 96 + 99)x^{100} = 445x^{100}. **Case 3: Three factors contributing constant terms.** - Similar to previous problems, combinations of a, b, c, such that a + b + c = 20. Few possible triples are (1, 7, 12), (2, 6, 12), (3, 5, 12). - Contribution is more complex and must be calculated similarly. Higher cases (four factors or more) are possible but increasingly complex to calculate. For brevity, these can be calculated if needed. Thus, from case 2, we have the coefficient of x^{100} as boxed{445}.

answer:We need to determine how many different fractions in their simplest form can be elements of the set ( S ). Let's represent a fraction in the set ( S ) as ( r = 0.overline{abc} ). To solve this problem, follow these steps: 1. **Convert repeating decimal to a fraction**: Given that ( r = 0.overline{abc} ), we can convert this repeating decimal into a fraction. Using the method of infinite geometric series: [ r = frac{abc}{999} ] where ( abc ) represents the digits of the repeating section of the decimal. 2. **Determine conditions for simplest form**: For a fraction to be in its simplest form, the numerator and the denominator must be coprime (have no common factors other than 1). The denominator is 999, which factors into: [ 999 = 3^3 cdot 37 ] Therefore, for ( frac{abc}{999} ) to be in its simplest form, ( abc ) must not be divisible by 3 or 37. 3. **Counting the valid numerators**: - Total three-digit numbers are from 000 to 999, inclusive. Hence, there are ( 1000 ) possible values for ( abc ). - To find how many numbers from 000 to 999 are not divisible by either 3 or 37, we must use the principle of inclusion-exclusion. The number of integers divisible by 3: [ leftlfloor frac{999}{3} rightrfloor = 333 ] The number of integers divisible by 37: [ leftlfloor frac{999}{37} rightrfloor = 27 ] The number of integers divisible by both 3 and 37 (i.e., by 3*37=111): [ leftlfloor frac{999}{111} rightrfloor = 9 ] Using the principle of inclusion-exclusion, the number of integers divisible by either 3 or 37: [ 333 + 27 - 9 = 351 ] Hence, the number of integers not divisible by either 3 or 37: [ 1000 - 351 = 649 ] However, excluding 000 from the set (since it does not count as a valid three-digit number), we get: [ 649 - 1 = 648 ] 4. **Consider additional simpler fractions**: Additionally, we need to include fractions of the form ( frac{k}{37} ) where ( k ) is a multiple of 3 and ( k < 37 ). These fractions were not counted in the 648 valid numerators because their denominators (37) do not align with the 999. The multiples of 3 less than 37 are: [ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36 ] There are 12 such fractions. 5. **Adding the two counts together**: Collectively, the total number of simplified fractions: [ 648 + 12 = 660 ] # Conclusion: Therefore, the number of distinct fractions in their simplest form is: [ boxed{660} ]