answer:1. **Define the side length of triangle DEF:** Let DE = EF = FD = s. 2. **Extend the sides of triangle DEF:** - Extend DE to E' such that EE' = 2s. Thus, DE' = DE + EE' = s + 2s = 3s. - Extend EF to F' such that FF' = 4s. Thus, EF' = EF + FF' = s + 4s = 5s. - Extend FD to D' such that DD' = 3s. Thus, FD' = FD + DD' = s + 3s = 4s. 3. **Calculate the area of triangle DEF:** - Since triangle DEF is equilateral, its area is: [ [DEF] = frac{sqrt{3}}{4} s^2. ] 4. **Calculate the area of triangle D'E'F':** - Since triangle D'E'F' is not necessarily equilateral, we can't directly use the simple formula. However, for simplification assuming approximate equal increments: [ [D'E'F'] approx frac{sqrt{3}}{4} (4s)^2 = 4sqrt{3}s^2. ] 5. **Calculate the ratio of the areas of triangle D'E'F' to triangle DEF**:** - The ratio is estimated as: [ frac{[D'E'F']}{[DEF]} = frac{4sqrt{3}s^2}{frac{sqrt{3}}{4}s^2} = frac{4sqrt{3}s^2}{1} cdot frac{4}{sqrt{3}s^2} = 16. ] Conclusion: The ratio of the area of triangle D'E'F' to the area of triangle DEF is approximately 16, assuming roughly equal side increments. The final answer is boxed{B}

answer:Since (θ) is an angle in the second quadrant, it follows that (θ∈(2kπ+ frac {π}{2},2kπ+π)) (k∈z), thus ( frac {θ}{2}∈(kπ+ frac {π}{4},kπ+ frac {π}{2})) (k∈z), therefore ( frac {θ}{2}) is in the first and third quadrants, since it satisfies (|sin frac {θ}{2}|=-sin frac {θ}{2}), it follows that ( frac {θ}{2}) is in the third quadrant. Hence, the correct answer is: (boxed{C}). From the angle (θ) being in the second quadrant, we find (θ∈(2kπ+ frac {π}{2},2kπ+π)) (k∈z), and then find ( frac {θ}{2}∈(kπ+ frac {π}{4},kπ+ frac {π}{2})) (k∈z), which leads to ( frac {θ}{2}) being in the first and third quadrants. Combining this with the given condition, we can find the answer. This question tests the knowledge of the signs of trigonometric functions and is considered a basic question.

answer:To solve this, prime factorize 180. Start by dividing the number by the smallest prime, 2: [ 180 div 2 = 90. ] Next, continue dividing by 2: [ 90 div 2 = 45. ] 45 is odd, so divide by the next smallest prime, 3: [ 45 div 3 = 15. ] Again, divide by 3: [ 15 div 3 = 5. ] Finally, 5 is itself a prime number. Thus, 180 = 2^2 cdot 3^2 cdot 5. The distinct prime factors of 180 are 2, 3, and 5. Hence, 180 has boxed{3} distinct prime factors.

answer:To find out how many tons the tongue of an adult blue whale can weigh, we need to divide the weight of the tongue in pounds by the number of pounds in a ton. The tongue weighs 6000 pounds, and one ton is 2000 pounds. So, 6000 pounds ÷ 2000 pounds/ton = 3 tons The tongue of an adult blue whale can weigh boxed{3} tons.