answer:To find the number of different paths from A to D, we analyze: 1. **Paths from A to B**: There are 2 ways (upper and lower arrow). 2. **Paths from B to C**: There are 2 ways (upper and lower arrow). 3. **Paths from C to D**: There are 2 ways (upper and lower arrow). 4. **Direct path from A to D** Using these paths: - From A to B to C to D, there are (2 (A to B) times 2 (B to C) times 2 (C to D) = 8) paths. - There is 1 direct path from A to D. Thus, there is a total of boxed{9} different paths from A to D.

answer:1. Let's observe the properties of the rectangular coordinates in the given problem. We are given that (C K perp B D). 2. Define the midpoints and relationships. Given that (K) is the midpoint of (AD), we have: [ AK = KD ] 3. Consider the angles formed by the intersection of lines (CK) and (BD). Specifically, (angle A B D) and (angle B C K) are equal due to the perpendicularity of their respective sides (AB) and (BC) with (BD) and (CK) respectively. Therefore: [ angle ABD = angle BCK ] 4. Since (A) and (H) lie on a circle with diameter (BK), we use the property of diameters and circles. Consequently: [ triangle AHB = triangle AKB ] This suggests that (angle AHB) is subtended by the same arc (BK). 5. Using triangle properties and symmetry, since (angle AHB = angle AKB), we have the triangles (CKD) and (BKA) being congruent. This is due to the equality of angles formed by parallel lines and a transversal: [ angle AKB = angle CKD ] where (angle CKD = angle BCK) because they are corresponding angles and (angle BCK) intersects parallel lines (BC) and (AD). 6. Therefore, based on the above reasoning, ( angle AHB = angle ABD ). Here, (angle ABD) was the original angle formed from the perpendicular bisectors and intersections. This confirms their equality: [ angle AHB = angle ABD ] 7. With (HB = AB) and their contained angles being equal, this hence proves that (triangle A H B) is isosceles, as required. Conclusion: ( boxed{text{Triangle } A H B text{ is isosceles.}} )

answer:Given: - ( AB parallel EF parallel DC ) - ( AC + BD = 250 ) - ( BC = 100 ) - ( EC + ED = 150 ) We need to find ( CF ). 1. By using the properties of parallel lines and similar triangles, we can set up the proportion involving the segments ( AC ), ( BD ), ( BC ), ( EC ), ( DE ), and ( CF ). According to the given ratios: [ frac{CE}{CF} = frac{AC}{BC} quad text{and} quad frac{DE}{CF} = frac{BD}{BC} ] 2. Let's start by finding the combined proportion: [ frac{CE + DE}{CF} = frac{AC}{BC} + frac{BD}{BC} ] Given that ( AC + BD = 250 ) and ( BC = 100 ): [ frac{AC + BD}{BC} = frac{250}{100} ] 3. Therefore, we have: [ frac{150}{CF} = frac{250}{100} ] 4. Simplifying the right-hand side of the equation: [ frac{250}{100} = 2.5 ] 5. Now, solving for ( CF ): [ frac{150}{CF} = 2.5 implies 150 = 2.5 times CF implies CF = frac{150}{2.5} ] 6. Dividing 150 by 2.5: [ CF = 60 ] # Conclusion: [ boxed{60} ]

answer:We are tasked with identifying numbers where the sum of digits equals 5, with no digit being zero, to ascertain if each is a counterexample (non-prime number). 1. **Set ({1,1,1,1,1})**: - The number formed is (N = 11111). - The factorization of (11111): [ 11111 = 41 times 271 ] - Since (11111) is not prime, we have a counterexample. 2. **Set ({1,1,1,2})**: - Possible numbers are (N = 1112, 1121, 1211, 2111). - All are even except (2111), hence not prime. - Check (N = 2111); by trial division, (2111) is not divisible by any prime number up to (sqrt{2111} approx 45.9). Thus, (2111) is prime. 3. **Set ({1,1,3})**: - Possible numbers are (N = 113, 131, 311). - (N = 113) and (N = 131) are both primes. - (N = 311) is also prime. 4. **Set ({1,4})**: - Possible numbers are (N = 14) and (N = 41). - (N = 14) is even, hence not prime. - (N = 41) is a prime number. 5. **Set ({2,3})**: - Possible numbers are (N = 23) and (N = 32). - Both (23) and (32) are prime numbers (consider (32 = 2^5), hence not prime). 6. **Set ({5})**: - The number formed is (N = 5). - 5 is prime. Considering the analysis, the counterexamples where the number is not prime are (N = 11111, N = 1112, N = 1121, N = 1211, N = 14, N = 32). There are (6) counterexamples. The final answer is boxed{The final answer, given the choices, is (boxed{text{D) 6}}).}