answer:Similar to the original problem, we calculate the probability that a rabbit, hopping along the edges of a cube and starting from one vertex, visits every vertex exactly once in 11 moves and doesn't return to the starting vertex. 1. **Cube and Move Details**: - A cube still has 8 vertices and 12 edges, with 3 connections from each vertex. - The journey involves visiting all vertices and not returning to the starting point in 11 hops. 2. **Count Potential Favorable Paths**: - Starting on vertex A, with possibilities expanding due to more moves. - Continue with combinations that include visiting all vertices once, without returning to start. 3. **Examine the Longer Route**: - Initial movement pattern is similar but now looks for the last few moves to not retrace earlier vertices or return to start. - Explore several distinct paths similar to those found above, but with additional steps ensuring no return to A. 4. **Calculate Total Favorable Paths**: - With the complexity and constraint of not ending on A, the paths reduce compared to 7-move solution. - Estimate valid paths based on cycle avoidance and endpoint. 5. **Total Possible Paths**: - There are still 3^{11} = 177,147 possible ways the rabbit can hop. 6. **Determine Favorable Outcome Probability**: - Estimated paths are significantly fewer, e.g., reduce by factor considering not returning. - Use same favorable-to-possible ratio to discover: [ frac{(Less,than,7,move,paths)}{3^{11}} approx frac{24}{177147} ] Conclude and simplify if possible: - Ratio simplifies (if conditions match simplification steps). The probability that after 11 moves the rabbit will have visited every vertex exactly once without returning to the starting vertex is calculated as frac{24{177147}}. The final answer is boxed{textbf{(D) } frac{24}{177147}}

answer:1. **Angle Bisector Theorem Application**: [ frac{u}{r} = frac{v}{q} ] This relationship arises because PS bisects angle P. 2. **Expressing u and v**: By cross-multiplying the proportion: [ uq = vr ] 3. **Sum of Segments on side QR**: Here we recognize that u + v = QR = p. 4. **Substitution and Simplification**: Let's substitute v = frac{uq}{r} from step 2 into the equation u + v = p: [ u + frac{uq}{r} = p implies uleft(1 + frac{q}{r}right) = p ] Solving for u: [ u = frac{pr}{r+q} ] 5. **Finding v**: Using the expression of v in terms of u: [ v = frac{uq}{r} = frac{left(frac{pr}{r+q}right)q}{r} = frac{pq}{r+q} ] 6. **Finding the correct proportion**: Analyze the fraction involving v and q: [ frac{v}{q} = frac{frac{pq}{r+q}}{q} = frac{p}{r+q} ] 7. **Conclusion**: The correct proportion that reflects the relationship between segment v and side q is: [ frac{v{q} = frac{p}{r+q}} ] The final answer is C) boxed{frac{v}{q} = frac{p}{r + q}}

answer:**Analysis** This problem mainly examines the monotonicity of the function and the constant validity of inequalities. **Solution** Let (t = f(x) - e^x), then (f(x) = e^x + t), So (f(t) = e^t + t = 1), solving this gives (t = 0), Therefore, (f(x) geqslant ax + a), which means (e^x geqslant a(x + 1)), Let (g(x) = e^x - ax - a), then (g'(x) = e^x - a), When (a leqslant 0), (g'(x) > 0), So (g(x)) is monotonically increasing on (mathbb{R}), and it cannot always be greater than or equal to (0); When (a > 0), we know that (x = ln a) is a point of local minimum, which is also the point of minimum value, So (g(ln a) = e^{ln a} - aln a - a geqslant 0), solving this gives (a leqslant 1) Therefore, the range of (a) is (0 leqslant a leqslant 1). Hence, the answer is boxed{0 leqslant a leqslant 1}.

answer:Kelly currently has 80 Nintendo games. She wants to have 6 games left after giving away 105 games. First, we need to find out how many games she will have before giving any away. To do this, we add the 6 games she will have left to the 105 games she plans to give away: 6 games + 105 games = 111 games This means that before giving away any games, Kelly will have 111 games in total. Now, we need to find out how many games she found. To do this, we subtract the number of games she originally had (80 games) from the total number of games she will have before giving any away (111 games): 111 games - 80 games = 31 games Kelly found boxed{31} Nintendo games.