answer:1. **From Sarah's house to the southwest corner of Central Park:** - Sarah needs to travel 3 blocks west and 2 blocks south. She has to choose any order for 3 W’s and 2 S’s. - Calculate possible combinations: binom{5}{2} = frac{5!}{2!3!} = 10 ways. 2. **From the northeast corner of Central Park to the library:** - The library is 3 blocks east and 3 blocks north. - Calculate combinations for 3 E’s and 3 N’s: binom{6}{3} = frac{6!}{3!3!} = 20 ways. 3. **Through Central Park:** - There is still one diagonal path, so 1 way. 4. **Total possible routes:** - Multiplying the segments’ ways: 10 times 1 times 20 = 200. Conclusion: The total number of different routes Sarah can take is 200. The final answer is boxed{textbf{(D)} 200}

answer:1. Let d be the common difference of the sequence. We have the system of equations: begin{cases} (a_{1}+2d)(a_{1}+7d)=3(a_{1}+10d) dfrac{3a_{1}+3d}{2}=9 end{cases} Solving the system, we find two possible solutions: - d=0 and a_{1}=3 (which is discarded as it leads to equal terms in the sequence) - d=1 and a_{1}=2 Therefore, the general term formula for the sequence is: a_{n}=2+(n-1)times1=n+1 2. Observe that: frac{1}{a_{n}a_{n+1}}=frac{1}{(n+1)(n+2)}=frac{1}{n+1}-frac{1}{n+2} Hence, T_{n}=frac{1}{a_{1}a_{2}}+frac{1}{a_{2}a_{3}}+ldots+frac{1}{a_{n}a_{n+1}} =left(frac{1}{2}-frac{1}{3}right)+left(frac{1}{3}-frac{1}{4}right)+ldots+left(frac{1}{n+1}-frac{1}{n+2}right) =frac{1}{2}-frac{1}{n+2}=boxed{frac{n}{2(n+2)}}

answer:To solve this problem, let's break it down step by step: 1. First, calculate three times the number of cows We the People has: [3 times 17 = 51] cows. 2. Next, find out how many cows Happy Good Healthy Family has, which is two more than three times the number of cows We the People has: [51 + 2 = 53] cows. 3. Finally, to find out how many cows will be in the ranch altogether when their cows are taken to graze together, we add the number of cows from both families: [53 + 17 = 70] cows. Therefore, the total number of cows in the ranch will be boxed{70}.

answer:Since a and b are positive real numbers, i.e., a>0, b>0; because a+3b=7, therefore a+1+3(b+2)=14 Then frac {a+1}{14}+ frac {3(b+2)}{14}=1, Thus: (frac {1}{1+a} + frac {4}{2+b})(frac {a+1}{14}+ frac {3(b+2)}{14}) = frac {1}{14}+ frac {12}{14}+ left( frac {4(a+1)}{14(2+b)}+ frac {3(b+2)}{14(a+1)}right) geq frac {13}{14}+2times frac { sqrt {12}}{14} = frac {13+4 sqrt {3}}{14} Equality holds if and only if 2(a+1)= sqrt {3}(b+2), that is, when the equality sign holds. therefore The minimum value of frac {1}{1+a} + frac {4}{2+b} is: boxed{frac {13+4 sqrt {3}}{14}}, Hence, the answer is: boxed{frac {13+4 sqrt {3}}{14}}. This problem can be solved by constructing a basic inequality and utilizing the "multiply by 1" method along with the properties of basic inequalities. This question examines the idea of constructing inequalities, using the "multiply by 1" method and the properties of basic inequalities, and is considered a medium-level problem.