answer:For the first nine months, the interest rate is frac{8 times 9}{12} = 6%. Thus, the initial investment grows to: [ 15000 times 1.06 = 15900. ] Let the annual interest rate for the second term deposit be r %. Since the interest is compounded annually, the compounded amount for nine months computes using the formula for compound interest for partial years via the equation for exponential growth: [ 15900 times left(1 + frac{r}{100}right)^{0.75} = 17010. ] Solving for r, we have: [ left(1 + frac{r}{100}right)^{0.75} = frac{17010}{15900} approx 1.07, ] [ 1 + frac{r}{100} = 1.07^{4/3}, ] [ frac{r}{100} = 1.07^{4/3} - 1, ] [ r = 100 times (1.07^{4/3} - 1). ] By calculation (calculating 1.07^{4/3} approx 1.096): [ r approx 100 times (1.096 - 1) = 9.6. ] Hence, the annual interest rate for the second term deposit is r = boxed{9.6%}.

answer:**Analysis** Given that the line ax+by-2=0 (a,b > 0) passes through the center of the circle (1,1), we have a+b=2. From this, we can use the property of the basic inequality to find the minimum value of dfrac{1}{2a}+dfrac{1}{b}. This question tests the method of finding the minimum value of the sum of two numbers, which is a medium-level problem. When solving, it is important to carefully read the question and appropriately apply the properties of the circle and the basic inequality. **Solution** Since the line ax+by-2=0 (a,b > 0) always bisects the circumference of the circle x^{2}+y^{2}-2x-2y=2, it follows that the line ax+by-2=0 (a,b > 0) passes through the center of the circle (1,1), thus a+b-2=0, that is, a+b=2, since a > 0, b > 0, we have dfrac{1}{2a}+dfrac{1}{b} = dfrac{(a+b)}{2} left(dfrac{1}{2a}+dfrac{1}{b}right) = dfrac{1}{4}+ dfrac{b}{4a}+ dfrac{a}{2b}+ dfrac{1}{2} geqslant dfrac{3}{4}+2sqrt{dfrac{b}{4a}cdotdfrac{a}{2b}} = dfrac{3+2sqrt{2}}{4}, therefore, the minimum value of dfrac{1}{2a}+dfrac{1}{b} is dfrac{3+2sqrt{2}}{4}. Hence, the correct choice is boxed{text{D}}.

answer:Since the statement "For any real number x, the inequality x^2 + 2x + a > 0 is always true" is false, there must exist a real number x for which x^2 + 2x + a leq 0 holds true. Considering the quadratic equation x^2 + 2x + a = 0 related to the inequality above, let's analyze its discriminant: [ Delta = b^2 - 4ac = 2^2 - 4 cdot 1 cdot a = 4 - 4a. ] For the original inequality to ever be non-positive, the discriminant Delta must be greater than or equal to zero. Thus, we solve for a: [ Delta geq 0 4 - 4a geq 0 -4a geq -4 a leq 1. ] Therefore, the range of a is (-infty, 1], and the corresponding answer is: boxed{D}

answer:To solve this problem, we need to calculate the number of ways to choose a team of 5 students from 7 boys and 10 girls with the condition that there are at least 2 girls on the team. We can approach this by considering different cases based on the number of girls: 1. **Case 1**: 2 girls and 3 boys. - The number of ways to choose 2 girls out of 10 is binom{10}{2}. - The number of ways to choose 3 boys out of 7 is binom{7}{3}. - So, the number of combinations for this case is binom{10}{2} times binom{7}{3}. 2. **Case 2**: 3 girls and 2 boys. - The number of ways to choose 3 girls out of 10 is binom{10}{3}. - The number of ways to choose 2 boys out of 7 is binom{7}{2}. - So, the number of combinations for this case is binom{10}{3} times binom{7}{2}. 3. **Case 3**: 4 girls and 1 boy. - The number of ways to choose 4 girls out of 10 is binom{10}{4}. - The number of ways to choose 1 boy out of 7 is binom{7}{1}. - So, the number of combinations for this case is binom{10}{4} times binom{7}{1}. 4. **Case 4**: 5 girls and 0 boys. - The number of ways to choose 5 girls out of 10 is binom{10}{5}. - There are no boys, so this case is just binom{10}{5}. Summing all these cases: binom{10}{2} times binom{7}{3} + binom{10}{3} times binom{7}{2} + binom{10}{4} times binom{7}{1} + binom{10}{5} = 45 times 35 + 120 times 21 + 210 times 7 + 252 = 1575 + 2520 + 1470 + 252 = boxed{5817}. Conclusion: The number of ways to select a team of 5 students that includes at least 2 girls from a group of 7 boys and 10 girls is boxed{5817}.