answer:We can solve this problem by calculating the compound interest year by year. For the deposit made on January 1, 2011: - The value of this deposit on January 1, 2012, is a(1+r). - The value of this deposit on January 1, 2013, is a(1+r)^2. - Continuing this pattern, the value of this deposit on January 1, 2015, is a(1+r)^4. For the deposit made on January 1, 2012: - It will have earned interest for 3 years by January 1, 2015, so its value will be a(1+r)^3. The deposit made on January 1, 2013, will be worth a(1+r)^2 by January 1, 2015, and the deposit made on January 1, 2014, will be worth a(1+r). The deposit to be made on January 1, 2015, is accessed immediately without accruing interest, so it remains a. Summing these up, we get the total amount he can withdraw on January 1, 2015: [ a(1+r)^4 + a(1+r)^3 + a(1+r)^2 + a(1+r) + a ] [ = a[(1+r)^4 + (1+r)^3 + (1+r)^2 + (1+r) + 1] ] [ = a frac{(1+r)[1-(1+r)^4]}{1-(1+r)} ] (This is the sum of a geometric series) [ = a frac{(1+r)[1-(1+r)^4]}{-r} ] (Simplifying the denominator) [ = frac{a}{r}[(1+r)^5 - (1+r)] ] (Further simplification) So, the correct answer is boxed{B: frac{a}{r}[(1+r)^5 - (1+r)]}.

answer:To solve for sin angle DAE in an equilateral triangle ABC where points D and E trisect overline{BC}, we proceed as follows: 1. **Assign a Length to the Sides of the Triangle**: Assume without loss of generality that the sides of the equilateral triangle ABC have a length of 6. This assumption simplifies calculations without affecting the generality of the solution. 2. **Identify and Analyze Triangle ACM**: - Let M be the midpoint of overline{DE}. By the properties of an equilateral triangle and its segments, triangle ACM forms a 30^circ-60^circ-90^circ right triangle. - In triangle ACM, since MC = 3 (half the length of BC because D and E trisect overline{BC}), we can deduce the lengths of the other sides based on the ratios in a 30^circ-60^circ-90^circ triangle: AC = 6 (given) and AM = 3sqrt{3}. 3. **Determine the Length of AE**: - Triangle AME is a right triangle with AM = 3sqrt{3} and ME = 1 (half the distance between D and E since M is the midpoint). - Using the Pythagorean Theorem, AE = sqrt{AM^2 + ME^2} = sqrt{(3sqrt{3})^2 + 1^2} = sqrt{27 + 1} = sqrt{28} = 2sqrt{7}. 4. **Calculate the Area of Triangle DAE in Two Ways**: - **Using Base DE and Height AM**: The area can be calculated as frac{1}{2} cdot DE cdot AM = frac{1}{2} cdot 2 cdot 3sqrt{3} = 3sqrt{3}. - **Using Sides AD, AE, and sin angle DAE**: The area can also be expressed as frac{1}{2} cdot AD cdot AE cdot sin angle DAE = frac{1}{2} cdot 6 cdot 2sqrt{7} cdot sin angle DAE = 6sqrt{7} cdot sin angle DAE. 5. **Equating the Two Expressions for the Area and Solving for sin angle DAE**: - Equating the two expressions for the area, we get 3sqrt{3} = 6sqrt{7} cdot sin angle DAE. - Solving for sin angle DAE, we find sin angle DAE = frac{3sqrt{3}}{6sqrt{7}} = frac{3sqrt{3}}{14}. Therefore, sin angle DAE = boxed{frac{3 sqrt{3}}{14}}.

answer:To solve the problem, we calculate the total hours worked by Davida in Weeks 1 and 2, and then in Weeks 3 and 4. After that, we find the difference in hours worked between these two periods. 1. Calculate the total hours worked in Weeks 1 and 2: [ text{Week 1} + text{Week 2} = 35 + 35 = 70 , text{hours} ] 2. Calculate the total hours worked in Weeks 3 and 4: [ text{Week 3} + text{Week 4} = 48 + 48 = 96 , text{hours} ] 3. Find the difference in hours worked between Weeks 3 and 4 and Weeks 1 and 2: [ text{Weeks 3 and 4} - text{Weeks 1 and 2} = 96 - 70 = 26 , text{hours} ] Therefore, Davida worked 26 more hours in Weeks 3 and 4 than in Weeks 1 and 2. [ boxed{26} ]

answer:Using the binomial theorem, - binom{100}{100} means selecting all 100 items from a set of 100 items. - By the property of binomial coefficients, binom{n}{n} = 1 for any non-negative integer n. Therefore, binom{100}{100} = 1. Thus, the solution to the problem is boxed{1}.