answer:To calculate the total amount Julia spent on food for her animals, we start by breaking down the costs for each animal. 1. The weekly cost for the rabbit's food is given as 12. 2. Since the total weekly cost for both animals is 30, we subtract the cost for the rabbit's food from this total to find the cost for the parrot's food: 30 - 12 = 18. Now, we calculate the total spent on each animal: 1. For the parrot, Julia spent 18 per week for 3 weeks, which totals to 18 times 3 = 54. 2. For the rabbit, Julia spent 12 per week for 5 weeks, which totals to 12 times 5 = 60. Adding these amounts gives us the total spent on both animals: - Total spent = 54 (for the parrot) + 60 (for the rabbit) = 114. Therefore, Julia spent boxed{114} on food for her animals.

answer:1. **Define the sequences:** Let ( t_{2k+1} ) represent the number of ways to cover a ( 3 times (2k+1) ) chessboard with one corner square removed, and let ( s_{2k} ) represent the number of ways to cover a ( 3 times 2k ) chessboard with no corner squares removed. 2. **Establish recurrence relations:** From the problem, we have the following recurrence relations: [ t_{2k+1} = s_{2k} + t_{2k-1} ] [ s_{2k} = s_{2k-2} + 2t_{2k-1} ] 3. **Initial conditions:** From visual inspection, we have: [ t_1 = 1 quad text{and} quad t_3 = 4 ] 4. **Derive a simplified recurrence relation:** By manipulating the two equations, we can derive: [ t_{2k+1} = 4t_{2k-1} - t_{2k-3} ] Define the sequence ( {u_n} ) by ( u_n = t_{2n-1} ). Thus, we need to find ( u_{1011} mod 19 ). 5. **Restate the recurrence relation:** The sequence ( {u_n} ) follows: [ u_{n+1} = 4u_n - u_{n-1} ] with initial conditions: [ u_1 = 1 quad text{and} quad u_2 = 4 ] 6. **Calculate the sequence modulo 19:** We calculate the first few terms modulo 19: [ begin{aligned} u_1 & = 1, u_2 & = 4, u_3 & = 4u_2 - u_1 = 4 cdot 4 - 1 = 15, u_4 & = 4u_3 - u_2 = 4 cdot 15 - 4 = 60 - 4 = 56 equiv 18 mod 19, u_5 & = 4u_4 - u_3 = 4 cdot 18 - 15 = 72 - 15 = 57 equiv 0 mod 19, u_6 & = 4u_5 - u_4 = 4 cdot 0 - 18 = -18 equiv 1 mod 19, u_7 & = 4u_6 - u_5 = 4 cdot 1 - 0 = 4 mod 19. end{aligned} ] We observe that the sequence starts repeating with ( u_6 = 1 ) and ( u_7 = 4 ). 7. **Identify the period of the sequence:** The sequence ( {u_n} ) repeats every 6 terms: [ u_1 equiv 1, quad u_2 equiv 4, quad u_3 equiv 15, quad u_4 equiv 18, quad u_5 equiv 0, quad u_6 equiv 1 mod 19. ] 8. **Find ( u_{1011} mod 19 ):** Since the sequence repeats every 6 terms, we find the position of ( u_{1011} ) in the cycle: [ 1011 mod 6 = 5. ] Therefore, ( u_{1011} equiv u_5 mod 19 ). 9. **Conclusion:** From the sequence, ( u_5 equiv 0 mod 19 ). The final answer is ( boxed{0} ).

answer:To solve the given problem, let's break down the expression left(2k+3right)^{2}-4k^{2} step by step: 1. First, expand the square term: [ left(2k+3right)^{2} = 4k^{2} + 12k + 9 ] 2. Subtract 4k^{2} from the expanded form: [ 4k^{2} + 12k + 9 - 4k^{2} = 12k + 9 ] 3. Factor out the common factor of 3: [ 12k + 9 = 3(4k + 3) ] Since the expression simplifies to 3(4k + 3), it is evident that it can always be divided by 3, regardless of the integer value of k. This is because the factor of 3 is explicitly present in the simplified form of the expression. Therefore, the correct choice is: [ boxed{text{B: Divided by } 3} ]

answer:Let's denote the weight of the grandmother as G, the weight of the daughter as D, and the weight of the child as C. According to the information given: 1. The combined weight of the grandmother, daughter, and child is 110 kg. G + D + C = 110 2. The child weighs 1/5th of her grandmother's weight. C = 1/5 * G 3. The daughter weighs 50 kg. D = 50 We need to find the combined weight of the daughter and her child, which is D + C. First, let's use the information about the child's weight to express the grandmother's weight in terms of the child's weight: G = 5 * C Now, let's substitute the values of D and G into the first equation: 5 * C + 50 + C = 110 Combine like terms: 6 * C + 50 = 110 Subtract 50 from both sides to isolate the terms with C: 6 * C = 60 Divide both sides by 6 to solve for C: C = 10 Now that we have the weight of the child, we can find the combined weight of the daughter and her child: D + C = 50 + 10 = 60 The combined weight of the daughter and her child is boxed{60} kg.