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question:A pet store regularly sells pet food at a discount of 10 percent to 30 percent from the manufacturer's suggested retail price. If during a sale, the store discounts an additional amount from the discount price, the lowest possible price of a container of pet food that had a manufacturer's suggested retail price of 45.00 would be 25.2. What is the additional discount the store offers during a sale?
answer:Let's first calculate the range of discounts the store regularly offers on the manufacturer's suggested retail price (MSRP) of 45.00. The regular discount range is 10% to 30%. The lowest possible regular discount price would be after applying the maximum regular discount of 30%. So, we calculate 30% of the MSRP: 30% of 45.00 = 0.30 * 45.00 = 13.50 Now, we subtract this discount from the MSRP to find the lowest regular discount price: 45.00 - 13.50 = 31.50 During the sale, the store offers an additional discount that brings the price down to 25.20. To find the additional discount percentage, we need to calculate the difference between the lowest regular discount price and the sale price, and then find what percentage this difference is of the lowest regular discount price. Difference between the lowest regular discount price and the sale price: 31.50 - 25.20 = 6.30 Now, we find what percentage 6.30 is of 31.50: (6.30 / 31.50) * 100 = 0.20 * 100 = 20% So, the additional discount the store offers during a sale is boxed{20%} .
question:Arrange the numbers 1, 2, 3, 4, 5, 6 in a row, denoting the ith number as a_i (i=1,2,ldots,6). If a_1 neq 1, a_3 neq 3, a_5 neq 5, and a_1 < a_3 < a_5, then the number of different arrangements is ___ (answer with a number).
answer:This problem is a step-by-step counting problem. There are two steps: 1. First, arrange a_1, a_3, a_5. - When a_1=2, there are 2 possibilities; - When a_1=3, there are 2 possibilities; - When a_1=4, there is 1 possibility; - In total, there are 5 possibilities. 2. Next, arrange a_2, a_4, a_6. There are A_3^3=6 possibilities. Therefore, the total number of different arrangements is 5 times 6 = 30. Hence, the answer is boxed{30}. **Analysis:** This problem is a step-by-step counting problem. First, arrange a_1, a_3, a_5 for the cases when a_1=2, a_1=3, and a_1=4, calculating the number of outcomes for these cases. In the second step, arrange a_2, a_4, a_6, and calculate the number of outcomes. The result is obtained by applying the principle of step-by-step counting.
question:Translate the given mathematical problem into English and format it using Markdown or LaTeX: Given the line (x y + 1 = 0) and the circle (x 2)^2 + (y 1)^2 = 1, find the shortest distance to translate the line such that it is tangent to the circle. A: sqrt{2} 1 B: 2 sqrt{2} C: sqrt{2} D: sqrt{2} + 1
answer:Let the equation of the line be (x - y + c = 0). The distance between the center of the circle (2, 1) and the line (x - y + c = 0) is given by frac{|1 + c|}{sqrt{2}} = 1. Solving for c, we get c = -1 pm sqrt{2}. Now, we need to find the shortest distance of translation. This will be the distance between the original line and the translated line, which is frac{|2 - sqrt{2}|}{sqrt{2}} = sqrt{2} - 1. So, the correct answer is A: boxed{sqrt{2} - 1}. To solve this problem, we first set up the equation of the line, then find the distance between the center of the circle and the line using the distance formula. After obtaining the values for c, we calculate the shortest distance of translation. This problem tests the understanding of the relationship between a line and a circle, the distance formula between a point and a line, and the distance between two lines. It primarily assesses computational skills and is relatively basic.
question:The difference between the minimum value of the sum of the squares of ten different odd numbers and the remainder of this minimum value divided by 4 is qquad. (Note: The product of a natural number with itself is called the square of the number, such as 1 times 1 = 1^2, 2 times 2 = 2^2, 3 times 3 = 3^3, and so on).
answer:1. Let's first identify the smallest sum of the squares of ten different odd numbers. The smallest ten different odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19. 2. We need to calculate the sum of the squares of these numbers. The equation for this sum is: [ sum_{i=1}^{10} (2i-1)^2 = 1^2 + 3^2 + 5^2 + 7^2 + 9^2 + 11^2 + 13^2 + 15^2 + 17^2 + 19^2 ] 3. Next, we calculate each square and then find the total sum: [ begin{array}{ll} 1^2 & = 1, 3^2 & = 9, 5^2 & = 25, 7^2 & = 49, 9^2 & = 81, 11^2 & = 121, 13^2 & = 169, 15^2 & = 225, 17^2 & = 289, 19^2 & = 361. end{array} ] 4. Adding these values together, we get: [ 1 + 9 + 25 + 49 + 81 + 121 + 169 + 225 + 289 + 361 = 1330 ] 5. Now, we find the remainder when 1330 is divided by 4. We perform the division: [ 1330 div 4 = 332 text{ with a remainder of } 2 ] 6. Thus, the remainder is 2. We subtract this remainder from the initial sum: [ 1330 - 2 = 1328 ] # Conclusion: [ boxed{1328} ]