answer:To find the month with the fewest relatively prime dates, we need to find the month with the fewest numbers that are relatively prime to that month. For each month, we count the number of days that are relatively prime to the number of the month. January has 31 days, and all the numbers from 1 to 31 are relatively prime to 1, so January has 31 relatively prime dates. February has 28 days (or 29 in a leap year), and all the odd numbers from 1 to 28 (or 29) are relatively prime to 2, so February has 14 (or 15) relatively prime dates. March has 31 days, and all the numbers from 1 to 31 are relatively prime to 3, so March has 31 relatively prime dates. April has 30 days, and the numbers relatively prime to 4 are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, and 29. So April has 15 relatively prime dates. May has 31 days, and all the numbers from 1 to 31 are relatively prime to 5, so May has 31 relatively prime dates. June has 30 days, and the numbers relatively prime to 6 are 1, 5, 7, 11, 13, 17, 19, 23, 25, and 29. So June has 10 relatively prime dates. July has 31 days, and all the numbers from 1 to 31 are relatively prime to 7, so July has 31 relatively prime dates. August has 31 days, and all the numbers from 1 to 31 are relatively prime to 8, so August has 31 relatively prime dates. September has 30 days, and the numbers relatively prime to 9 are 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, and 28. So September has 19 relatively prime dates. October has 31 days, and all the numbers from 1 to 31 are relatively prime to 10, so October has 31 relatively prime dates. November has 30 days, and the numbers relatively prime to 11 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, and 29. So November has 28 relatively prime dates. December has 31 days, and all the numbers from 1 to 31 are relatively prime to 12, so December has 31 relatively prime dates. Therefore, the month with the fewest relatively prime dates is June with 10 dates.The answer is: 10

answer:The line y = 2x + 5 has a slope of 2. A line perpendicular to it will have a slope that is the negative reciprocal of 2, which is -frac{1}{2}. So, the equation of the perpendicular line passing through (5, 5) is of the form y = -frac{1}{2}x + b. To find b, we substitute the coordinates of the point (5, 5) into the equation: 5 = -frac{1}{2}(5) + b 5 = -frac{5}{2} + b b = frac{15}{2} So, the equation of the perpendicular line is y = -frac{1}{2}x + frac{15}{2}. To find the point of intersection between the two lines, we set their equations equal to each other and solve for x: 2x + 5 = -frac{1}{2}x + frac{15}{2} 4x + 10 = -x + 15 5x = 5 x = 1 Substituting this value back into either of the original equations, we find that y = 2(1) + 5 = 7. So, the point of intersection is boxed{(1, 7)}.The answer is: (1,7)

answer:Suki bought 6.5 bags of coffee beans and each bag weighs 22 kilograms, so she bought a total of 6.5 * 22 = 143 kilograms of coffee beans. Jimmy bought 4.5 bags of coffee beans and each bag weighs 18 kilograms, so he bought a total of 4.5 * 18 = 81 kilograms of coffee beans. Together, they bought a total of 143 + 81 = 224 kilograms of coffee beans. To repackage the coffee beans into 8-kilogram containers, they would need 224 / 8 = 28 containers. Therefore, they used 28 containers to repack their coffee beans. 28 The answer is: 28

answer:We want to find the value of X in the given situation. The number of ways to draw 3 balls from X balls, in order, without replacement is given by the formula: Number of ways = X choose 3 = frac{X!}{3!(X-3)!} We are given that the number of ways is 1320, so we can write: 1320 = frac{X!}{3!(X-3)!} To solve for X, we can simplify the equation. Since 3! = 3 * 2 * 1, we can rewrite the equation as: 1320 = frac{X!}{6(X-3)!} Multiplying both sides of the equation by 6 to eliminate the fraction, we get: 7920 = X!/(X-3)! Now, we need to find the value of X that satisfies this equation. We can start by trying different values of X and calculating the factorial. Let's start with X = 15: 7920 = 15!/(15-3)! 7920 = 15!/12! 7920 = 15 * 14 * 13 7920 = 2730 The equation does not hold true for X = 15. Let's try another value, X = 12: 7920 = 12!/(12-3)! 7920 = 12!/9! 7920 = 12 * 11 * 10 7920 = 1320 The equation holds true for X = 12. Therefore, the value of X is 12. The answer is: 12