answer:- By the Pythagorean theorem, for the triangle to also be a right triangle, we should first calculate whether 8 and 15 could form legs. 8^2 + 15^2 = 64 + 225 = 289 which is 17^2, so if the triangle is right-angled, the hypotenuse must be 17 when 8 and 15 are legs. - For different configurations using the sides 8 and 15 where a new side, say x, makes the triangle still a right triangle, we apply the Triangle Inequality: - The sum of any two sides must be greater than the third: - x + 8 > 15 implies x > 7 - x + 15 > 8 implies x > -7 (this condition is automatically satisfied since x > 7) - 8 + 15 > x implies x < 23 - So x must be more than 7 and less than 23. - The only additional candidate values for x to maintain a right triangle besides 17 (which is already found) are those where the new triangle formed with 8, 15, and x adheres to the Pythagorean Theorem. The next step is to test integer values in this range: - Test for x = 9 to 22; none of these will satisfy the Pythagorean theorem with 8 and 15 except for the already identified hypotenuse, 17. Consequently, the only integer side length that can complete a right triangle with sides 8 and 15 is boxed{17}.

answer:Let's denote the total capacity of the pool as ( C ) gallons. We are told that adding 300 gallons of water will fill the pool to 75% of its capacity. This means that before adding the 300 gallons, the pool was filled to ( 75% - 30% = 45% ) of its capacity. So, the amount of water that was initially in the pool is ( 45% ) of ( C ), which can be written as ( 0.45C ). Adding 300 gallons increases the amount of water in the pool to ( 75% ) of its capacity, which can be written as ( 0.75C ). Therefore, the equation to represent this situation is: [ 0.45C + 300 = 0.75C ] Now, let's solve for ( C ): [ 0.75C - 0.45C = 300 ] [ 0.30C = 300 ] [ C = frac{300}{0.30} ] [ C = 1000 ] The total capacity of the pool is boxed{1000} gallons.

answer:To solve this problem, let's follow the steps based on the provided solution, breaking it down for clarity: 1. **Identify the Sample Space**: When rolling a die twice, each roll can result in any number from 1 to 6. Therefore, the total number of outcomes (sample space Omega) is the product of the number of outcomes for each roll, which is 6 times 6 = 36. These outcomes can be represented as ordered pairs (x, y) where x is the result of the first roll and y is the result of the second roll. Hence, Omega consists of 36 ordered pairs. 2. **Define the Event of Interest**: We are interested in the event A where the number obtained in the first roll can be divided by the number obtained in the second roll. To find the outcomes that satisfy this condition, we need to list all pairs where the first number is a divisor of the second number. 3. **List the Favorable Outcomes**: The favorable outcomes are those where the first number divides the second number without leaving a remainder. These are: - (1,1), (2,1), (3,1), (4,1), (5,1), (6,1) because 1, 2, 3, 4, 5, and 6 all divide 1. - (2,2), (4,2), (6,2) because 2 divides 2, 4, and 6. - (3,3), (6,3) because 3 divides 3 and 6. - (4,4) because 4 divides 4. - (5,5) because 5 divides 5. - (6,6) because 6 divides 6. Counting these, we find there are 14 favorable outcomes. 4. **Calculate the Probability**: The probability of event A happening is the ratio of the number of favorable outcomes to the total number of outcomes in the sample space. This can be calculated as: [ P(A) = frac{text{Number of favorable outcomes}}{text{Total number of outcomes}} = frac{14}{36} ] Simplifying the fraction, we get: [ P(A) = frac{7}{18} ] Therefore, the probability that the number of points obtained in the first roll can be divided by the number of points obtained in the second roll is boxed{frac{7}{18}}. Hence, the correct answer is boxed{text{A}}.

answer:To find the smallest three-digit number divisible by both 4 and 5, we look for the least common multiple (LCM) of 4 and 5. The LCM of 4 and 5 is 20. The smallest three-digit number divisible by 20 is the smallest three-digit multiple of 20. - The smallest three-digit number is 100. - Check if 100 is divisible by 20: (100 div 20 = 5), which is an integer, so 100 is divisible by 20. - Therefore, the smallest three-digit number divisible by both 4 and 5 is boxed{100}.