💡First, I recognize that the ages of Ashley and Mary are given in the ratio 4:7. This means that for every 4 years of Ashley's age, Mary is 7 years old.Since Ashley's age is provided as 8 years, I can set up a proportion to find Mary's age. By cross-multiplying, I can solve for Mary's age.Once I have both ages, I simply add them together to find the total sum of their ages.

💡Okay, so I have this problem: "How many positive integers less than 500 are divisible by 2, 3, 5, and 7?" Hmm, let's break this down step by step. First, I know that when a number is divisible by multiple numbers, it means that number is a multiple of each of those divisors. So, if a number is divisible by 2, 3, 5, and 7, it must be a multiple of all these numbers. Now, to find such numbers, I think I need to find the least common multiple (LCM) of these numbers. The LCM of a set of numbers is the smallest number that is a multiple of each of them. Since 2, 3, 5, and 7 are all prime numbers, their LCM should just be their product, right? Let me check that.Calculating the LCM:- 2 is a prime number.- 3 is a prime number.- 5 is a prime number.- 7 is a prime number.Since they're all primes, there are no common factors other than 1. So, the LCM is simply 2 × 3 × 5 × 7. Let me compute that:2 × 3 = 6,6 × 5 = 30,30 × 7 = 210.So, the LCM is 210. That means 210 is the smallest number that is divisible by 2, 3, 5, and 7. Now, any multiple of 210 will also be divisible by these numbers. So, the numbers we're looking for are multiples of 210.Next, I need to find how many multiples of 210 are less than 500. To do this, I can divide 500 by 210 and see how many whole times 210 fits into 500.Let's do that division:500 ÷ 210 ≈ 2.38095.Since we're looking for whole numbers, we take the integer part of this result, which is 2. This means there are 2 multiples of 210 less than 500.To double-check, let's list the multiples of 210:- 210 × 1 = 210,- 210 × 2 = 420,- 210 × 3 = 630.Wait, 630 is greater than 500, so we stop at 420. So, the multiples are 210 and 420, which are both less than 500. That confirms there are 2 such numbers.I think that's it. So, the answer should be 2.

💡First, I need to calculate the values of (2^8) and (5^5).Calculating (2^8):[2^8 = 256]Calculating (5^5):[5^5 = 3125]Next, I add these two results together:[256 + 3125 = 3381]Now, I need to find the prime factors of 3381. I'll start by checking divisibility by smaller prime numbers.Checking divisibility by 3:[3381 div 3 = 1127]So, 3 is a prime factor.Next, I check if 1127 is divisible by 3:[1127 div 3 approx 375.666...]It is not divisible by 3.Checking divisibility by 7:[1127 div 7 = 161]So, 7 is another prime factor.Now, I check if 161 is divisible by 7:[161 div 7 = 23]Thus, 7 is a prime factor again.Finally, 23 is a prime number.Therefore, the prime factors of 3381 are 3, 7, 7, and 23. The greatest prime factor is 23.

💡First, I need to identify the coordinates of the parallelogram's vertices: (0, 0), (4, 0), (1, 5), and (5, 5).Next, I'll determine the vectors that define the sides of the parallelogram. The vector from (0, 0) to (4, 0) is (4, 0), and the vector from (0, 0) to (1, 5) is (1, 5).To find the area of the parallelogram, I can use the cross product of these two vectors. The cross product of (4, 0) and (1, 5) is calculated as follows:4 * 5 - 0 * 1 = 20 - 0 = 20The absolute value of this cross product gives the area of the parallelogram.Therefore, the area of the parallelogram is 20 square units.

💡To find the algebraic expression for a number that is 3 more than twice ( x ), follow these steps:1. **Identify the operation "twice ( x )":** This means multiplying ( x ) by 2, which gives ( 2x ).2. **Add 3 to this result:** To get "3 more than twice ( x )", add 3 to ( 2x ), resulting in ( 2x + 3 ).Now, compare this expression with the given options:- **A:** ( frac{1}{2}x - 3 )- **B:** ( frac{1}{2}x + 3 )- **C:** ( 2x - 3 )- **D:** ( 2x + 3 )The correct answer is **D:** ( 2x + 3 ).

💡First, I recognize that the number of bacteria triples every day, starting with 5 bacteria. This can be modeled by the formula ( 5 cdot 3^n ), where ( n ) represents the number of days.To determine when the colony will exceed 200 bacteria, I set up the inequality:[ 5 cdot 3^n > 200 ]Next, I simplify the inequality by dividing both sides by 5:[ 3^n > 40 ]Now, I need to find the smallest integer ( n ) that satisfies this inequality. I'll calculate the values of ( 3^n ) for successive values of ( n ):- ( 3^0 = 1 )- ( 3^1 = 3 )- ( 3^2 = 9 )- ( 3^3 = 27 )- ( 3^4 = 81 )Since ( 81 ) is greater than ( 40 ), the smallest value of ( n ) that meets the condition is ( 4 ).Therefore, it will take 4 days for the bacterial colony to exceed 200 bacteria.