Work out. 1 1/5 + 2 1/3 m = 2 x 3^a x 7^b and n = 3 x 72 x 11 Given that m > 126, find the smallest values of a and b which make m a multiple of 126. Explain why 33n is a perfect square.

Here are the solutions to the problems: 1. Work out 1(1)/(5) + 2(1)/(3) Step 1: Convert the mixed numbers to improper fractions. 1(1)/(5) = (1 × 5 + 1)/(5) = (6)/(5) 2(1)/(3) = (2 × 3 + 1)/(3) = (7)/(3) Step 2: Find a common denominator for the fractions, which is 15. (6)/(5) = (6 × 3)/(5 × 3) = (18)/(15) (7)/(3) = (7 × 5)/(3 × 5) = (35)/(15) Step 3: Add the fractions. (18)/(15) + (35)/(15) = (18 + 35)/(15) = (53)/(15) Step 4: Convert the improper fraction back to a mixed number. (53)/(15) = 3(8)/(15) The final answer is 3(8)/(15). 2. Given m = 2 × 3^a × 7^b and n = 3 × 7^2 × 11 a) Given that m > 126, find the smallest values of a and b which make m a multiple of 126. Step 1: Find the prime factorization of 126. 126 = 2 × 63 = 2 × 9 × 7 = 2^1 × 3^2 × 7^1 Step 2: For m = 2 × 3^a × 7^b to be a multiple of 126 = 2^1 × 3^2 × 7^1, the exponents of its prime factors must be greater than or equal to those of 126. For the prime factor 2: 2^1 in m matches 2^1 in 126. For the prime factor 3: 3^a in m means a 2. For the prime factor 7: 7^b in m means b 1. Step 3: Consider the condition m > 126. If we take the smallest possible values from Step 2 (a=2, b=1), then m = 2 × 3^2 × 7^1 = 2 × 9 × 7 = 126. This value of m is not strictly greater than 126, so we need to increase m. To do this, we must increase either a or b by at least 1. Step 4: Find the smallest a and b that satisfy a 2, b 1, and m > 126. Let's test the smallest possible value for a, which is a=2. If a=2, then m = 2 × 3^2 × 7^b = 18 × 7^b. We need 18 × 7^b > 126. Divide by 18: 7^b > (126)/(18) = 7. So, 7^b > 7^1, which implies b > 1. Since b must be an integer and b 1, the smallest integer value for b that satisfies b > 1 is b=2. Therefore, the smallest values are a=2 and b=2. Let's check this: m = 2 × 3^2 × 7^2 = 2 × 9 × 49 = 18 × 49 = 882. 882 is a multiple of 126 (882 = 7 × 126) and 882 > 126. The final answer is a=2, b=2. b) Explain why 33n is a perfect square. Step 1: Substitute the expression for n into 33n. Given n = 3 × 7^2 × 11. 33n = 33 × (3 × 7^2 × 11) Step 2: Express 33 as a product of its prime factors. 33 = 3 × 11 Step 3: Substitute the prime factors of 33 into the expression for 33n and combine like terms. 33n = (3 × 11) × (3 × 7^2 × 11) 33n = 3^1 × 11^1 × 3^1 × 7^2 × 11^1 33n = 3^1+1 × 7^2 × 11^1+1 33n = 3^2 × 7^2 × 11^2 Step 4: Explain why the result is a perfect square. A number is a perfect square if all the exponents in its prime factorization are even. In the expression 3^2 × 7^2 × 11^2, all the exponents (2, 2, and 2) are even. Therefore, 33n is a perfect square. It can also be written as (3 × 7 × 11)^2 = (231)^2. The final answer is Because the prime factorization of 33n is 3^2 × 7^2 × 11^2, where all prime factors have even exponents. Send me the next one 📸