This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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QUESTION 3. Write the following as product of primes raised to powers. a) 200 Step 1: Find the prime factorization of 200. 200 = 2 × 100 100 = 2 × 50 50 = 2 × 25 25 = 5 × 5 Step 2: Combine the prime factors. 200 = 2 × 2 × 2 × 5 × 5 = 2^3 × 5^2 The final answer is 2^3 × 5^2. b) 2(1)/(4) Step 1: Convert the mixed number to an improper fraction. 2(1)/(4) = (2 × 4 + 1)/(4) = (9)/(4) Step 2: Find the prime factorization of the numerator and the denominator. 9 = 3^2 4 = 2^2 Step 3: Write the fraction as a product of primes raised to powers. (9)/(4) = (3^2)/(2^2) = 3^2 × 2^-2 The final answer is 3^2 × 2^-2. QUESTION 4. Write the following as simply as possible without using negative indices and as a single fraction. a) (2x^-3)^3 y^-2x Step 1: Apply the power rule (ab)^n = a^n b^n and (a^m)^n = a^mn to the numerator. (2x^-3)^3 = 2^3 (x^-3)^3 = 8x^-9 Step 2: Substitute this back into the expression. 8x^-9 y^-2x Step 3: Simplify using the division rule a^m ÷ a^n = a^m-n. 8x^-9 y^-2x^1 = 8x^-9-1 y^-2 = 8x^-10 y^-2 Step 4: Rewrite the expression without negative indices and as a single fraction using a^-n = (1)/(a^n). 8x^-10 y^-2 = (8)/(x^10)y^2 The final answer is (8)/(x^10)y^2. b) x(2^-1y + y^-1x^-1) + ((xy)^2)/(x^2) Step 1: Simplify the terms with negative exponents inside the parenthesis. 2^-1 = (1)/(2) y^-1 = (1)/(y) x^-1 = (1)/(x) So, the expression inside the parenthesis becomes (1)/(2)y + (1)/(y)(1)/(x) = (y)/(2) + (1)/(xy). Step 2: Distribute x into the parenthesis. x((y)/(2) + (1)/(xy)) = x · (y)/(2) + x · (1)/(xy) = (xy)/(2) + (1)/(y) Step 3: Simplify the second part of the expression. ((xy)^2)/(x^2) = (x^2 y^2)/(x^2) = y^2 Step 4: Combine the simplified parts. (xy)/(2) + (1)/(y) + y^2 Step 5: Express as a single fraction by finding a common denominator, which is 2y. (xy · y)/(2 · y) + (1 · 2)/(y · 2) + (y^2 · 2y)/(1 · 2y) = (xy^2)/(2y) + (2)/(2y) + (2y^3)/(2y) Step 6: Add the fractions. (xy^2 + 2 + 2y^3)/(2y) The final answer is (xy^2 + 2y^3 + 2)/(2y). 3 done, 2 left today. You're making progress.