Simplify the following expressions: 5d) 3/(2x3) - sqrt((8x2y3)/(2y)); 5f) 3/(2x3) - sqrt((32y8)/(9y2)) + sqrt((98y6)/(2))

You're on a roll — here are the solutions for questions 5d and 5f, simplified for a Grade 9 learner. 5d) The expression is: 3/(2x^3) - sqrt((8x^2y^3)/(2y)) Step 1: Simplify the fractions inside each radical. For the first term: (54x^6y^3)/(2x^3) = (54)/(2) · (x^6)/(x^3) · y^3 = 27x^6-3y^3 = 27x^3y^3 For the second term: (8x^2y^3)/(2y) = (8)/(2) · x^2 · (y^3)/(y) = 4x^2y^3-1 = 4x^2y^2 Now the expression looks like: [3]27x^3y^3 - sqrt(4x^2y^2) Step 2: Take the cube root and the square root. For the first term, [3]27x^3y^3: [3]27 · [3]x^3 · [3]y^3 = 3 · x · y = 3xy For the second term, sqrt(4x^2y^2): sqrt(4) · sqrt(x^2) · sqrt(y^2) = 2 · x · y = 2xy Step 3: Combine the simplified terms. 3xy - 2xy = xy The final answer is xy. 5f) The expression is: 3/(2x^3) - sqrt((32y^8)/(9y^2)) + sqrt((98y^6)/(2)) Step 1: Simplify the fractions inside each radical. For the first term: (16x^3y^9)/(2x^3) = (16)/(2) · (x^3)/(x^3) · y^9 = 8y^9 For the second term: (32y^8)/(9y^2) = (32)/(9) · (y^8)/(y^2) = (32y^6)/(9) For the third term: (98y^6)/(2) = (98)/(2) · y^6 = 49y^6 Now the expression looks like: [3]8y^9 - sqrt((32y^6)/(9)) + sqrt(49y^6) Step 2: Take the cube root and square roots. For the first term, [3]8y^9: [3]8 · [3]y^9 = 2 · y^9/3 = 2y^3 For the second term, sqrt((32y^6)/(9)): sqrt(32y^6)sqrt(9) = sqrt(16 · 2 · y^6)3 = sqrt(16) · sqrt(2) · sqrt(y^6)3 = 4sqrt(2)y^33 For the third term, sqrt(49y^6): sqrt(49) · sqrt(y^6) = 7 · y^6/2 = 7y^3 Step 3: Combine the simplified terms. 2y^3 - 4sqrt(2)y^33 + 7y^3 Combine the terms with y^3: (2y^3 + 7y^3) - 4sqrt(2)y^33 = 9y^3 - 4sqrt(2)y^33 To combine these, find a common denominator for the coefficients: (9 · 3)/(3)y^3 - 4sqrt(2)3y^3 = (27)/(3)y^3 - 4sqrt(2)3y^3 = 27 - 4sqrt(2)3y^3 The final answer is y^3(27 - 4sqrt(2))3. What's next?