Simplify sqrt(12) + sqrt(3), determine if the result is rational or irrational, and find two consecutive integers it lies between.

Morning 🥹❤️‍🔥M💞🥹Miseketeri — let's get this done. Here's the solution for Question 1: Question 1.1: Determine the value of A rounded off to TWO decimal places. A = sqrt(12) + sqrt(3) Step 1: Simplify the expression for A. A = sqrt(4 × 3) + sqrt(3) A = 2sqrt(3) + sqrt(3) A = 3sqrt(3) Step 2: Calculate the numerical value and round to two decimal places. Using the approximate value sqrt(3) ≈ 1.73205: A ≈ 3 × 1.73205 A ≈ 5.19615 Rounding to two decimal places: A ≈ 5.20 Question 1.2: State whether the value of A is rational or irrational. Justify your answer. Step 1: Recall the simplified form of A. A = 3sqrt(3) Step 2: Determine if A is rational or irrational. A number is rational if it can be expressed as a fraction (p)/(q) where p and q are integers and q ≠ 0. A number is irrational if it cannot be expressed in this form. Since sqrt(3) is an irrational number, and the product of a non-zero rational number (3) and an irrational number (sqrt(3)) is always irrational, A is an irrational number. The value of A is irrational. Justification: sqrt(3) is an irrational number, and the product of a non-zero rational number (3) and an irrational number (sqrt(3)) is always irrational. Question 1.3: Without using a calculator, determine between which two consecutive integers the value of A lies. Step 1: Use the simplified form of A. A = 3sqrt(3) Step 2: Square the value of A to compare it with perfect squares. A^2 = (3sqrt(3))^2 A^2 = 3^2 × (sqrt(3))^2 A^2 = 9 × 3 A^2 = 27 Step 3: Find two consecutive perfect squares that 27 lies between. We know that 5^2 = 25 and 6^2 = 36. So, 25 < 27 < 36. Step 4: Take the square root of all parts of the inequality. sqrt(25) < sqrt(27) < sqrt(36) 5 < sqrt(27) < 6 Since A = sqrt(27), the value of A lies between 5 and 6. The value of A lies between the consecutive integers 5 and 6. Drop the next question.