Solve the following equations: Q.5: 3x + 4 = 19 - 2x; Q.6: Find the zeroes of 2x2 - 7x + 3; Q.7: Find the hypotenuse of a right-angled triangle with legs 6cm and 8cm.

Fresh day BILIJITIN, let's solve. Q.5 Step 1: Gather terms involving x on one side and constant terms on the other side. The given equation is 3x + 4 = 19 - 2x. Add 2x to both sides: 3x + 2x + 4 = 19 5x + 4 = 19 Subtract 4 from both sides: 5x = 19 - 4 5x = 15 Step 2: Solve for x. Divide both sides by 5: x = (15)/(5) x = 3 The final answer is x=3. Q.6 Step 1: Set the quadratic polynomial to zero to find its zeroes. The polynomial is 2x^2 - 7x + 3. 2x^2 - 7x + 3 = 0 Step 2: Factor the quadratic expression. We look for two numbers that multiply to 2 × 3 = 6 and add up to -7. These numbers are -1 and -6. Rewrite the middle term: 2x^2 - 6x - x + 3 = 0 Factor by grouping: 2x(x - 3) - 1(x - 3) = 0 (2x - 1)(x - 3) = 0 Step 3: Set each factor to zero and solve for x. 2x - 1 = 0 2x = 1 x = (1)/(2) x - 3 = 0 x = 3 The zeroes of the polynomial are (1)/(2) and 3. The final answer is x=(1)/(2), x=3. Q.7 Step 1: Identify the given information for the right-angled triangle. In ABC, it is right-angled at B. The lengths of the two legs are AB = 6 cm and BC = 8 cm. We need to find the length of the hypotenuse AC. Step 2: Apply the Pythagorean theorem. For a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides (legs). AC^2 = AB^2 + BC^2 Step 3: Substitute the given values and calculate AC. AC^2 = (6 cm)^2 + (8 cm)^2 AC^2 = 36 cm^2 + 64 cm^2 AC^2 = 100 cm^2 Take the square root of both sides: AC = sqrt(100 cm)^2 AC = 10 cm The final answer is AC=10 cm. Q.8 Step 1: Recall the standard trigonometric value for 45^. The tangent of an angle in a right-angled triangle is the ratio of the length of the opposite side to the length of the adjacent side. For 45^, the opposite and adjacent sides are equal. 45^ = 1 The final answer is 45^ = 1. What's next?