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To help with these math problems, let's go through them one by one: 1. Solve the equations: a. \( 14x^2 - 9x = 0 \) Factor out the common term: \[ x(14x - 9) = 0 \] Solutions: \[ x = 0 \quad \text{or} \quad 14x - 9 = 0 \implies x = \frac{9}{14} \] b. \( 2x^2 - 11x + 12 = 0 \) Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 2 \), \( b = -11 \), and \( c = 12 \). Calculate the discriminant: \[ b^2 - 4ac = (-11)^2 - 4 \cdot 2 \cdot 12 = 121 - 96 = 25 \] Solutions: \[ x = \frac{11 \pm \sqrt{25}}{4} = \frac{11 \pm 5}{4} \] \[ x = \frac{16}{4} = 4 \quad \text{or} \quad x = \frac{6}{4} = \frac{3}{2} \] c. \( 16x^2 - 49 = 0 \) Recognize as a difference of squares: \[ (4x - 7)(4x + 7) = 0 \] Solutions: \[ 4x - 7 = 0 \implies x = \frac{7}{4} \quad \text{or} \quad 4x + 7 = 0 \implies x = -\frac{7}{4} \] 2. One of the two natural numbers is 5 greater than 3. Find these numbers. Since one number is 5 greater than 3, it is: \[ 3 + 5 = 8 \] The numbers are 3 and 8. 3. Find the roots of the quadratic equation: \( x^2 + bx + 20 = 0 \), where one root is \( \text{-4} \). Use Vieta's formulas: Sum of roots is \(-b = x_1 + x_2 = -4 + x_2\). Product of roots is \(20 = x_1 \cdot x_2 = -4 \cdot x_2\). Solving for \(x_2\): \[ -4 \cdot x_2 = 20 \implies x_2 = -5 \] Thus, roots are \(-4\) and \(-5\) and \(b = -(x_1 + x_2) = 9\). 4. A rectangular piece of cardboard that is 40 cm longer than it is wide is open at the top. Its volume is 1360 cm³. Find the width and the length. Let the width be \( x \) cm. Then the length will be \( x + 40 \) cm. The height (when the box is open at the top) is represented in terms of the base. Volume of the box is given by: \[ x(x + 40)h = 1360 \] Solve the equation for width \( x \). Once the width is known, the length can be found using \( x + 40 \). Since no specific details about solving the last equation are provided in terms of manipulability, further algebraic manipulations or constraints will be needed based on the problem's context, or usage of approximation/numeric solvers for exact dimensions.