Here are the solutions to the problems: 1. Solve the equation $2x^2 - 13x + 20 = 0$. Step 1: Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a=2$, $b=-13$, $c=20$. $$x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(2)(20)}}{2(2)}$$ Step 2: Simplify the expression under the square root. $$x = \frac{13 \pm \sqrt{169 - 160}}{4}$$ $$x = \frac{13 \pm \sqrt{9}}{4}$$ Step 3: Calculate the square root. $$x = \frac{13 \pm 3}{4}$$ Step 4: Find the two possible values for $x$. $$x_1 = \frac{13 + 3}{4} = \frac{16}{4} = 4$$ $$x_2 = \frac{13 - 3}{4} = \frac{10}{4} = \frac{5}{2}$$ The solutions are: $$\boxed{x = 4 \text{ or } x = \frac{5}{2}}$$ 2. A delivery company models fuel consumption using the rule: $F = \frac{64a^7b^9}{8a^3b^3}$. Simplify using index laws. Step 1: Separate the numerical coefficients and the variables. $$F = \left(\frac{64}{8}\right) \times \left(\frac{a^7}{a^3}\right) \times \left(\frac{b^9}{b^3}\right)$$ Step 2: Simplify the numerical coefficient and apply the index law $\frac{x^m}{x^n} = x^{m-n}$ for the variables. $$F = 8 \times a^{7-3} \times b^{9-3}$$ Step 3: Perform the subtractions in the exponents. $$F = \boxed{8a^4b^6}$$ 3. If $a=2$ and $b=3$, find the numerical value of the simplified expression from Problem 2. Step 1: Substitute $a=2$ and $b=3$ into the simplified expression $F = 8a^4b^6$. $$F = 8(2)^4(3)^6$$ Step 2: Calculate the powers. $$F = 8(16)(729)$$ Step 3: Multiply the values. $$F = 128 \times 729$$ $$F = \boxed{93312}$$ 4. A student plans to fence a rectangular kitchen garden whose area is given by $A = x^2 - 4x - 21$. Factorize. Step 1: Find two numbers that multiply to -21 and add up to -4. These numbers are 3 and -7. Step 2: Rewrite the quadratic expression using these numbers. $$A = x^2 + 3x - 7x - 21$$ Step 3: Factor by grouping. $$A = x(x + 3) - 7(x + 3)$$ Step 4: Factor out the common binomial factor. $$A = \boxed{(x + 3)(x - 7)}$$ Got more? Send 'em!