1. **Problem 7: Solve for all values of $x$ to the nearest hundredth:** Given equation: $$\frac{x^2 - 3x + 2}{x + 3} = 5(x + 3)$$ 2. **Step 1: Multiply both sides by $x + 3$ to eliminate the denominator:** $$x^2 - 3x + 2 = 5(x + 3)^2$$ 3. **Step 2: Expand the right side:** $$(x + 3)^2 = x^2 + 6x + 9$$ So, $$x^2 - 3x + 2 = 5(x^2 + 6x + 9)$$ 4. **Step 3: Distribute 5:** $$x^2 - 3x + 2 = 5x^2 + 30x + 45$$ 5. **Step 4: Bring all terms to one side:** $$x^2 - 3x + 2 - 5x^2 - 30x - 45 = 0$$ Simplify: $$-4x^2 - 33x - 43 = 0$$ 6. **Step 5: Multiply both sides by $-1$ to make the leading coefficient positive:** $$4x^2 + 33x + 43 = 0$$ 7. **Step 6: Use the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=4$, $b=33$, $c=43$. Calculate discriminant: $$b^2 - 4ac = 33^2 - 4(4)(43) = 1089 - 688 = 401$$ 8. **Step 7: Calculate roots:** $$x = \frac{-33 \pm \sqrt{401}}{8}$$ $$\sqrt{401} \approx 20.024984$$ So, $$x_1 = \frac{-33 + 20.024984}{8} = \frac{-12.975016}{8} = -1.62$$ $$x_2 = \frac{-33 - 20.024984}{8} = \frac{-53.024984}{8} = -6.63$$ --- 2. **Problem 8: Find the average rate of change over the interval $2 \leq t \leq 8$** Given points: At $t=2$, distance = 110 miles At $t=8$, distance = 350 miles Average rate of change formula: $$\text{Average rate} = \frac{\text{Change in distance}}{\text{Change in time}} = \frac{f(8) - f(2)}{8 - 2}$$ Calculate: $$= \frac{350 - 110}{6} = \frac{240}{6} = 40$$ So, the average rate of change is 40 miles per hour. --- 3. **Problem 9: Solve for all values of $x$:** Given: $$3x^3 - 2x^2 - 27x + 18 = 0$$ 4. **Step 1: Group terms:** $$(3x^3 - 2x^2) - (27x - 18) = 0$$ 5. **Step 2: Factor each group:** $$x^2(3x - 2) - 9(3x - 2) = 0$$ 6. **Step 3: Factor out common binomial:** $$(x^2 - 9)(3x - 2) = 0$$ 7. **Step 4: Factor difference of squares:** $$(x + 3)(x - 3)(3x - 2) = 0$$ 8. **Step 5: Solve each factor:** $$x + 3 = 0 \Rightarrow x = -3$$ $$x - 3 = 0 \Rightarrow x = 3$$ $$3x - 2 = 0 \Rightarrow x = \frac{2}{3}$$ --- 4. **Problem 10: Solve for all values of $x$:** Given: $$x^4 + 6x^2 - 27 = 0$$ 5. **Step 1: Substitute $y = x^2$ to reduce degree:** $$y^2 + 6y - 27 = 0$$ 6. **Step 2: Use quadratic formula for $y$:** $$y = \frac{-6 \pm \sqrt{6^2 - 4(1)(-27)}}{2(1)} = \frac{-6 \pm \sqrt{36 + 108}}{2} = \frac{-6 \pm \sqrt{144}}{2}$$ $$= \frac{-6 \pm 12}{2}$$ 7. **Step 3: Calculate $y$ values:** $$y_1 = \frac{-6 + 12}{2} = 3$$ $$y_2 = \frac{-6 - 12}{2} = -9$$ 8. **Step 4: Recall $y = x^2$, so solve:** For $y_1 = 3$: $$x^2 = 3 \Rightarrow x = \pm \sqrt{3} \approx \pm 1.73$$ For $y_2 = -9$: $$x^2 = -9$$ No real solutions since square of real number cannot be negative. **Final solutions:** $$x = \pm 1.73$$