1. **State the problem:** We have the function $$x = \frac{1}{6}t^3 - 5t$$ and need to find: a) A value of $$t$$ such that $$\frac{dx}{dt} = 27$$. b) The corresponding value of $$x$$ at that $$t$$. 2. **Find the derivative:** The derivative $$\frac{dx}{dt}$$ represents the rate of change of $$x$$ with respect to $$t$$. Using the power rule: $$\frac{dx}{dt} = \frac{d}{dt}\left(\frac{1}{6}t^3 - 5t\right) = \frac{1}{6} \times 3t^2 - 5 = \frac{3}{6}t^2 - 5 = \frac{1}{2}t^2 - 5$$ 3. **Set the derivative equal to 27 and solve for $$t$$:** $$\frac{1}{2}t^2 - 5 = 27$$ Add 5 to both sides: $$\frac{1}{2}t^2 = 32$$ Multiply both sides by 2: $$\cancel{\frac{1}{2}} \times 2 t^2 = 32 \times 2$$ $$t^2 = 64$$ Take the square root: $$t = \pm 8$$ 4. **Find the corresponding $$x$$ values for $$t = 8$$ and $$t = -8$$:** For $$t=8$$: $$x = \frac{1}{6} \times 8^3 - 5 \times 8 = \frac{1}{6} \times 512 - 40 = 85.3333 - 40 = 45.3333$$ Rounded to 1 decimal place: $$x = 45.3$$ For $$t=-8$$: $$x = \frac{1}{6} \times (-8)^3 - 5 \times (-8) = \frac{1}{6} \times (-512) + 40 = -85.3333 + 40 = -45.3333$$ Rounded to 1 decimal place: $$x = -45.3$$ **Final answers:** - Values of $$t$$ where $$\frac{dx}{dt} = 27$$ are $$t = 8$$ and $$t = -8$$. - Corresponding $$x$$ values are $$x = 45.3$$ when $$t=8$$ and $$x = -45.3$$ when $$t=-8$$.