Let's break down the last problem (the balloon radius) into **two steps**: --- ## **Step 1: Set up the equation for the radius** Given: - \( r(V) = \sqrt[3]{\frac{V}{4\pi}} \) - \( V(t) = 10 \cdot 20t = 200t \) So, \[ r(V(t)) = \sqrt[3]{\frac{200t}{4\pi}} \] We want the radius to reach 10 inches: \[ 10 = \sqrt[3]{\frac{200t}{4\pi}} \] **Explanation:** Here, we substituted the volume function into the radius formula and set it equal to 10 to find when the radius reaches 10 inches. --- ## **Step 2: Solve for \( t \)** Cube both sides to remove the cube root: \[ 10^3 = \frac{200t}{4\pi} \] \[ 100 = \frac{200t}{4\pi} \] Multiply both sides by \( 4\pi \): \[ 100 \times 4\pi = 200t \] \[ 400\pi = 200t \] Divide both sides by 200: \[ t = \frac{400\pi}{200} \] \[ t = 20\pi \] **Explanation:** We solved for \( t \) by algebraic manipulation: cubing both sides and isolating \( t \). --- ## **Final Answer** **The exact time when the radius reaches 10 inches is:** \[ \boxed{20\pi} \] (seconds)

Let's solve the entire problem step by step. --- ## **Part 1: Finding the Discounted Price** ### **Step 1: Set up the equation for price after discounts** Given: - Original price: \( x \) - Discounts: 30% and 15% The discounted price \( P(x) \) can be calculated as: \[ P(x) = (1 - 0.30)(1 - 0.15)x \] This simplifies to: \[ P(x) = (0.70)(0.85)x \] **Explanation:** Here, we first convert the percentage discounts into decimal form (30% is 0.30 and 15% is 0.15) and calculate the remaining price after applying both discounts. --- ### **Step 2: Calculate the final price** Calculating the final price: \[ P(x) = 0.70 \times 0.85 \times x = 0.595x \] **Explanation:** We multiplied the factors to find that the final price after both discounts is 59.5% of the original price. --- ## **Final Answer for Part 1** The final price after applying the discounts is: \[ \boxed{0.595x} \] --- ## **Part 2: Area of the Ripple as a Function of Time** ### **Step 1: Set up the equation for the area of the ripple** Given: - Radius as a function of time: \( r(t) = 25\sqrt{t} \) - Area formula: \( A = \pi r^2 \) Substituting \( r(t) \): \[ A(t) = \pi (25\sqrt{t})^2 = \pi \cdot 625t = 625\pi t \] **Explanation:** We used the area formula and substituted the radius function to express the area as a function of time. --- ### **Step 2: Calculate the area at \( t = 2 \)** Plugging \( t = 2 \) into the area function: \[ A(2) = 625\pi \cdot 2 = 1250\pi \] **Explanation:** We evaluated the area function at \( t = 2 \) to find the area of the ripple at that specific time. --- ## **Final Answer for Part 2** The area of the ripple at \( t = 2 \) is: \[ \boxed{1250\pi} \text{ square inches} \] --- ## **Part 3: Finding Time for Balloon Radius to Reach 10 Inches** ### **Step 1: Set up the equation for the radius** Given: - Volume function: \( V(t) = 200t \) - Radius formula: \( r(V) = \sqrt[3]{\frac{V}{4\pi}} \) Substituting the volume into the radius formula: \[ r(V(t)) = \sqrt[3]{\frac{200t}{4\pi}} \] Setting the radius to 10: \[ 10 = \sqrt[3]{\frac{200t}{4\pi}} \] **Explanation:** We expressed the radius in terms of time and set it equal to 10 inches. --- ### **Step 2: Solve for \( t \)** Cubing both sides: \[ 10^3 = \frac{200t}{4\pi} \implies 1000 = \frac{200t}{4\pi} \] Multiply both sides by \( 4\pi \): \[ 1000 \cdot 4\pi = 200t \implies 4000\pi = 200t \] Divide both sides by 200: \[ t = \frac{4000\pi}{200} = 20\pi \] **Explanation:** We manipulated the equation algebraically to isolate \( t \). --- ## **Final Answer for Part 3** The exact time when the radius reaches 10 inches is: \[ \boxed{20\pi} \text{ seconds} \]