Another day — let's solve it. 1. The volume of a spherical ball is $5,000 \text{ cm}^3$. What is the radius of the ball correct to one decimal place? (Use $\pi = 3.14$) Step 1: Write the formula for the volume of a sphere. $$V = \frac{4}{3}\pi r^3$$ Step 2: Substitute the given values into the formula. $$5000 = \frac{4}{3} \times 3.14 \times r^3$$ Step 3: Isolate $r^3$. $$r^3 = \frac{5000 \times 3}{4 \times 3.14}$$ $$r^3 = \frac{15000}{12.56}$$ $$r^3 \approx 1194.2675$$ Step 4: Calculate $r$ by taking the cube root. $$r = \sqrt[3]{1194.2675}$$ $$r \approx 10.610 \text{ cm}$$ Step 5: Round the radius to one decimal place. $$r \approx 10.6 \text{ cm}$$ The radius of the ball is $\boxed{\text{10.6 cm}}$. 2. A cone has a diameter of $70 \text{ mm}$ and a height of $100 \text{ mm}$. What is its volume to the nearest $\text{mm}^3$? (Use $\pi = 3.14$) Step 1: Calculate the radius from the given diameter. $$r = \frac{\text{diameter}}{2} = \frac{70 \text{ mm}}{2} = 35 \text{ mm}$$ Step 2: Write the formula for the volume of a cone. $$V = \frac{1}{3}\pi r^2 h$$ Step 3: Substitute the given values into the formula. $$V = \frac{1}{3} \times 3.14 \times (35 \text{ mm})^2 \times 100 \text{ mm}$$ $$V = \frac{1}{3} \times 3.14 \times 1225 \text{ mm}^2 \times 100 \text{ mm}$$ $$V = \frac{1}{3} \times 384650 \text{ mm}^3$$ $$V \approx 128216.666... \text{ mm}^3$$ Step 4: Round the volume to the nearest $\text{mm}^3$. $$V \approx 128217 \text{ mm}^3$$ The volume of the cone is $\boxed{\text{128217 mm}^3}$. 3. Find the surface area of a box with dimensions $125 \text{ mm} \times 64 \text{ mm} \times 84 \text{ mm}$ to the nearest $\text{mm}^2$. Step 1: Identify the dimensions of the box. Length $l = 125 \text{ mm}$ Width $w = 64 \text{ mm}$ Height $h = 84 \text{ mm}$ Step 2: Write the formula for the surface area of a rectangular box (cuboid). $$SA = 2(lw + lh + wh)$$ Step 3: Substitute the dimensions into the formula. $$SA = 2((125 \text{ mm} \times 64 \text{ mm}) + (125 \text{ mm} \times 84 \text{ mm}) + (64 \text{ mm} \times 84 \text{ mm}))$$ $$SA = 2(8000 \text{ mm}^2 + 10500 \text{ mm}^2 + 5376 \text{ mm}^2)$$ $$SA = 2(23876 \text{ mm}^2)$$ $$SA = 47752 \text{ mm}^2$$ Step 4: The surface area is already an integer, so no rounding is needed to the nearest $\text{mm}^2$. The surface area of the box is $\boxed{\text{47752 mm}^2}$. Send me the next one 📸