What is the diameter of a hemisphere with a volume of \( 8582 \mathrm{~m}^{3} \), to the nearest tenth of a meter?

To find the diameter of a hemisphere given its volume, we can use the formula for the volume of a hemisphere: \[ V = \frac{2}{3} \pi r^3 \] where \( V \) is the volume and \( r \) is the radius. We can rearrange this formula to solve for the radius \( r \): \[ r^3 = \frac{3V}{2\pi} \] Once we find the radius, we can calculate the diameter \( d \) using the formula: \[ d = 2r \] Now, let's substitute the given volume \( V = 8582 \, \mathrm{m}^3 \) into the equation and calculate the radius. Then, we will find the diameter. First, we will calculate \( r^3 \): \[ r^3 = \frac{3 \times 8582}{2\pi} \] Now, let's perform the calculation. Calculate the value by following steps: - step0: Calculate: \(\frac{\left(3\times 8582\right)}{2\pi }\) - step1: Remove the parentheses: \(\frac{3\times 8582}{2\pi }\) - step2: Multiply the numbers: \(\frac{25746}{2\pi }\) - step3: Reduce the fraction: \(\frac{12873}{\pi }\) Calculate or simplify the expression \( 4097.603165^{1/3} \). Calculate the value by following steps: - step0: Calculate: \(4097.603165^{\frac{1}{3}}\) - step1: Convert the expressions: \(\left(\frac{819520633}{200000}\right)^{\frac{1}{3}}\) - step2: Use the properties of exponents: \(\frac{819520633^{\frac{1}{3}}}{200000^{\frac{1}{3}}}\) - step3: Rewrite the expression: \(\frac{\sqrt[3]{819520633}}{20\sqrt[3]{25}}\) - step4: Multiply by the Conjugate: \(\frac{\sqrt[3]{819520633}\times \sqrt[3]{25^{2}}}{20\sqrt[3]{25}\times \sqrt[3]{25^{2}}}\) - step5: Simplify: \(\frac{\sqrt[3]{819520633}\times 5\sqrt[3]{5}}{20\sqrt[3]{25}\times \sqrt[3]{25^{2}}}\) - step6: Multiply the numbers: \(\frac{5\sqrt[3]{4097603165}}{20\sqrt[3]{25}\times \sqrt[3]{25^{2}}}\) - step7: Multiply the numbers: \(\frac{5\sqrt[3]{4097603165}}{500}\) - step8: Reduce the fraction: \(\frac{\sqrt[3]{4097603165}}{100}\) Calculate or simplify the expression \( 2 * 16.002087 \). Calculate the value by following steps: - step0: Calculate: \(2\times 16.002087\) - step1: Multiply the numbers: \(32.004174\) We have calculated the radius \( r \) of the hemisphere as approximately \( 16.0 \, \mathrm{m} \). Now, we can find the diameter \( d \): \[ d = 2r \approx 2 \times 16.0 \approx 32.0 \, \mathrm{m} \] Thus, the diameter of the hemisphere, to the nearest tenth of a meter, is: \[ \boxed{32.0 \, \mathrm{m}} \]