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Home » \(\boxed{\frac2603 \pi}\) - ClickBalance

\(\boxed{\frac2603 \pi}\) - ClickBalance

📅 May 18, 2026 👤 scraface

May 18, 2026

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📌 V_{\text{hemisphere}} = \frac{1}{2} \cdot \frac{4}{3}\pi (3r)^3 = \frac{2}{3}\pi \cdot 27r^3 = 18\pi r^3

📌 Now, the ratio of the volume of the nanoparticle to the volume of the carrier is:

📌 \frac{V_{\text{sphere}}}{V_{\text{hemisphere}}} = \frac{\frac{4}{3}\pi r^3}{18\pi r^3} = \frac{4}{3} \cdot \frac{1}{18} = \frac{4}{54} = \frac{2}{27}

📌 Final answer: $ \boxed{\frac{2}{27}} $

📌 Question: A 5 cm by 8 cm rectangle is inscribed in a circle. What is the circumference of the circle in centimeters? Express your answer in terms of $ \pi $.

📌 Using the Pythagorean theorem, the diagonal $ d $ is:

📌 So, the diameter of the circle is $ \sqrt{89} $ cm, and the radius is $ \frac{\sqrt{89}}{2} $ cm.

📌 The circumference of the circle is:

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