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Question Number 60545 by Tawa1 last updated on 21/May/19

Answered by MJS last updated on 22/May/19

$$\mathrm{diameter}2\mathrm{volume}{2}^3=8$$$$\mathrm{diameter}3\mathrm{volume}{3}^3=27$$$$\mathrm{diameter}4\mathrm{volume}{4}^3=64$$$$\mathrm{new}\mathrm{sphere}\mathrm{has}\mathrm{got}\mathrm{volume}8+27+64=99$$$$\Rightarrow {\mathrm{it}}^{\prime }\mathrm{s}\mathrm{diameter}\mathrm{is}\sqrt[3]{99}$$

Commented by MJS last updated on 22/May/19

$$\mathrm{we}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{need}\mathrm{the}\mathrm{exact}\mathrm{formula}$$$$V=\frac{4\pi }{3}{r}^3=\frac{\pi }{6}{d}^3$$$$\mathrm{because}\mathrm{the}\mathrm{factor}\frac{\pi }{6}\mathrm{will}\mathrm{be}\mathrm{omitted}\mathrm{at}\mathrm{the}$$$$\mathrm{end}:$$$$\frac{\pi }{6}{d}_1^3+\frac{\pi }{6}{d}_2^3+\frac{\pi }{6}{d}_3^3=\frac{\pi }{6}{d}_4^3\Rightarrow d_4=\sqrt[3]{d_1^3+d_2^3+d_3^3}$$

Commented by Tawa1 last updated on 22/May/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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