Ushtrimi 6 | 5.2A | Matematika 10 - 11: Pjesa II | Zgjidhje Ushtrimesh

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$\dfrac{4}{3}\pi \times 6^3 + \dfrac{2}{3}\pi \times 3^2$
$\dfrac{1}{3}\pi \times 9^2 \times 5 - \dfrac{1}{3}\pi(\dfrac{18}{5})^2 \times 2$
$\dfrac{1}{2}(4\pi \times 11^2 + 4\pi \times 9^2)$
$16\pi + 4\pi\sqrt{6^2 + 4^2}$
$\dfrac{8\pi}{5} + \dfrac{3\pi}{4}$
$2\dfrac{1}{4}\pi + 1\dfrac{1}{2} \times 4\dfrac{2}{3}\pi$
$\dfrac{5}{6}\pi - 2\sqrt{2} + \dfrac{7}{9}\pi + \sqrt{28}$
$\dfrac{\sqrt{15\pi} \times 4 \times \sqrt{3\pi}}{6\sqrt{3}}$

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$\dfrac{4}{\cancel{3}}\pi \times \cancel{216} + \dfrac{2}{\cancel{3}}\pi \times \cancel{9} = 72 \times 4\pi + 2\pi \times 3 = 288\pi + 6\pi = 294\pi$
$\dfrac{1}{\cancel{3}}\pi \times \cancel{81} \times 5 - \dfrac{1}{\cancel{3}}\pi \times \dfrac{\cancel{324}}{25} \times 2 = 27\pi \times 5 - \dfrac{108}{25}\pi \times 2 = 135\pi - \dfrac{216}{25}\pi = \dfrac{3375 - 216}{25}\pi = \dfrac{3159}{25}\pi$ $=$ $126\dfrac{9}{25}\pi$
$\dfrac{1}{2}(484\pi + 324\pi) = \dfrac{1}{\cancel2} \times \cancel{808}\pi = 404\pi$
$16\pi + 4\pi \sqrt{52} \rArr 16\pi+4\pi \times 2\sqrt{13} \rArr 8\pi(2+\sqrt{13})$
$\dfrac{32\pi + 15\pi}{20} = \dfrac{47\pi}{20}$ $=$ $2\dfrac{7}{20}\pi$
$\dfrac{9}{4}\pi + \dfrac{\cancel{3}}{\cancel{2}} \times \dfrac{\cancel{14}}{\cancel{3}}\pi = \dfrac{9}{4}\pi + 7\pi = \dfrac{37}{4}\pi$ $=$ $9\dfrac{1}{4}\pi$
$\dfrac{5}{6}\pi + \dfrac{7}{9}\pi - 2\sqrt{2} + 2\sqrt{7} = \dfrac{15 + 14}{18}\pi - 2\sqrt{2} + 2\sqrt{7} = \dfrac{29}{18}\pi - 2 (\sqrt{2} - \sqrt{7})$ $=$ $1\dfrac{11}{18}\pi - 2(\sqrt2 - \sqrt7)$
$\dfrac{\sqrt{15\pi \times 3\pi} \times 4}{6\sqrt{3}} = \dfrac{\sqrt{45\pi^2} \times 4}{6\sqrt{3}} = \dfrac{\cancel{12}\pi\sqrt{5}}{\cancel{6}\sqrt{3}} = \dfrac{2\pi\sqrt{5}}{\sqrt{3}}$ $=$ $\dfrac{2\pi \sqrt{5}}{\sqrt{3}} \times \dfrac{\sqrt3}{\sqrt3}$ $=$ $\dfrac{2\pi \sqrt{15}}{3}$