A solid cone of radius 6 cm and height 21 cm is melted and made into 2 spheres of different sizes.The radius of the first sphere is 4 cm , find the radius of the second sphere.

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A solid cone of radius 6 cm and height 21 cm is melted and made into 2 spheres of different sizes.The radius of the first sphere is 4 cm , find the radius of the second sphere.

Mathematics
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hint: the volume of the 2 spheres has to be equal to the volume of the starting cone, so we can write this equation: \[\Large \frac{{4\pi }}{3}{R^3} + \frac{{4\pi }}{3}{x^3} = \frac{{\pi {r^2}h}}{3}\] where x is the requested radius, R=4 cm, h=21 cm, and r= 6 cm
???

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please you have to solve that equation for x
i didnt how you got yhe equation *the
the sum of the volumes of the 2 spheres, has to be equal to the volume of the starting cone
\[\frac{ 4 }{ 3 }\pi (R^3 + r^3) = \frac{ 1 }{ 3 }\pi(r^2h)\]
not exactly, better is: \[\Large \frac{{4\pi }}{3}\left( {{R^3} + {x^3}} \right) = \frac{{\pi {r^2}h}}{3}\] where x is the requested radius, and r is the radius of the starting cone, namely r= 6 cm
3 and pi get cancelled?
yes!
r^h = 4(R^3 + x^3)?
more precisely: \[\Large 4\left( {{R^3} + {x^3}} \right) = {r^2}h\]
what next?
we have to divide both sides by 4
(R^3 +x^3) = r^2h/4
ok! now we have subtract R^3 at both sides
x^3 = r^2h/4 - R^3
finally we have to take the 3-rd root of both sides
???
you should get this: \[\Large x = \sqrt[3]{{\frac{{{r^2}h}}{4} - {R^3}}}\]
ok!
\[\Large x = \sqrt[3]{{\frac{{{r^2}h}}{4} - {R^3}}} = \sqrt[3]{{\frac{{{6^2} \cdot 21}}{4} - {4^3}}} = ...?\]
5?
that's right! x= 5 cm
thanks.
:)

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