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if volume of sphere increases by 72.8% what happen to the surface area?

Mathematics
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V = 4/3 pi r^3 . this is a calculus differential question>
maybe not
i didnt get it mam

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Other answers:

ok , an increase of 72.8% is 172.8% of the original volume. do you agree ?
just like a 100% increase is actually double the original amount (200% of original amount)
yuppp
ok so New Volume = 1.728 * Old volume
let the old volume be 4/3 pi * r^3. New Volume = 1.728 * 4/3 pi * r^3
yup but wht abt its surface area ?
we didnt get to that. one step at a time
who is that a picture of?
ok
so ... New volume = 4/3 pi [(1.728)^(1/3) * r ]^3
see how i brought in the 1.728, by taking the cube root (and then cubing it )
anybody know what a cluster point is ? please help
??
since volume is directly proportional to r^3 hence if new volume is (100%+72.8%)=1.728 times the old volume then new r =cube root of (1.728) times old radius hence new r=1.2 times old r as surface area is directly proportional to r^2 hence new surface area =(1.2)^2 times old surface area=1.44 times old surface area hence increase in surface area =.44 times=44% increase
Volume and surface area both depends on radius........so, you need to understand and calculate what happens to radius when volume increase by that much
thank you so much to all...
??
welcome
The volume of a sphere is proportional to the radius cubed: \[volume=\frac{4}{3}\pi r ^{3}\] The surface area of a sphere is proportional to the radius squared: \[Surface\ area=4\pi r ^{2}\] Let the original radius = 1 unit Then for the volume to increase by 72.8%, the cube of the radius must increase from 1unit cubed up to 1.728 units cubed. This means the radius has increased from 1 unit up to \[\sqrt[3]{1.726}=1.2\ units\] Since the surface area is proportional to the radius squared, the square of the radius will increase from 1 unit squared up to \[1.2^{2}=1.44\] Therefore the surface area has increased by \[\frac{1.44-1}{1}\times 100=44\%\]

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