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Mathematics 23 Online
OpenStudy (anonymous):

Anyone ready for a math question.

OpenStudy (anonymous):

OpenStudy (anonymous):

For a question like this, I like to plug in a random number, do the suggestion, and then observe the end result.

OpenStudy (anonymous):

I did this question before, but I forgot the answer.

geerky42 (geerky42):

Just use formula: it is known that \(V = \dfrac{4}{3}\pi r^3\) and \(SA = 4\pi r^2\) Just plug 1/4 r in r, then simplify it

geerky42 (geerky42):

replace r to 1/4 r

OpenStudy (anonymous):

OKay, give me a minute to get the answer.

OpenStudy (anonymous):

The volume of a sphere is V=(4/3)Πr². Let's plug in 4 to make it easy. You get 66.99 as your volume. Do you follow it so far?

OpenStudy (anonymous):

I got (pi)/12

OpenStudy (anonymous):

Now, multiply 4 by 1/4. You get 1. When you plug it back in, you get 4.19.

OpenStudy (anonymous):

1

OpenStudy (anonymous):

You multiplied 16 (r²) by 3/4, not 4/3. If you multiply it by 4/3, you get 21.33.

OpenStudy (anonymous):

I thought for volume I use 4/3?

OpenStudy (anonymous):

You do. You multiplied it by 3/4 though.

OpenStudy (anonymous):

What would be r?

geerky42 (geerky42):

\[V = \dfrac{3}{4}\pi r^3\]Plug in \(\dfrac{1}{4}\); \[\large V_{new} = \dfrac{3}{4}\pi \left(\dfrac{1}{4}r\right)^3 = \left(\dfrac{1}{4}\right)^3\left(\dfrac{3}{4}\pi r^3\right) = \left(\dfrac{1}{4}\right)^3(V)\]Do you know what (1/4)^3 is?

geerky42 (geerky42):

Plug in \(\dfrac{1}{4}r\) **

OpenStudy (anonymous):

1/64

geerky42 (geerky42):

Yes, so \(\Large V_{new} = \dfrac{1}{64}V\) So that means volume is multiplied by 1/64 does that make sense?

OpenStudy (anonymous):

So the volume is 1/64 and the surface area is 1/16?

geerky42 (geerky42):

Yes

geerky42 (geerky42):

you squared 1/4 in surface area, so you will get 1/16.

OpenStudy (anonymous):

Alright, thank you so much. Do you have time for another question?

geerky42 (geerky42):

Yes of course

OpenStudy (anonymous):

I'll post a as a new question. I'll tag you.

geerky42 (geerky42):

alright

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