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anonymous

  • one year ago

Given that the surface area of a sphere is 31 \pi cm^2, find its volume.

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  1. Jhannybean
    • one year ago
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    \[\sf SA ~=~31\pi~cm^2\] Surface area of a sphere = \(\sf 4\pi r^2\) Volume of a sphere = \(\sf\dfrac{4}{3}\pi r^3\) So if we take what we know, the SA of the sphere, and plug it into the formula for SA, we can find \(\sf r^2\)

  2. Jhannybean
    • one year ago
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    \[\sf 31\pi = 4\pi r^2\]\[\sf \color{red}{r^2 = \frac{31\pi}{4\pi} = \frac{31}{4}}\] \[\sf V = \frac{4}{3}\pi r^3=\frac{4}{3}\pi\color{red}{ r^2} \cdot r \]\[\sf V =\frac{4}{3} \pi \left(\frac{31}{4}\right)\cdot r = \frac{31}{3}\pi r\]

  3. Jhannybean
    • one year ago
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    Do you understand how this wrks? @Kimes

  4. anonymous
    • one year ago
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    yeah I'm just trying to rewrite everything

  5. Jhannybean
    • one year ago
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    Im not sure if \(\sf r\) is art of your answer, or we have to reduce it even more.

  6. anonymous
    • one year ago
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    hmm its saying its wrong, so i guess keep reducing it

  7. Jhannybean
    • one year ago
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    Okay.

  8. Jhannybean
    • one year ago
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    We already have \[\sf r^2=\frac{31}{4} ~~~~\text{which can imply} \implies \color{red}{r=\sqrt{\frac{31}{4}} =\frac{\sqrt{31}}{2}}\] Plug this into what we have already found for V. \[\sf V = \frac{31}{3}\pi \left(\frac{\sqrt{31}}{2}\right)= \frac{31\sqrt{31}}{6}\pi =\frac{\sqrt{31^3}}{6}\pi =\frac{31^{3/2}}{6}\pi\]

  9. Jhannybean
    • one year ago
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    Im 95% sure that that is your simplest, most reduced form.

  10. anonymous
    • one year ago
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    so i got 16. 23 cm^3

  11. Jhannybean
    • one year ago
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    You could leave it as \(\sf \dfrac{31\sqrt{31}}{6}\pi\) or even \(\dfrac{31^{3/2}}{6}\pi\)

  12. Jhannybean
    • one year ago
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    I'm getting \(\approx\) 90.37 cm\(^3\)

  13. anonymous
    • one year ago
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    oh ok i just got that too

  14. Jhannybean
    • one year ago
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    inputting it into your calculator, I did : ( (31\(\sf\sqrt{31}\) ) / 6 ) * \(\pi\)

  15. Jhannybean
    • one year ago
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    Oh, okay. Cool!

  16. anonymous
    • one year ago
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    thank you so much!

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