Mathematics
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OpenStudy (lizz123):
Help and MEDAL
http://prntscr.com/75ahtu
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OpenStudy (astrophysics):
What's the question?
OpenStudy (lizz123):
With everything?
OpenStudy (astrophysics):
What? All you gave us is a picture but no question.
OpenStudy (lizz123):
Find the surface area of each figure.
OpenStudy (astrophysics):
So we know it's a cylinder right, we have height and radius so we can use \[SA = 2 \pi r h + 2 \pi r^2\]
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OpenStudy (lizz123):
So \[SA=2*3.14*6*5+2*3.14*6^{2}\]
OpenStudy (astrophysics):
Plug and chug at this point, surface area is just finding area of each 2 dimension figure of the drawing.
OpenStudy (astrophysics):
and adding them
OpenStudy (astrophysics):
Yeah that seems reasonable
OpenStudy (lizz123):
So
2*3.14=6.28*6=37.68*5=188.4
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OpenStudy (anonymous):
YAASS NAILED IT!!! <3
OpenStudy (lizz123):
2*3.14=6.28*36=226.08
OpenStudy (lizz123):
So what do I do after
OpenStudy (astrophysics):
Add them
OpenStudy (astrophysics):
And that's your surface area
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OpenStudy (lizz123):
okay so, 188.4+226.08=414.48
OpenStudy (astrophysics):
Sounds good :)
OpenStudy (lizz123):
Therefore, my answer is 414.48.
Thanks
OpenStudy (astrophysics):
Np
OpenStudy (anonymous):
a=
6.283185hr+6.283185r2
s
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OpenStudy (anonymous):
a=
6.283185hr+6.283185r2/s
OpenStudy (lizz123):
@JayisSmart I just added again and I got the answer of 454.48
OpenStudy (lizz123):
And before I got 414.48
OpenStudy (anonymous):
you are right with 454.48
OpenStudy (lizz123):
nvm I understand now
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OpenStudy (lizz123):
and thanks for your help
OpenStudy (anonymous):
yep