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Mathematics 11 Online
OpenStudy (lizz123):

Help and MEDAL http://prntscr.com/75ahtu

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 \[SA = 2 \pi r h + 2 \pi r^2\]

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

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

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

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

OpenStudy (lizz123):

and thanks for your help

OpenStudy (anonymous):

yep

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