Dana plans to cover this model of a half moon completely with silver foil.
How many square centimeters of silver foil would Dana need to cover the model completely?
402.2
536.2
603.4
150.8

- anonymous

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- anonymous

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- anonymous

Dude, this is a homework help site, do not ask for emails :P

- anonymous

its not and ur a player

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## More answers

- anonymous

true @trackzz

- campbell_st

well the surface area of a hemisphere is
\[SA = \frac{1}{2}(4pir^2) + \pi r^2\]
and you have a radius of 8 cm
just substitute and solve.

- anonymous

i dont know how to do that

- anonymous

cause i just see leters

- anonymous

@campbell_st

- campbell_st

ok... do you have a calculator ?

- anonymous

yesss

- campbell_st

ok you need to type in
\[0.5 \times 4 \times \pi \times 8^2 + \pi \times 8^2\]
that will give the answer to covering the hemisphere in foil..

- anonymous

Alrii so basically what campbell has show you is the surface are of a sphere. Any problem in which they ask for something that wraps around a specific shape involves surface are.
The surface area of a sphere is: 4 pi r^2
Where r is the radius of the sphere.

- anonymous

what is p 12.18 or 12.14

- anonymous

If you're familiar with the button pi, it is the value of 3.14.
Now for this particular question, it is a hemisphere. So half the surface are of the sphere.
1/2*(4pi r^2) = SA

- anonymous

thnks

- anonymous

@crystal1 what are you talking about>

- anonymous

you said:" what is p 12.18 or 12.14"
what do you mean?

- anonymous

my bad sorry

- anonymous

so you got the answer?

- anonymous

wait

- anonymous

my answer is 5048.1152

- campbell_st

nope... thats a long way from being correct

- campbell_st

did you type in the equation I gave you as its written...?

- anonymous

Shall we break it down Campbell?
I think it's the BEDMAS

- anonymous

Alright crystal, I want you to do r^2 first. Find the value of the radius squared.

- anonymous

yes i did @campbell_st

- campbell_st

and tracks... I think you possible solution is missing the circular base... as the question talks about competely covering... it

- anonymous

you guys lost me now im confuse

- campbell_st

ok... lets to it in little bits can you calculate
\[0.5 \times 4 \times \pi \times 8^2\]

- anonymous

Ok my bad, alrii yeah you're right. I'll leave this to Campbell. Dont wanna confuse you,

- campbell_st

@crystal1 did you get an answer for the new calculation..?

- anonymous

b

- campbell_st

what was the number..?

- anonymous

i dont reaally know im sorry im dumb i dont know how to start this thats why i need help

- anonymous

You're not dumb. Tell yourself you're smart. You'll get the answer. Crystal, you got this. Wake up, look at what campbell wrote for you.

- anonymous

the answer is a

- anonymous

you're not finished yet. Add the pi r^2 to what you got to get the answer

- campbell_st

nope thats only the 1st part of the problem
the curved surface has an area of 402.2 cm^2
now you need to find the area of the circular base...
\[\pi \times 8^2\]

- campbell_st

you need to calculate this number...

- campbell_st

then the final answer will be
\[402.2 + \pi \times 8^2 = \]

- anonymous

6485.44

- campbell_st

wow... nope not it
lets make it easier
what is
\[\pi \times 8^2 =\]

- anonymous

75.36

- campbell_st

nope still not right...
\[\pi \times 8^2 = 201.1\]

- anonymous

how do you get thaat i tough that 8x2 its = 24

- campbell_st

do you know how to square a number...

- anonymous

no

- campbell_st

ok... then perhaps you need to ask you teacher... how to do the problem is you don't know the basics...

- anonymous

thnks you @campbell_st

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