anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
anonymous
  • anonymous
Dude, this is a homework help site, do not ask for emails :P
anonymous
  • anonymous
its not and ur a player

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More answers

anonymous
  • anonymous
true @trackzz
campbell_st
  • 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
  • anonymous
i dont know how to do that
anonymous
  • anonymous
cause i just see leters
anonymous
  • anonymous
@campbell_st
campbell_st
  • campbell_st
ok... do you have a calculator ?
anonymous
  • anonymous
yesss
campbell_st
  • 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
  • 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
  • anonymous
what is p 12.18 or 12.14
anonymous
  • 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
  • anonymous
thnks
anonymous
  • anonymous
@crystal1 what are you talking about>
anonymous
  • anonymous
you said:" what is p 12.18 or 12.14" what do you mean?
anonymous
  • anonymous
my bad sorry
anonymous
  • anonymous
so you got the answer?
anonymous
  • anonymous
wait
anonymous
  • anonymous
my answer is 5048.1152
campbell_st
  • campbell_st
nope... thats a long way from being correct
campbell_st
  • campbell_st
did you type in the equation I gave you as its written...?
anonymous
  • anonymous
Shall we break it down Campbell? I think it's the BEDMAS
anonymous
  • anonymous
Alright crystal, I want you to do r^2 first. Find the value of the radius squared.
anonymous
  • anonymous
yes i did @campbell_st
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
  • anonymous
you guys lost me now im confuse
campbell_st
  • campbell_st
ok... lets to it in little bits can you calculate \[0.5 \times 4 \times \pi \times 8^2\]
anonymous
  • anonymous
Ok my bad, alrii yeah you're right. I'll leave this to Campbell. Dont wanna confuse you,
campbell_st
  • campbell_st
@crystal1 did you get an answer for the new calculation..?
anonymous
  • anonymous
b
campbell_st
  • campbell_st
what was the number..?
anonymous
  • anonymous
i dont reaally know im sorry im dumb i dont know how to start this thats why i need help
anonymous
  • 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
  • anonymous
the answer is a
anonymous
  • anonymous
you're not finished yet. Add the pi r^2 to what you got to get the answer
campbell_st
  • 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
  • campbell_st
you need to calculate this number...
campbell_st
  • campbell_st
then the final answer will be \[402.2 + \pi \times 8^2 = \]
anonymous
  • anonymous
6485.44
campbell_st
  • campbell_st
wow... nope not it lets make it easier what is \[\pi \times 8^2 =\]
anonymous
  • anonymous
75.36
campbell_st
  • campbell_st
nope still not right... \[\pi \times 8^2 = 201.1\]
anonymous
  • anonymous
how do you get thaat i tough that 8x2 its = 24
campbell_st
  • campbell_st
do you know how to square a number...
anonymous
  • anonymous
no
campbell_st
  • campbell_st
ok... then perhaps you need to ask you teacher... how to do the problem is you don't know the basics...
anonymous
  • anonymous
thnks you @campbell_st

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