## crystal1 Group Title 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 one year ago one year ago

1. crystal1 Group Title

2. trackzz Group Title

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

3. crystal1 Group Title

its not and ur a player

4. crystal1 Group Title

true @trackzz

5. campbell_st Group Title

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.

6. crystal1 Group Title

i dont know how to do that

7. crystal1 Group Title

cause i just see leters

8. crystal1 Group Title

@campbell_st

9. campbell_st Group Title

ok... do you have a calculator ?

10. crystal1 Group Title

yesss

11. campbell_st Group Title

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..

12. trackzz Group Title

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.

13. crystal1 Group Title

what is p 12.18 or 12.14

14. trackzz Group Title

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

15. crystal1 Group Title

thnks

16. trackzz Group Title

@crystal1 what are you talking about>

17. trackzz Group Title

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

18. crystal1 Group Title

19. trackzz Group Title

20. crystal1 Group Title

wait

21. crystal1 Group Title

22. campbell_st Group Title

nope... thats a long way from being correct

23. campbell_st Group Title

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

24. trackzz Group Title

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

25. trackzz Group Title

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

26. crystal1 Group Title

yes i did @campbell_st

27. campbell_st Group Title

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

28. crystal1 Group Title

you guys lost me now im confuse

29. campbell_st Group Title

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

30. trackzz Group Title

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

31. campbell_st Group Title

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

32. crystal1 Group Title

b

33. campbell_st Group Title

what was the number..?

34. crystal1 Group Title

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

35. trackzz Group Title

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.

36. crystal1 Group Title

37. trackzz Group Title

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

38. campbell_st Group Title

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... $\pi \times 8^2$

39. campbell_st Group Title

you need to calculate this number...

40. campbell_st Group Title

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

41. crystal1 Group Title

6485.44

42. campbell_st Group Title

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

43. crystal1 Group Title

75.36

44. campbell_st Group Title

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

45. crystal1 Group Title

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

46. campbell_st Group Title

do you know how to square a number...

47. crystal1 Group Title

no

48. campbell_st Group Title

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

49. crystal1 Group Title

thnks you @campbell_st