## anonymous one year ago Help!!! Scientific Notation. Document below

1. anonymous

2. anonymous

I need help with A B C and D

3. anonymous

@Michele_Laino

4. anonymous

@MikeyMaximum @Nnesha @phi

5. Michele_Laino

question #1 we need to compute the distance Venus-Earth

6. anonymous

ok

7. anonymous

108,000,000 and 150,000,000

8. Michele_Laino

from your table, we can write this: $\Large d = 150 \cdot {10^6} - 108 \cdot {10^6} = \left( {150 - 108} \right) \cdot {10^6} = ...?$

9. anonymous

32 x 1,000,000

10. anonymous

so then 32,000,000 is the difference/what they have to travel from venus to earth?

11. phi

42, not 32

12. Michele_Laino

better is: $\Large \begin{gathered} d = 150 \cdot {10^6} - 108 \cdot {10^6} = \left( {150 - 108} \right) \cdot {10^6} = \hfill \\ = 42 \cdot {10^6} = 4.2 \cdot {10^7} \hfill \\ \end{gathered}$

13. anonymous

oh ok

14. anonymous

so then I have to keep on going until jupiter and then add all of them together right?

15. Michele_Laino

question B we have to compute the sum of the distance of Mercury, Venus and Earth, first. Such sum is: $\large d = 57 \cdot {10^6} + 108 \cdot {10^6} + 150 \cdot {10^6} = \left( {57 + 108 + 150} \right) \cdot {10^6} = ...?$

16. Michele_Laino

$d = 57 \cdot {10^6} + 108 \cdot {10^6} + 150 \cdot {10^6} = \left( {57 + 108 + 150} \right) \cdot {10^6} = ...?$

17. anonymous

315 x 10^6?

18. Michele_Laino

correct! second we have to compare that distance with the distance of Neptune, which is: $\Large {d_1} = 4.5 \cdot {10^9} = 4500 \cdot {10^6}$

19. anonymous

? whats d^1??

20. Michele_Laino

d1 is the distance of Neptune from the Sun Is the distance of Neptune from the Sun greater or less than the sum of distances above?

21. anonymous

greater?

22. Michele_Laino

correct since: $\Large {d_1} = 4.5 \cdot {10^9} = 4500 \cdot {10^6} > 42 \cdot {10^6}$

23. anonymous

ohh

24. anonymous

can you type that in the document please and then send it to me through here??

25. anonymous

26. Michele_Laino

oops.. since : $\Large {d_1} = 4.5 \cdot {10^9} = 4500 \cdot {10^6} > 315 \cdot {10^6}$ sorry for my typo

27. anonymous

its all right :D

28. Michele_Laino

now, let's go to question C)

29. anonymous

lol ok

30. Michele_Laino

if I multiply the distance of the Earth from the Sun, by 10, I get this new distance: $\Large L = 150 \cdot {10^6} \cdot 10 = 150 \cdot {10^7}$

31. anonymous

ok

32. Michele_Laino

and that distance is very close to the distance of Saturn which is: $\Large {L_1} = 1.43 \cdot {10^9} = 143 \cdot {10^7}$ am I right?

33. anonymous

i think so

34. Michele_Laino

so what is the right name of planet?

35. anonymous

saturn

36. anonymous

right?

37. Michele_Laino

correct!

38. Michele_Laino

now, let's go on question D)

39. anonymous

haha ok

40. Michele_Laino

first, we have to compute the distance Saturn-Earth, which is: $\Large D = 1430 \cdot {10^6} - 150 \cdot {10^6} = ...?$

41. anonymous

1280 x 10^6?

42. Michele_Laino

correct! now the requested time is: time=distance over speed and the speed of Shuttle is: $\Large v = 28000 = 28 \cdot {10^3}{\text{Km/h}}$ so the requested time is: $\Large t = \frac{{{\text{distance}}}}{{{\text{speed}}}} = \frac{{1280 \cdot {{10}^6}}}{{28 \cdot {{10}^3}}} = \frac{{1280}}{{28}}{10^{6 - 3}} = ...hours$

43. anonymous

45.71 x 10^3?

44. Michele_Laino

that's right!

45. anonymous

so thats it?

46. Michele_Laino

yes!

47. anonymous

thanks!! i just have a question... could you help solve the rest of part a??

48. Michele_Laino

yes! sure!

49. anonymous

ok i just have trouble calculating the distances in scientific notation.. showing my work

50. Michele_Laino

we need to know the distance Venus-Jupiter, which is: $\Large d = 779 \cdot {10^6} - 108 \cdot {10^6} = \left( {779 - 108} \right) \cdot {10^6} = ...?$

51. anonymous

yea we already did the one from venus to earth so now i guess we have to calculate from earth to mars

52. Michele_Laino

sorry are you asking for question A) right?

53. anonymous

yes

54. Michele_Laino

ok! then we have to compute the distance Venus-Jupiter

55. anonymous

yea

56. Michele_Laino

and such distance is: $\Large d = 779 \cdot {10^6} - 108 \cdot {10^6} = \left( {779 - 108} \right) \cdot {10^6} = ...?$

57. anonymous

671 x 10^6?

58. Michele_Laino

yes! correct! :)

59. anonymous

is that all?

60. Michele_Laino

yes! we have finished!

61. anonymous

wow! yay thanks!! :D

62. Michele_Laino

:) :)

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