A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Help!!! Scientific Notation. Document below

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I need help with A B C and D

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Michele_Laino

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @MikeyMaximum @Nnesha @phi

  5. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    108,000,000 and 150,000,000

  8. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    32 x 1,000,000

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  11. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    42, not 32

  12. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  15. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    315 x 10^6?

  18. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ? whats d^1??

  20. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    greater?

  22. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  23. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ohh

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  25. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    for b? please

  26. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  27. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    its all right :D

  28. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    now, let's go to question C)

  29. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol ok

  30. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

  32. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i think so

  34. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so what is the right name of planet?

  35. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    saturn

  36. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right?

  37. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    correct!

  38. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    now, let's go on question D)

  39. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    haha ok

  40. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  41. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1280 x 10^6?

  42. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    45.71 x 10^3?

  44. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    that's right!

  45. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so thats it?

  46. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes!

  47. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  48. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes! sure!

  49. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  50. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sorry are you asking for question A) right?

  53. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  54. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  55. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yea

  56. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  57. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    671 x 10^6?

  58. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes! correct! :)

  59. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is that all?

  60. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes! we have finished!

  61. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wow! yay thanks!! :D

  62. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    :) :)

  63. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.