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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I need help with A B C and D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@MikeyMaximum @Nnesha @phi

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1question #1 we need to compute the distance VenusEarth

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0108,000,000 and 150,000,000

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1from your table, we can write this: \[\Large d = 150 \cdot {10^6}  108 \cdot {10^6} = \left( {150  108} \right) \cdot {10^6} = ...?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so then 32,000,000 is the difference/what they have to travel from venus to earth?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1better is: \[\Large \begin{gathered} d = 150 \cdot {10^6}  108 \cdot {10^6} = \left( {150  108} \right) \cdot {10^6} = \hfill \\ = 42 \cdot {10^6} = 4.2 \cdot {10^7} \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so then I have to keep on going until jupiter and then add all of them together right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1question B we have to compute the sum of the distance of Mercury, Venus and Earth, first. Such sum is: \[\large d = 57 \cdot {10^6} + 108 \cdot {10^6} + 150 \cdot {10^6} = \left( {57 + 108 + 150} \right) \cdot {10^6} = ...?\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\[d = 57 \cdot {10^6} + 108 \cdot {10^6} + 150 \cdot {10^6} = \left( {57 + 108 + 150} \right) \cdot {10^6} = ...?\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1correct! second we have to compare that distance with the distance of Neptune, which is: \[\Large {d_1} = 4.5 \cdot {10^9} = 4500 \cdot {10^6}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1d1 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?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1correct since: \[\Large {d_1} = 4.5 \cdot {10^9} = 4500 \cdot {10^6} > 42 \cdot {10^6}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you type that in the document please and then send it to me through here??

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1oops.. since : \[\Large {d_1} = 4.5 \cdot {10^9} = 4500 \cdot {10^6} > 315 \cdot {10^6}\] sorry for my typo

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, let's go to question C)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1if I multiply the distance of the Earth from the Sun, by 10, I get this new distance: \[\Large L = 150 \cdot {10^6} \cdot 10 = 150 \cdot {10^7}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1and 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?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so what is the right name of planet?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, let's go on question D)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1first, we have to compute the distance SaturnEarth, which is: \[\Large D = 1430 \cdot {10^6}  150 \cdot {10^6} = ...?\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1correct! 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: \[\Large t = \frac{{{\text{distance}}}}{{{\text{speed}}}} = \frac{{1280 \cdot {{10}^6}}}{{28 \cdot {{10}^3}}} = \frac{{1280}}{{28}}{10^{6  3}} = ...hours\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks!! i just have a question... could you help solve the rest of part a??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok i just have trouble calculating the distances in scientific notation.. showing my work

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we need to know the distance VenusJupiter, which is: \[\Large d = 779 \cdot {10^6}  108 \cdot {10^6} = \left( {779  108} \right) \cdot {10^6} = ...?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea we already did the one from venus to earth so now i guess we have to calculate from earth to mars

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1sorry are you asking for question A) right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! then we have to compute the distance VenusJupiter

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1and such distance is: \[\Large d = 779 \cdot {10^6}  108 \cdot {10^6} = \left( {779  108} \right) \cdot {10^6} = ...?\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! correct! :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! we have finished!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wow! yay thanks!! :D
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