anonymous
  • anonymous
Help!!! Scientific Notation. Document below
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
I need help with A B C and D
anonymous
  • anonymous
@Michele_Laino

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anonymous
  • anonymous
@MikeyMaximum @Nnesha @phi
Michele_Laino
  • Michele_Laino
question #1 we need to compute the distance Venus-Earth
anonymous
  • anonymous
ok
anonymous
  • anonymous
108,000,000 and 150,000,000
Michele_Laino
  • Michele_Laino
from your table, we can write this: \[\Large d = 150 \cdot {10^6} - 108 \cdot {10^6} = \left( {150 - 108} \right) \cdot {10^6} = ...?\]
anonymous
  • anonymous
32 x 1,000,000
anonymous
  • anonymous
so then 32,000,000 is the difference/what they have to travel from venus to earth?
phi
  • phi
42, not 32
Michele_Laino
  • Michele_Laino
better is: \[\Large \begin{gathered} d = 150 \cdot {10^6} - 108 \cdot {10^6} = \left( {150 - 108} \right) \cdot {10^6} = \hfill \\ = 42 \cdot {10^6} = 4.2 \cdot {10^7} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
so then I have to keep on going until jupiter and then add all of them together right?
Michele_Laino
  • Michele_Laino
question B we have to compute the sum of the distance of Mercury, Venus and Earth, first. Such sum is: \[\large d = 57 \cdot {10^6} + 108 \cdot {10^6} + 150 \cdot {10^6} = \left( {57 + 108 + 150} \right) \cdot {10^6} = ...?\]
Michele_Laino
  • Michele_Laino
\[d = 57 \cdot {10^6} + 108 \cdot {10^6} + 150 \cdot {10^6} = \left( {57 + 108 + 150} \right) \cdot {10^6} = ...?\]
anonymous
  • anonymous
315 x 10^6?
Michele_Laino
  • Michele_Laino
correct! second we have to compare that distance with the distance of Neptune, which is: \[\Large {d_1} = 4.5 \cdot {10^9} = 4500 \cdot {10^6}\]
anonymous
  • anonymous
? whats d^1??
Michele_Laino
  • 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?
anonymous
  • anonymous
greater?
Michele_Laino
  • Michele_Laino
correct since: \[\Large {d_1} = 4.5 \cdot {10^9} = 4500 \cdot {10^6} > 42 \cdot {10^6}\]
anonymous
  • anonymous
ohh
anonymous
  • anonymous
can you type that in the document please and then send it to me through here??
anonymous
  • anonymous
for b? please
Michele_Laino
  • Michele_Laino
oops.. since : \[\Large {d_1} = 4.5 \cdot {10^9} = 4500 \cdot {10^6} > 315 \cdot {10^6}\] sorry for my typo
anonymous
  • anonymous
its all right :D
Michele_Laino
  • Michele_Laino
now, let's go to question C)
anonymous
  • anonymous
lol ok
Michele_Laino
  • Michele_Laino
if I multiply the distance of the Earth from the Sun, by 10, I get this new distance: \[\Large L = 150 \cdot {10^6} \cdot 10 = 150 \cdot {10^7}\]
anonymous
  • anonymous
ok
Michele_Laino
  • 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?
anonymous
  • anonymous
i think so
Michele_Laino
  • Michele_Laino
so what is the right name of planet?
anonymous
  • anonymous
saturn
anonymous
  • anonymous
right?
Michele_Laino
  • Michele_Laino
correct!
Michele_Laino
  • Michele_Laino
now, let's go on question D)
anonymous
  • anonymous
haha ok
Michele_Laino
  • Michele_Laino
first, we have to compute the distance Saturn-Earth, which is: \[\Large D = 1430 \cdot {10^6} - 150 \cdot {10^6} = ...?\]
anonymous
  • anonymous
1280 x 10^6?
Michele_Laino
  • 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: \[\Large t = \frac{{{\text{distance}}}}{{{\text{speed}}}} = \frac{{1280 \cdot {{10}^6}}}{{28 \cdot {{10}^3}}} = \frac{{1280}}{{28}}{10^{6 - 3}} = ...hours\]
anonymous
  • anonymous
45.71 x 10^3?
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
so thats it?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
thanks!! i just have a question... could you help solve the rest of part a??
Michele_Laino
  • Michele_Laino
yes! sure!
anonymous
  • anonymous
ok i just have trouble calculating the distances in scientific notation.. showing my work
Michele_Laino
  • Michele_Laino
we need to know the distance Venus-Jupiter, which is: \[\Large d = 779 \cdot {10^6} - 108 \cdot {10^6} = \left( {779 - 108} \right) \cdot {10^6} = ...?\]
anonymous
  • anonymous
yea we already did the one from venus to earth so now i guess we have to calculate from earth to mars
Michele_Laino
  • Michele_Laino
sorry are you asking for question A) right?
anonymous
  • anonymous
yes
Michele_Laino
  • Michele_Laino
ok! then we have to compute the distance Venus-Jupiter
anonymous
  • anonymous
yea
Michele_Laino
  • Michele_Laino
and such distance is: \[\Large d = 779 \cdot {10^6} - 108 \cdot {10^6} = \left( {779 - 108} \right) \cdot {10^6} = ...?\]
anonymous
  • anonymous
671 x 10^6?
Michele_Laino
  • Michele_Laino
yes! correct! :)
anonymous
  • anonymous
is that all?
Michele_Laino
  • Michele_Laino
yes! we have finished!
anonymous
  • anonymous
wow! yay thanks!! :D
Michele_Laino
  • Michele_Laino
:) :)

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