purplemexican
  • purplemexican
On the Moon, a falling object falls just 2.65 feet in the first second after being dropped. Each second it falls 5.3 feet farther than in the previous second. How far would an object fall in the first ten seconds after being dropped?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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welshfella
  • welshfella
this is a an arithmetic series where the first term is 2.65 and each subsequent term is obtained by adding the common difference 35.
purplemexican
  • purplemexican
a. 270 ft c. 313 ft b. 692 ft d. 265 ft
welshfella
  • welshfella
nth term = a1 + ( n- 1)d where a1 = first term , n = number of terms and d = common difference so tenth term = 2.65 + 9*5.3 (common difference = 5.3 NOT 35 as i said in my first post)

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welshfella
  • welshfella
so in the second second it traveled 2.65 + 5.3 = 7.95 feet the total distance = sum of n terms = (n/2)[ a1 + L) where L = 10th term so plug in the values of a1 , n (=10) , and L and work it out
purplemexican
  • purplemexican
what the formula
welshfella
  • welshfella
i've just given it
purplemexican
  • purplemexican
im confused please state it again if you don't mind
welshfella
  • welshfella
plug in a1 = 2.65 , n = 10 and L = 10th term = 2.65 + 9*5.3 = 50.35
purplemexican
  • purplemexican
so L=50.35 correct?
welshfella
  • welshfella
thats the distance travelled in the 10 th second . You want the total distance travelled after 10 seconds have passed Use the formula to work that out.
welshfella
  • welshfella
Total distance = (n/2) * [a1 + L)
welshfella
  • welshfella
work ouy n/2 first then a1 + L then multiply the 2 results
purplemexican
  • purplemexican
i get 265
welshfella
  • welshfella
that's right
purplemexican
  • purplemexican
thank you
welshfella
  • welshfella
yw

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