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hodyreva [135]
3 years ago
15

in a plantation,there are 230 trees in a row. There are at least 3 papaya trees between any 2 rambutan trees. What is the larges

t possible number of rambutan trees in that row?​
Mathematics
1 answer:
Zinaida [17]3 years ago
7 0

Answer:

  58

Step-by-step explanation:

The total number of trees in a row that starts and ends with a rambutan tree is ...

  4n -3 . . . . for n rambutan trees

So, the number of rambutan trees that there can be in a row of 230 trees is ...

  4n -3 ≤ 230

  4n ≤ 233

  n ≤ 233/4 = 58 1/4

The largest possible number of rambutan trees in a row of 230 trees is 58.

_____

1 rambutan tree: R . . . n=1, tree count = 1

2 rambutan trees: RpppR . . . n=2, tree count = 5

3 rambutan trees: RpppRpppR . . . n=3, tree count = 9

The tree count is an arithmetic sequence with first term 1 and common difference 4. Then the n-th term of the sequence is ...

  tn = a1 +d(n -1) = 1 +4(n -1) = 4n -3

Term 58 is ...

  t58 = 4·58 -3 = 229

So, the row of trees might be ...

  RpppRppp ... RpppRpppRp . . . . with 58 R and 172 p.

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Answer:

Two answers:

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Point m is the midpoint of ab¯¯¯¯¯ . am=2x+9, and ab=8x−50. what is the length of am¯¯¯¯¯¯ ?
iogann1982 [59]
Since m is the midpoint of ab, then the following relationship is fulfilled:
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 We have then:
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 Then, the value of am, is given by substituting x in the expression:
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 Answer:
 
the length of am is:
 
am=43
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