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 child is being pushed on a swing by his father spotted at Imus Plaza, reaching a maximum height of 4 feet. The father stops pushing, and the maximum height of the swing decreases by 15% on each successive swing. What is the maximum height during the 5th swing?​

Respuesta :

jepessoa

Answer:

1.7748 feet

Step-by-step explanation:

maximum height after 1 swing = 4 feet x (1 - 15%) = 3.4 feet

maximum height after 2 swings = 3.4 feet x (1 - 15%) = 2.89 feet

maximum height after 3 swings = 2.89 feet x (1 - 15%) = 2.4565 feet

maximum height after 4 swings = 2.4565 feet x (1 - 15%) = 2.088025 feet

maximum height after 5 swings = 2.088025 feet x (1 - 15%) = 1.7748 feet

Using a geometric sequence to model this situation, it is found that the maximum height during the 5th swing is of 2.09 feet.

In a geometric sequence, the quotient between consecutive terms is always the same, and which is called common ratio.

The nth term of a sequence is given by:

[tex]a_n = a_1q^{n-1}[/tex]

In which:

  • [tex]a_1[/tex] is the first term.
  • q is the common ratio.

In this problem:

  • During the 1st swing, maximum height of 4 feet, thus [tex]a_1 = 4[/tex].
  • Each swing, the height decreases by 15%, thus, the maximum height is 100 - 15 = 85% of the previous swing, which means that [tex]q = 0.85[/tex].

Thus, the nth term is:

[tex]a_n = 4(0.85)^{n-1}[/tex]

For the 5th swing:

[tex]a_5 = 4(0.85)^{5-1} = 2.09[/tex]

The maximum height during the 5th swing is of 2.09 feet.

A similar problem is given at https://brainly.com/question/24138365

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