Question

A botanist measures a plant, in feet, at the beginning of each month and notices that the measurements form a
geometric sequence, as shown below.
2, 2.2, 2.42, 2.662, ...
If the first measurement was taken on September 1, about how tall was the plant three months before, on June 1,
assuming the same growth pattern?
0.9 ft
1.5 ft
1.7
2.0 ft

Answer 1

Answer:

1.5ft , B  for EDGE2020

Step-by-step explanation:

they are being multiplied by 1.1, so if you divide 1.1 down from 2 three times you get 1.5 roughly.

Answer 2

Using a geometric sequence, it is found that the plant's height on June 1 was of 1.5 ft.

What is a geometric sequence?

A geometric sequence is a sequence in which the result of the division of consecutive terms is always the same, called common ratio q.

The nth term of a geometric sequence is given by:

$a_n = a_1q^{n-1}$

In which $a_1$ is the first term.

In this problem, we have that the first term and the common ratio are given, respectively, by:

$a_1 = 2, q = \frac{2.2}{2} = 1.1$

Hence the rule for the nth term is given by:

$a_n = 2(1.1)^{n-1}$

June 1st would be equivalent to three months before the term 1, hence it is the term 1 - 3 = -2, then:

$a_{-2} = 2(1.1)^{-3} = 1.5$

The plant's height on June 1 was of 1.5 ft.

More can be learned about geometric sequences at https://brainly.com/question/11847927

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