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barxatty [35]
3 years ago
6

In a music stadium, there are 18 seats in the first row and 21 seats in the second row. The number of seats in a row continues t

o increase by 3 with each additional row.
(a) Write an explicit rule to model the sequence formed by the number of seats in each row. Show your work.

(b) Use the rule to determine which row has 120 seats. Show your work.
Mathematics
1 answer:
sineoko [7]3 years ago
3 0

Answer:

  (a) an = 18 +3(n -1)

  (b) row 35

Step-by-step explanation:

(a) The general term (an) of an arithmetic sequence can be expressed in terms of the first term (a1) and the common difference (d) as ...

  an = a1 +d(n -1)

The given sequence has a1 = 18 and d = 3, so the explicit formula is ...

  an = 18 + 3(n -1)

__

(b) We want to find n such that an = 120.

  120 = 18 + 3(n -1)

  40 = 6 + n - 1 . . . . . . . divide by 3

  35 = n . . . . . . . . . . . . . subtract 5

Row 35 has 120 seats.

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Pachacha [2.7K]

The capital formation of the investment function over a given period is the

accumulated  capital for the period.

  • (a) The capital formation from the end of the second year to the end of the fifth year is approximately <u>298.87</u>.

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Reasons:

(a) The given investment function is presented as follows;

I(t) = 100 \cdot e^{0.1 \cdot t}

(a) The capital formation is given as follows;

\displaystyle Capital = \int\limits {100 \cdot e^{0.1 \cdot t}} \, dt =1000 \cdot  e^{0.1 \cdot t}} + C

From the end of the second year to the end of the fifth year, we have;

The end of the second year can be taken as the beginning of the third year.

Therefore,  for the three years; Year 3, year 4, and year 5, we have;

\displaystyle Capital = \int\limits^5_3 {100 \cdot e^{0.1 \cdot t}} \, dt \approx 298.87

The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87

(b) When the capital stock exceeds $100,000, we have;

\displaystyle  \mathbf{\left[1000 \cdot  e^{0.1 \cdot t}} + C \right]^t_0} = 100,000

Which gives;

\displaystyle 1000 \cdot  e^{0.1 \cdot t}} - 1000 = 100,000

\displaystyle \mathbf{1000 \cdot  e^{0.1 \cdot t}}} = 100,000 + 1000 = 101,000

\displaystyle e^{0.1 \cdot t}} = 101

\displaystyle t = \frac{ln(101)}{0.1} \approx 46.15

The number of years before the capital stock exceeds $100,000 ≈ <u>46.15 years</u>.

Learn more investment function here:

brainly.com/question/25300925

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