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Crank
2 years ago
8

There are 30 homes in Neighborhood A. Each year, the number of homes increases by 20%. Just down the road, Neighborhood B has 45

homes. Each year, 3 new homes are built in Neighborhood B.
Part A: Write functions to represent the number of homes in Neighborhood A and Neighborhood B throughout the years. (4 points)
Part B: How many homes does Neighborhood A have after 5 years? How many does Neighborhood B have after the same number of years? (2 points)
Part C: After approximately how many years is the number of homes in Neighborhood A and Neighborhood B the same? Justify your answer mathematically. (4 points)
Mathematics
1 answer:
RoseWind [281]2 years ago
4 0
Neighborhood A: 30 homes and increases by 20% per year
Neighborhood B: 45 homes and increases by 3 per year.

N.A = 30 (1.20)^t
N.B = 45 + 3(t)

t = 5

N.A = 30(1.20)^5 = 30(2.48832) = 74.65 or 75
<span>N.B = 45 + 3(5) = 45 + 15 = 60</span>
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Gekata [30.6K]

Answer:

\large\boxed{\dfrac{\boxed{8}}{81}}

Step-by-step explanation:

\text{If}\ a_1,\ a_2,\ a_3,\ a_4,\ ...,\ a_n\ \text{is the geometric sequence, then}\\\\\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=...=\dfrac{a_n}{a_{n-1}}=constant=r-\text{common ration}.\\\\\text{We have}\ \dfrac{1}{2},\ \dfrac{1}{3},\ \dfrac{2}{9},\ \dfrac{4}{27},\ ...\\\\\dfrac{\frac{1}{3}}{\frac{1}{2}}=\dfrac{1}{3}\cdot\dfrac{2}{1}=\dfrac{2}{3}\\\\\dfrac{\frac{2}{9}}{\frac{1}{3}}=\dfrac{2}{9}\cdot\dfrac{3}{1}=\dfrac{2}{3}\\\\\dfrac{\frac{4}{27}}{\frac{2}{9}}=\dfrac{4}{27}\cdot\dfrac{9}{2}=\dfrac{2}{3}

\bold{CORRECT}\\\\\dfrac{x}{\frac{4}{27}}=\dfrac{2}{3}\qquad\text{multiply both sides by}\ \dfrac{4}{27}\\\\x=\dfrac{2}{3}\cdot\dfrac{4}{27}\\\\x=\dfrac{8}{81}

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