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fiasKO [112]
3 years ago
10

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.
Part B: How many homes does Neighborhood A have after 5 years? How many does Neighborhood B have after the same number of years?
Part C: After approximately how many years is the number of homes in Neighborhood A and Neighborhood B the same? Justify your answer mathematically.
Mathematics
2 answers:
tatyana61 [14]3 years ago
7 0
Part A
y=number of years (in both equations)
F(a)=30(1+r/n)^y
r=.2(20%)
n=1

F(b)=45+(3y)

Part B
A=74.6496 (personally I'd round to 74 because we are talking about an item that's not complete/useable yet, vs just a number if rounding is even required)
B=60

Part C
y=3.32 or approx 3 and 1/3 years
solve both and each neighborhood has about 55 houses 
54.95 for a
54.96 for b

ryzh [129]3 years ago
6 0
A.
this is assuming it grows by 20% of the houses it has at that time (compound interest)
A(x)=30(1.2)ˣ⁻¹ or A(x)=25(1.2)ˣ
B(x)=3(x-1)+45 or B(x)=3x+42





B.
A(5)=62.28 or about 62 homes
B(5)=60


C.
this is tricky so I think it should have been only 20% of 30







start over

A.
20% of 30=6
A(x)=6(x-1)+30 or A(x)=6x+24
B(x)=3(x-1)+45 or B(x)=3x+42





B.
A(5)=54 homes
B(5)=60


C.
6x+24=3x+42
minus 3x
3x+24=42
minus 24 both sides
3x=18
divide by 3
x=6

after 6 years
both will have 63 homes
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<u>ANSWER: </u>

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<u>SOLUTION: </u>

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