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Was part A and B to come up with the equations to model both those situations?
Part A:Neighborhood A:y=30(1.2)^x Neighborhood B: y=40+3x Part B: I just substitute x for 5. Neighborhood A's work: y=30(1.2)^5 y=30*2.48832 y=74.6494, or around 75 houses. Neighborhood B's work: y=40+3*5 y=40+15 y=55, so 55 houses @DanJS my answers^
Part C wants you to figure the years X, when both are equal # houses
30(1.2)^x = 40 + 3x
I know but I dont know what number to start at
Oh, you have 40 for neighborhood B, it is 45 30(1.2)^x = 45 + 3x
So recalculate part B for that one
you sure? Okay so final houses=60
"Neighborhood B has 45 homes. Each year, 3 new homes are built ", less the prob is typed wrong
its right; 45
/?What number should I START with
20% of 30 is 6 so you could just use process of elimination. 3*5=15 15+45=60 5*6=30 30+30=60 so after 5 years they will be the same
30(1.2)^5 does not equal 60
where did you get 30(1.2)^5??
The equation from part one
idk i just went off what your initial question was
well the equations are: Neighborhood A:y=30(1.2)^x Neighborhood B: y=45+3x
the intersection is the point you need
okay so when x=3.32, the two points intersect
3.321 i meant
yes, when 3.32 years have passed , both have same number of houses at near 55
i just asked a question about that form of equation...