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mezya [45]
3 years ago
7

50 POINTS!!! PLEASE HELP!!!

Mathematics
2 answers:
Alisiya [41]3 years ago
6 0

Let <em>t</em> be the time in years, t=0 is now.


Part A


A(t) = 30 (1.20^t)


B(t) = 45 + 3t


Part B


A(5)=  30(1.20^5) = 74.6496  \approx 75


B(5) = 45 + 3(5) = 45+14= 60


Part C


45+3t = 30(1.20^t)


Graphically for positive t these intersect around


t = 3.3 \textrm{ years}




Aleks04 [339]3 years ago
6 0

As with any involved word problem, we're gonna want to figure out what we have, what we're looking for, and find a path to get from our knowns to our unknowns.


We'll start by giving our most important numbers labels - I'll include some justifications, too:


\alpha (t) - The number of homes in neighborhood A (α) after t years

\beta (t) - The number of homes in neighborhood B (β) after t years


Why should we go straight for the functions here? Well, a function, in extremely basic terms, is a machine that takes in a number (an <em>input</em>)and produces another one (an <em>output</em>). In part B, we're given an input - 5 - and we want to see what output we get for our machines, α and β. In part C, we're told something about the outputs: we want to see when they're the same;<em> </em>what does t have to be to make \alpha (t)=\beta (t)?


Before we do any of that, we have to give our machines some rules, though, so let's lock that down first.


For \alpha (t), we're given this: <em>"There are 30 homes in Neighborhood A. Each year, the number of homes increases by 20%." </em>Here are what the first few years look like from that information:


\alpha (0)=30\\\alpha (1)=30 +30(0.2)=30(1+0.2)=30(1.2)\\a(2)=30(1.2) +30(1.2)(0.02)=30(1.2)(1+0.2)=30(1.2)(1.2)=30(1.2)^2


The pattern emerging here is <em>exponential</em> - we start with 30 homes at the beginning (\alpha (0)), and we multiply that starting value by 120% (1.2) again and again to get the number of houses for the following years. In general, we can say that


\alpha (t)=30(1.2)^t


For \beta (t), we're told: <em>"Just down the road, Neighborhood B has 45 homes. Each year, 3 new homes are built in Neighborhood B." </em>The first few years look like this:


\beta (0)=45\\\beta (1)=45 + 3\\\beta (2)=45 + 3 + 3 = 45 + 3(2)


Or, in general:


\beta (t)=45+3t


Unlike \alpha (t), \beta (t)'s growth is <em>linear</em> or <em>constant</em>. While \alpha (t) grows faster as it gets bigger, \beta (t)'s never changes. This means that, in the long term, a(t) will shoot above \beta (t). We'll see roughly when one overtakes the other in just a moment.


Now that we have our functions, we can solve parts B and C:


Part B asks how many homes neighborhoods A and B have after 5 years. Since we've already defined our rules, let's use them:


\alpha (5)=30(1.2)^5=74.6496\approx74\\\beta (5)=45+3(5)=45+15=60


After 5 years, neighborhood A has 74 homes and neighborhood B has 60. You'll notice I rounded the decimal value for \alpha (5) <em>down</em>; we don't count fractions of houses/houses under construction, so in light of that, neighborhood A has 74 <em>complete </em>houses.


For part C, we want to find out when \alpha (t)=\beta (t). We could go for an exact answer, but since the question asks for an approximation, it's fine if we just test some values and make an informed estimate.


Fortunately, we have some information that can help us. Neighborhood A has fewer homes than B at the beginning, but as we just found, A will surpass B by year 5. That means that the point where one passes the other is <em>somewhere between 0-5 years</em>. Let's test some values:


\alpha (1)=30(1.2)=36\\\beta (1)=45 + 3(1) = 48\\\\\alpha (2)=30(1.2)^2=43.2\\\beta(2)=45+3(2)=51\\\\\alpha (3) = 30(1.2)^3=51.84\\\beta (3)=45+3(3)=54\\\\\alpha (4)=30(1.2)^4=62.206\\\beta (4)=45+3(4)=57


So a good estimate for part C might be 3 1/2 years, since \alpha (t) passed \beta (t) when t was somewhere between 3 and 4.

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