3 years ago

Time sensitive. Please help!!

SAT

1 answer:

11Alexandr11 [23.1K]3 years ago

8 0

athletics.

<h2>Hope that helps </h2>

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The population of two different villages are modeled by the equations shown below. The population (in thousands) is represented

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Answer:

1990 and 2010

$y_l_w(15)=315\hspace{3}thousands\\y_l_c(15)=315\hspace{3}thousands\\y_l_w(35)=715\hspace{3}thousands\\y_l_c(35)=715\hspace{3}thousands$

Explanation:

Let:

$y_l_w=Population\hspace{3}of\hspace{3}Lewiston\\y_l_c=Population\hspace{3}of\hspace{3}Lockport$

We need to know, in what year(s) the villages had the same population, mathematically this is:

$y_l_w=y_l_c$

So:

$x^2-30x+540=20x+15\\\\Subtract\hspace{3}20x\hspace{3}from\hspace{3}both\hspace{3}sides\\\\x^2-50x+540=15\\\\Subtract\hspace{3}15\hspace{3}from\hspace{3}both\hspace{3}sides\\\\x^2-50x+525=0$

Solving for x:

Factoring

$(x-15)(x-35)=0$

Hence:

$x=15\\\\or\\\\x=35$

Therefore the year(s) which the village had the same population are:

$1975+15=1990\\\\and\\\\1975+35=2010$

In order to find the population of both cities during the year(s) of equal population, just evalue the equations at x=15 and x=35:

$y_l_w(15)=315\hspace{3}thousands\\y_l_c(15)=315\hspace{3}thousands\\y_l_w(35)=715\hspace{3}thousands\\y_l_c(35)=715\hspace{3}thousands$

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3 years ago