document.write( "Question 1007192: The annual revenue R, in dollars, of a new company can be closely modeled by the logistic function
\n" ); document.write( "R(t) =
\n" ); document.write( "615,000/1 + 3.6e^−0.044t
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\n" ); document.write( "where the natural number t is the time, in years, since the company was founded.
\n" ); document.write( "(a) According to the model, what will be the company's annual revenue for its first year and its second year (t = 1 and t = 2) of operation? Round to the nearest $1000.
\n" ); document.write( "R(1) = $
\n" ); document.write( "R(2) = $ \r
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\n" ); document.write( "\n" ); document.write( "(b) According to the model, what will the company's annual revenue approach in the long-term future?
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Algebra.Com's Answer #623171 by fractalier(6550)\"\" \"About 
You can put this solution on YOUR website!
You merely plug in the values 1 and 2 in for t in the formula
\n" ); document.write( "\"R%28t%29+=+615%2C000%2F%281+%2B+3.6%2Ae%5E%28%26%238722%3B0.044t%29%29\" and get
\n" ); document.write( "R(1) = 615,000/(1 + 3.6*e^(−0.044)) = $138,356.64
\n" ); document.write( "R(2) = 615,000/(1 + 3.6*e^(−0.044*2)) = $143,131.80
\n" ); document.write( "Then we look at what would happen to R(t) if t gets large...you can plug in 1000 for t to see that...
\n" ); document.write( "R(1000) = 615,000/(1 + 3.6*e^(−0.044*1000)) = almost 615,000
\n" ); document.write( "As you can see, the revenue approaches $615,000.
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