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Algebra
Pregunta

A company's revenue comes in at a rate of \( R ( t...

A company's revenue comes in at a rate of \( R ( t ) = 10000 e ^ { 0.07 t } \) from the beginning of \( 2008 \) to the beginning \( 2019 \) where \( t = 0\) corresponds with \( 2008 \) . Find the total value at the beginning of \( 2013 \) . 

\(\$ 15,709.20 \) 

\( \$ 59,866.79 \) 

\( \$ 165,680.89 \) 

\( \$ 78,546.37\) 

Answer

option A is correct

Solución
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