The function m(n) = 20(1.2)n – 1 approximates the number of members in a club n years after it was started, and the function f(n) = 15 models the yearly fee each member pays during year n.

Which function approximates the total amount of money collected for yearly fees in year n?

t(n) = 360^n – 1
t(n) = 300(1.2)^n – 1

Question

The function m(n) = 20(1.2)n – 1 approximates the number of members in a club n years after it was started, and the function f(n) = 15 models the yearly fee each member pays during year n.

Which function approximates the total amount of money collected for yearly fees in year n?

t(n) = 360^n – 1
       t(n) = 300(1.2)^n – 1 <<<
       t(n) = 35(1.2)^n – 1
       t(n) = 20(1.2)^n – 1+ 15

solomonzelman · Answer

$\large\color{blue}{    m(n) = 20(1.2)^{n-1}   }$
$\large\color{blue}{    f(n) = 15  }$

$\large\color{blue}{    f(n) \times m(n)  }$

solomonzelman · Answer

f(n) is a cost per person, and m(n) is a number of people.
So the product of the two, would be the cost for all people in a certain year.

anonymous · Answer

@jim_thompson5910 @SolomonZelman So, I'm right? :)

solomonzelman · Answer

no.

solomonzelman · Answer

lets call what the m(n) equals, "D"
So, this is what you are saying.

n per D = n+D

solomonzelman · Answer

it is more like multiplication.

solomonzelman · Answer

yes, B is right (adited), smart choice...

solomonzelman · Answer

D wouldn't be right, but B is.