Create a recursive function to model the table of values from part a. in that function, the year 2013 is the first data point, so n = 1 for 2013 (meaning f(1) = 2,000 is the first term of the sequence) and r is the constant value by which the number increases every year. f(n)=

In year 1 (2013) we had 2000 trees. This is our starting point

In year 2 (2014) we add 200 new trees, bringing the total to 2200.

In year 3 200 2400.

and so on.

So: the number of trees we have in the nth year is the number of trees in the

previous year, plus 200. This is a recursive function.

a(n+1) = a(n) + 200, where a(1) =2000.

Note that a(2) = 2200

a(4) = 2600

a(5) = 2800

and so on.

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.