Question

the table below shows the salary that ellen earned at her job based on her years of experience.

years of experience	salary
5	$32,625
6	$33,350
7	$34,075
8	$34,800
9	$35,525
10	$36,250

which explicit formula can be used to determine the salary, f(n), that ellen earned with n years of experience?

Explanation:

Step1: Identify the sequence type

Check the difference between consecutive salaries. For \( n = 4 \), salary is \( 31900 \); \( n = 5 \), \( 32625 \). The difference is \( 32625 - 31900 = 725 \). Check another pair: \( 33350 - 32625 = 725 \). So it's an arithmetic sequence with common difference \( d = 725 \).

Step2: Recall arithmetic sequence formula

The explicit formula for an arithmetic sequence is \( f(n)=f(1)+(n - 1)d \), but here \( n \) starts at 4. Let's adjust. Let \( n \) be the years of experience. When \( n = 4 \), \( f(4)=31900 \). The formula can also be written as \( f(n)=f(4)+(n - 4)d \). Substitute \( f(4)=31900 \) and \( d = 725 \):

\( f(n)=31900+725(n - 4) \)

Simplify: \( f(n)=31900 + 725n-2900=725n + 29000 \) (Wait, let's check with \( n = 4 \): \( 725\times4+29000 = 2900 + 29000 = 31900 \), correct. \( n = 5 \): \( 725\times5+29000 = 3625+29000 = 32625 \), correct.)

Alternatively, the general form for arithmetic sequence with first term (when \( n = 4 \) as a starting point) can be modeled as \( f(n)=725n + c \). Plug \( n = 4 \), \( 31900 = 725\times4 + c \Rightarrow 31900 = 2900 + c \Rightarrow c = 29000 \). So \( f(n)=725n + 29000 \).

Answer:

The explicit formula is \( \boldsymbol{f(n)=725n + 29000} \) (or equivalent simplified form from the arithmetic sequence model).