so the first term is 36500.
after a year, he gets 2375 extra, so
1st year, 36500
2nd year 36500 + 2375
3rd year 36500 + 2375 + 2375
4th year 36500 + 2375 + 2375 + 2375
and so on
so the common difference is 2375, namely the number we add in order to get the following term.
![\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\[-0.5em] \hrulefill\\ a_1=36500\\ d=2375 \end{cases} \\\\\\ a_n=36500+(n-1)2375\implies a_n=36500+2375(n-1) \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{his salary after 15 years, \boxed{n = 15}}}{a_{15}=36500+2375(15-1)}\implies a_{15}=36500+2375(14) \\\\\\ a_{15}=36500+33250\implies a_{15}=69750](https://tex.z-dn.net/?f=%5Cbf%20n%5E%7Bth%7D%5Ctextit%7B%20term%20of%20an%20arithmetic%20sequence%7D%20%5C%5C%5C%5C%20a_n%3Da_1%2B%28n-1%29d%5Cqquad%20%5Cbegin%7Bcases%7D%20n%3Dn%5E%7Bth%7D%5C%20term%5C%5C%20a_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%20d%3D%5Ctextit%7Bcommon%20difference%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a_1%3D36500%5C%5C%20d%3D2375%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20a_n%3D36500%2B%28n-1%292375%5Cimplies%20a_n%3D36500%2B2375%28n-1%29%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bhis%20salary%20after%2015%20years%2C%20%5Cboxed%7Bn%20%3D%2015%7D%7D%7D%7Ba_%7B15%7D%3D36500%2B2375%2815-1%29%7D%5Cimplies%20a_%7B15%7D%3D36500%2B2375%2814%29%20%5C%5C%5C%5C%5C%5C%20a_%7B15%7D%3D36500%2B33250%5Cimplies%20a_%7B15%7D%3D69750)
Answer:
D) 
Step-by-step explanation:
It's going up so it has to be greater than sign, and it's dotted
F=x-48 over 4x. I hope that is right :)
Y - y₁ = m(x - x₁)
y - 5 = -8(x - 2)
y - 5 = -8(x) + 8(2)
y - 5 = -8x + 16
<u> + 5 + 5</u>
y = -8x + 21
I have attached the graph of y = -x + 1.5
The line goes through the points (0,1.5) and (1.5,0)
