Answer:
A. y = 1x + 4
B. 1
C. 4
Step-by-step explanation:
The point (0, 4) is the y-intercept. So b = 4
y goes up each time by 1. So m = 1
Plug into y = mx + b
y = 1x + 4
T = days passed
r = rate of growth
by 0 day, or t = 0, there are 2 folks sick,

by the third day, t = 3, there are 40 folks sick,
![\bf \qquad \textit{Amount for Exponential Growth} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\to &40\\ P=\textit{initial amount}\to &2\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\to &3\\ \end{cases} \\\\\\ 40=2(1+r)^3\implies 20=(1+r)^3\implies \sqrt[3]{20}=1+r \\\\\\ \sqrt[3]{20}-1=r\implies 1.7\approx r\qquad \boxed{A=2(2.7)^t}](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7BAmount%20for%20Exponential%20Growth%7D%0A%5C%5C%5C%5C%0AA%3DP%281%20%2B%20r%29%5Et%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D%5Ctextit%7Baccumulated%20amount%7D%5Cto%20%2640%5C%5C%0AP%3D%5Ctextit%7Binitial%20amount%7D%5Cto%20%262%5C%5C%0Ar%3Drate%5Cto%20r%5C%25%5Cto%20%5Cfrac%7Br%7D%7B100%7D%5C%5C%0At%3D%5Ctextit%7Belapsed%20time%7D%5Cto%20%263%5C%5C%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A40%3D2%281%2Br%29%5E3%5Cimplies%2020%3D%281%2Br%29%5E3%5Cimplies%20%5Csqrt%5B3%5D%7B20%7D%3D1%2Br%0A%5C%5C%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7B20%7D-1%3Dr%5Cimplies%201.7%5Capprox%20r%5Cqquad%20%5Cboxed%7BA%3D2%282.7%29%5Et%7D)
how many folks are there sick by t = 6?
The first 5 outputs are:

As you can see, the outputs keep doubling each time we increment x by 1.
This can be written formally, observing that if you know the value of
, the next value will be

So, again, we've shown that the next value is twice the previous one, so you have

5 divided by one half, would be 2.5, since its half of 5.
I believe the answer is 16/9