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?
Answer:
6/9
Step-by-step explanation:
I believe that is 275 because percent is just out of one hundred
Answer:
b.
Step-by-step explanation:
We have to look at sign changes in f(x) to determine the possible positive real roots.

There is only one sign change here, between the -8x and the +4. So that means there is only 1 possible real positive root.
Now we have to look at sign changes in f(-x) to determine the possible negative real roots.

There are 3 sign changes here. That means there are either 3 negative roots or 3-2 = 1 negative root. So we have:
1 positive
3 or 1 negative
We need to pair them up now with all the possible combinations.
If we have 1 positive and 1 negative, we have to have 2 imaginary
If we have 1 positive and 3 negative, we have to have 0 imaginary
Keep in mind that the total number or roots--positive, negative, imaginary--have to add up to equal the degree of the polynomial. This is a 4th degree polynomial, so we will have 4 roots.
A proportional relationship is described by the equation
... y = k·x
The point (x, y) = (0, 0) is <em>always</em> a solution to this equation.
_____
In short, if the relationship is proportional, its graph will go through the origin. If the graph does not go through the origin, the relationship is not proportional.
___
Note that this is true if the domain includes the origin. You can have y = kx <em>for x > 10 </em>and the graph will <em>not</em> go through the origin because the function is <em>not defined</em> there.