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?
It takes 0.15625 minutes to paint 1 square feet
<em><u>Solution:</u></em>
Given that, I can paint 96 sq ft in 15 minutes
To find: Time taken to paint 1 square feet
From given,
96 square feet is painted in 15 minutes
Let "x" be the time taken to paint 1 square feet
Then, we can say,
96 square feet = 15 minutes
1 square feet = x minutes
This forms a proportion and we can solve the sum by cross multiplying
Thus it takes 0.15625 minutes to paint 1 square feet
Answer:
Step-by-step explanation:
Only one of them has a < b in ax² + by² = c formula, making its graph horizontal
All the rest are vertical, see attached
The line of reflection is b
You can see it much easier if take away all the clutter. Redraw the diagram and take way a d and especially c.
You should be left with only line b and the two triangles. You can see it much easier if you do that. Then just look at the parallel lines of the triangles. They are right across from each other.
