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?
Easy peasy
the average rate of change in section A is the slope from (1,g(1)) to (2,g(2))
the average rate of chagne in section B is the slope from (3,g(3)) to (4,g(4))
A.
section A
g(1)=4(3)^1=12
g(2)=4(3)^2=4(9)=36
slope=(36-12)/(2-1)=24/1=24
section B
g(3)=4(3)^3=4(27)=108
g(4)=4(3)^4=4(81)=324
slope=(324-108)/(4-3)=216/1=216
section A has an average rate of change of 24
section B has an average rate of change of 216
Answer:
4.F(x) > 0 over the intervals (-0.7, 0.76) and (0.76, ∞).
Step-by-step explanation:
Answer:
Question
Match each expression with the equivalent expanded expression.
0.5 (2x + 3)
- 10x - 3
1/3 ( 9 +81)
8x - 4
-5 (2x + 3/5)
-2x + 3
2 (4x-2)
x + 1.5
-3 (2/3x - 1)
30
Answer:
Step-by-step explanation:
Step-by-step explanation:
Question
Match each expression with the equivalent expanded expression.
0.5 (2x + 3)
- 10x - 3
1/3 ( 9 +81)
8x - 4
-5 (2x + 3/5)
-2x + 3
2 (4x-2)
x + 1.5
-3 (2/3x - 1)
30
Answer:
Step-by-step explanation:
In a trapazoid the diagonals are the same so therefore the two equations equal echother
3x+7=5x-11 solve for x by combining like terms
18=2x then divide to unto multiplication
9=x
9 is the value of x
I hope I've helped!
