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:
x=1
y=s
z=1
Step-by-step explanation:
(x, y, z)=(1, 0, 1)
Substitute 0 for y

Confirming if t=0 satisfy the other equation
x = e^−2t cos 4t = e^−2(0)cos(4*0)
= e^(0)cos(0) = 1
z = e^−2t = e^−2(0) = 0
Therefore t=0 satisfies the other equation
Finding the tangent vector at t=0

The vector equation of the tangent line is
(1, 0, 1) +s(0,1,0)= (1, s, 1)
The parametric equations are:
x=1
y=s
z=1
Hello from MrBillDoesMath!
Answer:
5 x^3 + 15 x^2 + 15 x + 5 , none of the provided choices
Discussion:
f(x) = 5 x^3
g(x) = x+ 1
=>
(f•g)(x) =
f(g(x)) =
f(x+1) =
5 * (x+1)^3 =
5 x^3 + 15 x^2 + 15 x + 5
which is none of the provided answers.
Thank you,
MrB
Answer:
10.5
Step-by-step explanation:
First, divide 6 by 2. You get 3, right? So now you know that you multiply by three to get the answer. 3.5 * 3 will equal 10.5
“Times as fast” means you need to multiply the original speed (10^11) by how many times faster the faster processor works:
10^11 instructions/second • 10^3 faster = 10^14 instructions/second
The key is “___ times as fast”.