The city monitors the steady rise of CO from various sources annually. In the year "C: 2019"<span> (rounded off to the nearest integer) will the CO level exceed the permissible limit.
If this isn't the answer, let me know and i'll figure out what it is. But I believe this is it. :) </span>
Answer:
![\frac{dV}{dt}=-78,4\pi \frac{cm^{3} }{min}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D-78%2C4%5Cpi%20%5Cfrac%7Bcm%5E%7B3%7D%20%7D%7Bmin%7D)
Explanation:
Knowing that the volume of a sphere is V=(4/3)πr³ and ![\frac{dr}{dt}=-0.1\frac{cm}{min}](https://tex.z-dn.net/?f=%5Cfrac%7Bdr%7D%7Bdt%7D%3D-0.1%5Cfrac%7Bcm%7D%7Bmin%7D)
We must find
when r=14cm
V=(4/3)πr³ ⇒
![\frac{dV}{dt}=\frac{4}{3}\pi3r^{2}\frac{dr}{dt}\\\frac{dV}{dt}=4\pi r^{2}(-0.1\frac{cm}{min})](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi3r%5E%7B2%7D%5Cfrac%7Bdr%7D%7Bdt%7D%5C%5C%5Cfrac%7BdV%7D%7Bdt%7D%3D4%5Cpi%20r%5E%7B2%7D%28-0.1%5Cfrac%7Bcm%7D%7Bmin%7D%29)
and r=14cm then
![\frac{dV}{dt}=4\pi(14cm)^{2}(-0.1\frac{cm}{min})\\\frac{dV}{dt}=4\pi196cm^{2}(-0.1\frac{cm}{min})\\\frac{dV}{dt}=-78,4\pi \frac{cm^{3} }{min}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D4%5Cpi%2814cm%29%5E%7B2%7D%28-0.1%5Cfrac%7Bcm%7D%7Bmin%7D%29%5C%5C%5Cfrac%7BdV%7D%7Bdt%7D%3D4%5Cpi196cm%5E%7B2%7D%28-0.1%5Cfrac%7Bcm%7D%7Bmin%7D%29%5C%5C%5Cfrac%7BdV%7D%7Bdt%7D%3D-78%2C4%5Cpi%20%5Cfrac%7Bcm%5E%7B3%7D%20%7D%7Bmin%7D)
Note: as you can see the relationship of change of r with respect to t is negative because it is a decrease, and also its volume ratio
<span>a subatomic particle of a type including the baryons and mesons that can take part in the strong interaction</span>
10pi is the answer. i believe. i hope this helps.