we are given
differential equation as
![y'''-9y''+20y'=0](https://tex.z-dn.net/?f=y%27%27%27-9y%27%27%2B20y%27%3D0)
we are given
![y=e^{rx}](https://tex.z-dn.net/?f=y%3De%5E%7Brx%7D)
Firstly, we will find y' , y'' and y'''
those are first , second and third derivative
First derivative is
![y'=re^{rx}](https://tex.z-dn.net/?f=y%27%3Dre%5E%7Brx%7D)
Second derivative is
![y''=r*re^{rx}](https://tex.z-dn.net/?f=y%27%27%3Dr%2Are%5E%7Brx%7D)
![y''=r^2e^{rx}](https://tex.z-dn.net/?f=y%27%27%3Dr%5E2e%5E%7Brx%7D)
Third derivative is
![y'''=r^2*re^{rx}](https://tex.z-dn.net/?f=y%27%27%27%3Dr%5E2%2Are%5E%7Brx%7D)
![y'''=r^3e^{rx}](https://tex.z-dn.net/?f=y%27%27%27%3Dr%5E3e%5E%7Brx%7D)
now, we can plug these values into differential equation
and we get
![r^3 e^{rx}-9r^2 e^{rx}+20re^{rx}=0](https://tex.z-dn.net/?f=r%5E3%20e%5E%7Brx%7D-9r%5E2%20e%5E%7Brx%7D%2B20re%5E%7Brx%7D%3D0)
now, we can factor out common terms
![e^{rx}(r^3 -9r^2 +20r)=0](https://tex.z-dn.net/?f=e%5E%7Brx%7D%28r%5E3%20-9r%5E2%20%2B20r%29%3D0)
we can move that term on right side
![(r^3 -9r^2 +20r)=0](https://tex.z-dn.net/?f=%28r%5E3%20-9r%5E2%20%2B20r%29%3D0)
now, we can factor out
![r(r^2 -9r +20)=0](https://tex.z-dn.net/?f=r%28r%5E2%20-9r%20%2B20%29%3D0)
![r(r-5)(r-4)=0](https://tex.z-dn.net/?f=r%28r-5%29%28r-4%29%3D0)
now, we can set them equal
![r=0](https://tex.z-dn.net/?f=r%3D0)
![r-5=0](https://tex.z-dn.net/?f=r-5%3D0)
![r=5](https://tex.z-dn.net/?f=r%3D5)
![r-4=0](https://tex.z-dn.net/?f=r-4%3D0)
![r=4](https://tex.z-dn.net/?f=r%3D4)
so, we will get
...............Answer
Answer:
-5.457875
Step-by-step explanation:
Length should be 2/5yd:
18inch*1yd/36inch=.5yd=> depth
volume= 2(length*width+lenght*depth+depth*width)
let l to be the length.
volume= 2(l*2+2(.5)+.5*2)
volume=5l+2
4=5l+2
2=5l
l=2/5
A.<span>(x+3)(x+4)=<span>0
x=-3,-4
B.X=5,4
hope this helps
</span></span>