Step-by-step explanation:
2250mg = 2250/1000 = 2.25g
59.4kilolitres = 59.4 ×1000 = 59400litres
0.000072g = 0.000072x 1000000 = 72micrograms
3.98 x 10^6 in = 3.98 × 10^6 × 25.4 × 10^-6 = 1.01092 × 10^2 km
or
3.98 × 10^6 m = 3.98 × 10^6 × 10^-3 = 3.98 × 10^3
Answer:
x = 0
Step-by-step explanation:
![x^3-7x^2-x =0\\](https://tex.z-dn.net/?f=x%5E3-7x%5E2-x%20%3D0%5C%5C)
Take out common factor
![x(x^2-7x-1) = 0](https://tex.z-dn.net/?f=x%28x%5E2-7x-1%29%20%3D%200)
The factor inside the parenthesis cannot be factored
Since there is a common factor that we pulled out of the original equation,
set that x equal to zero and solve
: Therefore x = 0
Let
![y=C_1x+C_2x^3=C_1y_1+C_2y_2](https://tex.z-dn.net/?f=y%3DC_1x%2BC_2x%5E3%3DC_1y_1%2BC_2y_2)
. Then
![y_1](https://tex.z-dn.net/?f=y_1)
and
![y_2](https://tex.z-dn.net/?f=y_2)
are two fundamental, linearly independent solution that satisfy
![f(x,y_1,{y_1}',{y_1}'')=0](https://tex.z-dn.net/?f=f%28x%2Cy_1%2C%7By_1%7D%27%2C%7By_1%7D%27%27%29%3D0)
![f(x,y_2,{y_2}',{y_2}'')=0](https://tex.z-dn.net/?f=f%28x%2Cy_2%2C%7By_2%7D%27%2C%7By_2%7D%27%27%29%3D0)
Note that
![{y_1}'=1](https://tex.z-dn.net/?f=%7By_1%7D%27%3D1)
, so that
![x{y_1}'-y_1=0](https://tex.z-dn.net/?f=x%7By_1%7D%27-y_1%3D0)
. Adding
![y''](https://tex.z-dn.net/?f=y%27%27)
doesn't change this, since
![{y_1}''=0](https://tex.z-dn.net/?f=%7By_1%7D%27%27%3D0)
.
So if we suppose
![f(x,y,y',y'')=y''+xy'-y=0](https://tex.z-dn.net/?f=f%28x%2Cy%2Cy%27%2Cy%27%27%29%3Dy%27%27%2Bxy%27-y%3D0)
then substituting
![y=y_2](https://tex.z-dn.net/?f=y%3Dy_2)
would give
![6x+x(3x^2)-x^3=6x+2x^3\neq0](https://tex.z-dn.net/?f=6x%2Bx%283x%5E2%29-x%5E3%3D6x%2B2x%5E3%5Cneq0)
To make sure everything cancels out, multiply the second degree term by
![-\dfrac{x^2}3](https://tex.z-dn.net/?f=-%5Cdfrac%7Bx%5E2%7D3)
, so that
![f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y](https://tex.z-dn.net/?f=f%28x%2Cy%2Cy%27%2Cy%27%27%29%3D-%5Cdfrac%7Bx%5E2%7D3y%27%27%2Bxy%27-y)
Then if
![y=y_1+y_2](https://tex.z-dn.net/?f=y%3Dy_1%2By_2)
, we get
![-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0](https://tex.z-dn.net/?f=-%5Cdfrac%7Bx%5E2%7D3%280%2B6x%29%2Bx%281%2B3x%5E2%29-%28x%2Bx%5E3%29%3D-2x%5E3%2Bx%2B3x%5E3-x-x%5E3%3D0)
as desired. So one possible ODE would be
![-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0](https://tex.z-dn.net/?f=-%5Cdfrac%7Bx%5E2%7D3y%27%27%2Bxy%27-y%3D0%5Ciff%20x%5E2y%27%27-3xy%27%2B3y%3D0)
(See "Euler-Cauchy equation" for more info)
Answer:
43.96 is pretty sure
Step-by-step explanation:
C=2πr
c= 2π(7)
using 3.13 as pie
c= 2(3.14)(7)
43.96