Answer:
A. -12h² - 22h + 14
Step-by-step explanation:
(-4h +2)(3h +7) = -4h(3h +7) +2(3h +7) . . . . . . . (a +b)c = ac +bc
= (-4h)(3h) + (-4h)(7) + (2)(3h) + (2)(7) . . . . . . . a(b +c) = ab +ac . . . (twice)
= -12h² -28h +6h +14
= -12h² -22h +14 . . . . . . . . collect terms
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)
the answer to your question is A.
Answer:
The point
is not a solution of the system of inequalities
Step-by-step explanation:
we have
-----> inequality A
-----> inequality B
we know that
If a ordered pair is a solution of the system of inequalities
then
the ordered pair must be satisfy the inequalities of the system
Verify
For ![x=0, y=0](https://tex.z-dn.net/?f=x%3D0%2C%20y%3D0)
substitute the value of x and the value of y in the inequalkity A and in the inequality B
Inequality A
![0\leq 0^{2} -4](https://tex.z-dn.net/?f=0%5Cleq%200%5E%7B2%7D%20-4)
-------> is not true
therefore
The point
is not a solution of the system of inequalities