It's unlikely that it the dimensions are either 14 and 1 in front of the x or 6 and 1 for the 6, but this is math, so you never know. I would try 7 and 2 and 3 and 2 and see what comes of it.
(7x 2)(2x 3 ) That looks promising. We'll put the 3 with the 2x. That leaves us with the signs. We need 17 to be minus
(7x 2)(2x - 3) that gives you - 21 x So the 2 with the seven has to be plus.
(7x + 2)(2x - 3) should be the the factored answer.
7x + 2 = 0 has no meaning. Measurements cannot be minus.
2x - 3 = 0
2x = 3
x = 3/2
If there are choices, please list them. That's as far as is possible.
Answer:
no it is false.
Step-by-step explanation:
Because The left side 12 does not equal to the right side 16, which means that the given statement is false.
Answer:
![y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B-2x%7D%5B3cos%28%5Csqrt%7B6%7Dx%29%2B%5Cfrac%7B2%5Csqrt%7B6%7D%7D%7B3%7Dsin%28%5Csqrt%7B6%7Dx%29%5D)
(See attached graph)
Step-by-step explanation:
To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation 
where the values of 
are the roots:
Since the values of are complex conjugate roots, then the general solution is ![y(x)=e^{\alpha x}[C_1cos(\beta x)+C_2sin(\beta x)]](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B%5Calpha%20x%7D%5BC_1cos%28%5Cbeta%20x%29%2BC_2sin%28%5Cbeta%20x%29%5D)
where 
.
Thus, the general solution for our given differential equation is ![y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B-2x%7D%5BC_1cos%28%5Csqrt%7B6%7Dx%29%2BC_2sin%28%5Csqrt%7B6%7Dx%29%5D)
.
To account for both initial conditions, take the derivative of 
, thus, ![y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]](https://tex.z-dn.net/?f=y%27%28x%29%3D-2e%5E%7B-2x%7D%5BC_1cos%28%5Csqrt%7B6%7Dx%2BC_2sin%28%5Csqrt%7B6%7Dx%29%5D%2Be%5E%7B-2x%7D%5B-C_1%5Csqrt%7B6%7Dsin%28%5Csqrt%7B6%7Dx%29%2BC_2%5Csqrt%7B6%7Dcos%28%5Csqrt%7B6%7Dx%29%5D)
Now, we can create our system of equations given our initial conditions:
![y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(0)=e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]=3\\\\C_1=3](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B-2x%7D%5BC_1cos%28%5Csqrt%7B6%7Dx%29%2BC_2sin%28%5Csqrt%7B6%7Dx%29%5D%5C%5C%5C%5Cy%280%29%3De%5E%7B-2%280%29%7D%5BC_1cos%28%5Csqrt%7B6%7D%280%29%29%2BC_2sin%28%5Csqrt%7B6%7D%280%29%29%5D%3D3%5C%5C%5C%5CC_1%3D3)
![y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]\\\\y'(0)=-2e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]+e^{-2(0)}[-C_1\sqrt{6}sin(\sqrt{6}(0))+C_2\sqrt{6}cos(\sqrt{6}(0))]=-2\\\\-2C_1+\sqrt{6}C_2=-2](https://tex.z-dn.net/?f=y%27%28x%29%3D-2e%5E%7B-2x%7D%5BC_1cos%28%5Csqrt%7B6%7Dx%2BC_2sin%28%5Csqrt%7B6%7Dx%29%5D%2Be%5E%7B-2x%7D%5B-C_1%5Csqrt%7B6%7Dsin%28%5Csqrt%7B6%7Dx%29%2BC_2%5Csqrt%7B6%7Dcos%28%5Csqrt%7B6%7Dx%29%5D%5C%5C%5C%5Cy%27%280%29%3D-2e%5E%7B-2%280%29%7D%5BC_1cos%28%5Csqrt%7B6%7D%280%29%29%2BC_2sin%28%5Csqrt%7B6%7D%280%29%29%5D%2Be%5E%7B-2%280%29%7D%5B-C_1%5Csqrt%7B6%7Dsin%28%5Csqrt%7B6%7D%280%29%29%2BC_2%5Csqrt%7B6%7Dcos%28%5Csqrt%7B6%7D%280%29%29%5D%3D-2%5C%5C%5C%5C-2C_1%2B%5Csqrt%7B6%7DC_2%3D-2)
We then solve the system of equations, which becomes easy since we already know that 
:
Thus, our final solution is:
![y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B-2x%7D%5BC_1cos%28%5Csqrt%7B6%7Dx%29%2BC_2sin%28%5Csqrt%7B6%7Dx%29%5D%5C%5C%5C%5Cy%28x%29%3De%5E%7B-2x%7D%5B3cos%28%5Csqrt%7B6%7Dx%29%2B%5Cfrac%7B2%5Csqrt%7B6%7D%7D%7B3%7Dsin%28%5Csqrt%7B6%7Dx%29%5D)
Equation 1) -4x + 7y = 1
Equation 2) 25 = 2x + 5y
For both equations, move the variables all onto one side, preferably the left side.
To do so, for equation 2, subtract 2x from both sides, as well as subtracting 5y from both sides, and then subtracting 25 from both sides.
2) -2x - 5y = -25
1) -4x + 7y = 1
Multiply all of equation 2 by 2.
2) 2(-2x - 5y = -25)
2) -4x - 10y = -50
1) -4x + 7y = 1
Subtract the equations from each other.
-17y = -51
Divide both sides by -17.
y = 51/17
y = 3
Plug in 3 for y in the first equation.
-4x + 7y = 1
-4x + 7(3) = 1
-4x + 21 = 1
-4x = 1 - 21
-4x = -25
x = -25/-4
x = 25/4
x = 6 1/4
(6 1/4, 3)
~Hope I helped!~
