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vazorg [7]
3 years ago
12

HBHBVHBHEBCB PLEASE HELP WHAT IS : x/6= -2

Mathematics
1 answer:
Fiesta28 [93]3 years ago
7 0
I think x = -12 , if that’s what the questions asking
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Which point is a solution to the inequality shown in this graph?
kykrilka [37]

Answer:

The only solution can be (0,-3) point.

Step-by-step explanation:

We have to judge whether the points in options are the solution to the graphed inequality or not.

The first point is (5,-5) which not included in the shaded region of the graph. Hence, it can not be a solution.

The second point is (6,0) which not included in the shaded region of the graph. Hence, it can not be a solution.

The third point is (0,-5) which not included in the shaded region of the graph. Hence, it can not be a solution.

The fourth point is (0,-3). It is on the firm red line which is included in the shaded region of the graph. Hence, it is a solution.

Therefore, the only solution can be (0,-3) point. (Answer)

5 0
3 years ago
Solve y over negative 2 + 5 = 13.
sattari [20]
Y/(-2+5)=13
y/(3)=13
y=39
6 0
3 years ago
Help pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
tamaranim1 [39]

The answer is the third one: 32 inches

3 0
3 years ago
Find the quotient -88, 11
irinina [24]
The answer is -88/11=-8
4 0
3 years ago
Read 2 more answers
Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).
I am Lyosha [343]
Take the homogeneous part and find the roots to the characteristic equation:

y''+4y=0\implies r^2+4=0\implies r=\pm2i

This means the characteristic solution is y_c=C_1\cos2x+C_2\sin2x.

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form y_p=ax\cos2x+bx\sin2x. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With y_1=\cos2x and y_2=\sin2x, you're looking for a particular solution of the form y_p=u_1y_1+u_2y_2. The functions u_i satisfy

u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx

where W(y_1,y_2) is the Wronskian determinant of the two characteristic solutions.

W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2

So you have

u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx
u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x

u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx
u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x

So you end up with a solution

u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x

but since \cos2x is already accounted for in the characteristic solution, the particular solution is then

y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x

so that the general solution is

y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x
7 0
3 years ago
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