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Eduardwww [97]
1 year ago
13

Which ordered pairs are in the solution set of the system of linear inequalities?

Mathematics
1 answer:
Anna [14]1 year ago
3 0

The ordered pairs (5 , -2) , (3 , 1) , (4 , 2) are in the set of the solution (Option C)

<h3>What is an ordered pair?</h3>

An ordered pair is a composite of the x coordinate (abscissa) and the y coordinate (ordinate), with two values expressed between parenthesis in a predetermined order.

It aids visual comprehension by locating a point on the Cartesian plane.

<h3>How do we arrive at the solution?</h3>

The first line has negative slope and passing through points (0 , 0) and (4 , -2)

That is y > (-1/2)x

The second line has positive slope and passing through points (-2 , 0) and (2 , 2)

That is: y < (1/2)x + 1.

- Refer to the accompanying diagram to discover the common component of the solutions.

- The inequity is shown by the red shading.

- The inequity is shown by the blue shading.

- The two-colored shaded area reflects the common solutions to the two inequalities.

- Let's discover the ordered pairs in the system of linear inequalities' solution set.

- Points (-4, 2), (-3, 1), (4, -3) define the common shaded area.

- Points (5, -2), (3, 1), (3, -1) (4 , 2)

As a result, Point (5, -2) is in the darkened region.

As a result, Point (3, 1) is in the darkened area.

As a result, Point (4, 2) is in the darkened area.

As a result, the ordered pairings (5, -2), (3, 1), (4, 2) are in the solution set.

Learn more about ordered pair at;
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D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x
MA_775_DIABLO [31]
The solution depends on the value of k. To make things simple, assume k>0. The homogeneous part of the equation is

\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0

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which admits the characteristic solution y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}.

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be y_p=ae^{4x}+be^x. Then

\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x

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(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x

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b(1-16k)=30\implies b=\dfrac{30}{1-16k}

and so the general solution would be

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