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Aloiza [94]
3 years ago
5

Which figure correctly demonstrates using a straight line to determine that the graphed equation is not a function of x?

Mathematics
2 answers:
Ulleksa [173]3 years ago
3 0

Answer:

Lower left.

Step-by-step explanation:

To determine if a graph is a function the "vertical line test is used.  If any point on a graph has one x value go to two y values it is not a function.  

Witht he four pictures, the two on the right use a horizontal line, so it doesn't tell at all, on the leftthe upper one is vertical but it doesn't show that one x value has two corresponding y values.  So it must be the lower left one.

The lower left picture has a line at x=1, and it passes through the line of the graph at two different y values, 4 and -4.  So this has two y values with one x value.

Semmy [17]3 years ago
3 0

Answer:

option C

Step-by-step explanation:

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Use the laplace transform to solve the given system of differential equations. dx dt + 3x + dy dt = 1 dx dt − x + dy dt − y = et
blagie [28]

Let X(s) and Y(s) denote the Laplace transforms of x(t) and y(t).

Taking the Laplace transform of both sides of both equations, we have

\dfrac{dx}{dt} + 3x + \dfrac{dy}{dt} = 1 \implies \left(sX(s) - x(0)\right) + 3X(s) + \left(sY(s) - y(0)\right) = \dfrac1s \\\\ \implies (s+3) X(s) + s Y(s) = \dfrac1s

\dfrac{dx}{dt} - x + \dfrac{dy}{dt} = e^t \implies \left(sX(s) - x(0)\right) - X(s) + \left(sY(s) - y(0)\right) = \dfrac1{s-1} \\\\ \implies (s-1) X(s) + s Y(s) = \dfrac1{s-1}

Eliminating Y(s), we get

\left((s+3) X(s) + s Y(s)\right) - \left((s-1) X(s) + s Y(s)\right) = \dfrac1s - \dfrac1{s-1} \\\\ \implies X(s) = \dfrac14 \left(\dfrac1s - \dfrac1{s-1}\right)

Take the inverse transform of both sides to solve for x(t).

\boxed{x(t) = \dfrac14 (1 - e^t)}

Solve for Y(s).

(s - 1) X(s) + s Y(s) = \dfrac1{s-1} \implies -\dfrac1{4s} + s Y(s) = \dfrac1{s-1} \\\\ \implies s Y(s) = \dfrac1{s-1} + \dfrac1{4s} \\\\ \implies Y(s) = \dfrac1{s(s-1)} + \dfrac1{4s^2} \\\\ \implies Y(s) = \dfrac1{s-1} - \dfrac1s + \dfrac1{4s^2}

Taking the inverse transform of both sides, we get

\boxed{y(t) = e^t - 1 + \dfrac14 t}

7 0
1 year ago
Which equation shows the substitution method being used to solve the system of linear equations?
svet-max [94.6K]
The equation
(y + 5) + y = 6
shows the substitution method being used to solve the system of linear equations. So the correct option among all the options given in the question is option "D".
For a better explanation, let us first write down the two equations given in the question.
x + 5 = 6
and
x = y + 5
Now the thing that has been done is that the value of x in the first equation has been replaced by the value of x that is given in the second equation. this is actually the substitution method of solving equations.
6 0
3 years ago
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Flauer [41]

Answer:

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Step-by-step explanation:

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And D. is incorrect because one it is not the correct property to use and also is very inaccurate in the answer. Also, it is a positive outcome.

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Step-by-step explanation:

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