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2 years ago

6

Which represents the solution(s) of the graphed system of equations, y = –x2 x 2 and y = –x 3? (1, 2) (1, 2) and (0, 3) (–1, 0)

and (2, 0) (2, 1)

2 answers:

agasfer [191]2 years ago

8 0

Answer:

(1,2)

Step-by-step explanation:

:)

zlopas [31]2 years ago

4 0

The solution of the graphed system of equations y = -x² + x + 2 and y = -x + 3 is (1,2)

<h3>How to solve the equations?</h3>

The system of equations is given as:

y = -x² + x + 2

y = -x + 3

From the graph in the complete question, we have the following point of intersection between the equations

(x,y) = (1,2)

The point of intersection between the equations is the solution

Hence, the solution of the graphed system of equations is (1,2)

Read more about system of equations at:

brainly.com/question/14323743

#SPJ4

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3 years ago

Solve these linear equations in the form y=yn+yp with yn=y(0)e^at.

WINSTONCH [101]

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a) $y(t) = y_{0}e^{4t} + 2$ . It does not have a steady state

b) $y(t) = y_{0}e^{-4t} + 2$ . It has a steady state.

Step-by-step explanation:

a) $y' -4y = -8$

The first step is finding $y_{n}(t)$ . So:

$y' - 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r - 4 = 0$

$r = 4$

So:

$y_{n}(t) = y_{0}e^{4t}$

Since this differential equation has a positive eigenvalue, it does not have a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' -4(y_{p}) = -8$

$(C)' - 4C = -8$

C is a constant, so (C)' = 0.

$-4C = -8$

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{4t} + 2$

b) $y' +4y = 8$

The first step is finding $y_{n}(t)$ . So:

$y' + 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r + 4 =$

$r = -4$

So:

$y_{n}(t) = y_{0}e^{-4t}$

Since this differential equation does not have a positive eigenvalue, it has a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' +4(y_{p}) = 8$

$(C)' + 4C = 8$

C is a constant, so (C)' = 0.

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{-4t} + 2$

6 0

3 years ago