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3 years ago

What are the solutions to the system of equations graphed below?

Mathematics

2 answers:

Digiron [165]3 years ago

8 0

Answer: (2,0) (0,-4)

Step-by-step explanation:

Finger [1]3 years ago

7 0

You haven't provided the graph, therefore, I cannot give an exact answer.
However, I will help you with the concept.

A solution to a system of equations would be one or more order pairs that satisfy ALL equations given in the system.

There are two ways to get the solution of a system of equations:
1- algebraically: 
This would be by solving the equations in the system simultaneously using elimination/substitution,...etc

2- graphically:
This would be by graphing all the equations given in the system and finding the points of intersection between the graphs.
Note that:
If the given system has two equations, then we need to find the point(s) of intersection between the two graphs
If the given system has three equations, then we need to find the point(s) of intersection between the three graphs

Therefore, for your question, all you have to do is find the point(s) of intersection between your graphs.

Hope this helps:)

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The life spans, in years, of some plants in a nursery are shown below:

yKpoI14uk [10]

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Explanation:

7 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLIEST

Alja [10]

A. The pair of linear equations are: x + y = 55 and y = x + 25

B. Number of time she spend running everyday is: 15 minutes.

C. It is not possible.

<h3>How to Write a System of Linear Equations?</h3>

A system of equations consist of linear equations that have unknown variables whose values make the linear equations in the system true.

Part A: To write a pair of linear equations, do the following,

Let x represent the time Jackie runs

Let y represent the time Jackie dances

We are given that, Jackie runs and dances for a total of 55 minutes every day, therefore, the first linear equation would be:

x + y = 55 --> equation 1

We are also given that Jackie dances for 25 minutes longer than she runs, therefore, the second linear equation would be:

y = x + 25 --> equation 2

Therefore, the two pair of linear equations that shows the relationship between the number of minutes Jackie runs and dances everyday are:

x + y = 55

y = x + 25

Part B: Substitute y = (x + 25) into equation 1

x + y = 55 --> equation 1

x + x + 25 = 55

2x + 25 = 55

2x = 55 - 25

2x = 30

x = 15

Therefore, the number of time she spend running everyday is: 15 minutes.

Part C:

If she spent 45 minutes dancing (y = 45), and x = 15, substitute the values into x + y = 55 to find out if it is possible.

15 + 45 = 55

60 = 55 (not true)

Therefore, it is not possible.

Learn more about the system of equations on:

brainly.com/question/13729904

#SPJ1

8 0

2 years ago

Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term f here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$