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

Mathematics

2 answers:

Digiron [165]2 years ago

8 0

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

Step-by-step explanation:

Finger [1]2 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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Answer:

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Step-by-step explanation:

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3 years ago

Given the function f(x) = 4(x + 3) -5, solve for the inverse function when x=3

daser333 [38]

Answer: Your answer would be -120

Step-by-step explanation: Remove Parentheses (3 + 3) add them together and make it 6 and plug in the 6 into the equation. -4 x 6 x 5. Then, you multiply the numbers together and get -120. There is no inverse for -120 since both sides aren't equal.

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3 years ago

A circle has the equation 2x²+12x+2y²−16y−150=0.

KonstantinChe [14]

Answer: B. The coordinates of the center are (-3,4), and the length of the radius is 10 units.

Step-by-step explanation:

The equation of a circle in the center-radius form is:

$(x-h)^{2} +(y-k)^{2}=r^{2}$ (1)

Where $(h,k)$ are the coordinates of the center and $r$ is the radius.

Now, we are given the equation of this circle as follows:

$2x^{2}+12x+2y^{2}-16y-150=0$ (2)

And we have to write it in the format of equation (1). So, let's begin by applying common factor 2 in the left side of the equation:

$2(x^{2}+6x+y^{2}-8y-75)=0$ (3)

Rearranging the equation:

$x^{2}+6x+y^{2}-8y=75$ (4)

$(x^{2}+6x)+(y^{2}-8y)=75$ (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of $(a\pm b)^{2}=a^{2}\pm+2ab+b^{2}$ :