Mathematics: Which statements are true of functions? check all that apply. all functions have a dependent variable. all functions have an independent variable. the range of a function includes its domain. avertical line is an example of a functional relationship. a horizontal line is an example of a functional relationship. each output value of a function can correspond to only one input value.

Which statements are true of functions? check all

Ответ:

heatherlemaster

03.01.2022

Mathematics

All functions have a dependent variable. TRUE

All functions have an independent variable. TRUE

The range of a function includes its domain. FALSE

[Range refers to the set of outputs the function produces, domain the set of inputs the function accepts. They don't really have to have anything to do with each other.]

A vertical line is an example of a functional relationship. FALSE

[For a function, each input (x value) maps to at most one output (y value). A vertical line has lots of ys for a single x.]

A horizontal line is an example of a functional relationship. TRUE

[It's a constant function.]

Each output value of a function can correspond to only one input value. FALSE

[A function can have the same output given different inputs, for example f(x)=x^2 has f(1)=f(-1) ]

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Ответ:

izzyw8833

05.03.2023

Mathematics

The center is (3,1) and the radius is 3.

Step-by-step explanation:

The goal is to write in (x-h)^2+(y-k)^2=r^2 because this tells us the center (h,k) and the radius r.

So we are going to need to complete the square 2 times here, once for x and the other time for y.

I'm going to use this formula to help me to complete the square:

$x^2+bx+(\frac{b}{2})^2=(x+\frac{b}{2})^2$ .

So first step:

I'm going to group my x's together and my y's.

$x^2-6x+\text{ ___ }+y^2-2y+\text{ ___ }+1=0$

Second step:

I'm going to go ahead and subtract that one on both sides. Those blanks are there because I'm going to fill them in with a number so that I can write the x part and y part as a square. Remember whatever you add on one side you must add on the other. So I'm going to put 2 more blanks to fill in on the opposite side of the equation.

$x^2-6x+\text{ ___ }+y^2-2y+\text{ ___ }=-1+\text{ ___ }+\text{ ___}$

Third step:

Alright first blank I'm putting (-6/2)^2 due to my completing the square formula. That means this value will also go on the other side in on of those blanks.

In the second blank I'm going to put (-2/2)^2 due to the completing the square formula. This must also go on one of the blanks on the other side.

So we have:

$x^2-6x+(\frac{-6}{2})^2+y^2-2y+(\frac{-2}{2})^2=-1+(\frac{-6}{2})^2+(\frac{-2}{2})^2$