All categories

3 years ago

The function f (x) is given by the set of ordered pairs 1,0 (-10,2), (0,6) (3,17) (-2,-1) which equation is true

2 answers:

5 0

If you are given a function such as f(x), it means that f(x) is dependent on the value of x. When finding the points that would correspond to the given function, it is written as (x,f(x)). The f(0) = 6 is the same as the point (0, 6).

finlep [7]3 years ago

3 0

Yes the answer is C, f(0)=6

You might be interested in

How many solutions does the system have? Y=-2x-4 y=3x+3

Fudgin [204]

Answer:

Step-by-step explanation:

Since they are not parallel nor the same line, they have 1 intersecting point, which is the solution.

If they were parallel, they would have no solution.

If they were the same line, they would have infinite solutions.

5 0

4 years ago

What is the second step in sketching the graph of a rational function?

Elodia [21]

Answer:

C. Choosing test numbers to the left and right of each of the function's zeros and finding the value of the function at each test number

Step-by-step explanation:

Rational function: In mathematics, the term "rational function" is determined as any specific function that can be described through any "rational fraction", in other words, an algebraic fraction, involving both the denominator and the numerator are considered as polynomials. Therefore, the coefficients of specific polynomials are ought not to be in rational numbers and can be considered in the K field.

In the question above, the given statement represents option C is correct.

3 0

3 years ago

Plzz solve in 2 mins

Katen [24]

Answer:

A is the answer

Step-by-step explanation:

3×3 is 9 and 2×3 is 6

4 0

3 years ago

Read 2 more answers

Consider the three points ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 ) . Let ¯ x be the average x-coordinate of these points, and let ¯ y

loris [4]

Answer:

$m=\dfrac{3}{2}$

Step-by-step explanation:

Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )

The average of x-coordinate will be:

$\overline{x} = \dfrac{x_1+x_2+x_3}{\text{number of points}}$

1) Finding $(\overline{x},\overline{y})$

Average of the x coordinates:

$\overline{x} = \dfrac{1+2+3}{3}$

$\overline{x} = 2$

Average of the y coordinates:

similarly for y

$\overline{y} = \dfrac{3+3+6}{3}$

$\overline{y} = 4$

2) Finding the line through $(\overline{x},\overline{y})$ with slope m.

Given a point and a slope, the equation of a line can be found using:

$(y-y_1)=m(x-x_1)$

in our case this will be

$(y-\overline{y})=m(x-\overline{x})$

$(y-4)=m(x-2)$

$y=mx-2m+4$

this is our equation of the line!

3) Find the squared vertical distances between this line and the three points.

So what we up till now is a line, and three points. We need to find how much further away (only in the y direction) each point is from the line.

Distance from point (1,3)

We know that when x=1, y=3 for the point. But we need to find what does y equal when x=1 for the line?

we'll go back to our equation of the line and use x=1.

$y=m(1)-2m+4$

$y=-m+4$

now we know the two points at x=1: (1,3) and (1,-m+4)

to find the vertical distance we'll subtract the y-coordinates of each point.

$d_1=3-(-m+4)$

$d_1=m-1$

finally, as asked, we'll square the distance

$(d_1)^2=(m-1)^2$

Distance from point (2,3)

we'll do the same as above here:

$y=m(2)-2m+4$

$y=4$

vertical distance between the two points: (2,3) and (2,4)

$d_2=3-4$

$d_2=-1$

squaring:

$(d_2)^2=1$

Distance from point (3,6)

$y=m(3)-2m+4$

$y=m+4$

vertical distance between the two points: (3,6) and (3,m+4)

$d_3=6-(m+4)$

$d_3=2-m$

squaring:

$(d_3)^2=(2-m)^2$

3) Add up all the squared distances, we'll call this value R.

$R=(d_1)^2+(d_2)^2+(d_3)^2$

$R=(m-1)^2+4+(2-m)^2$

4) Find the value of m that makes R minimum.

Looking at the equation above, we can tell that R is a function of m:

$R(m)=(m-1)^2+4+(2-m)^2$

you can simplify this if you want to. What we're most concerned with is to find the minimum value of R at some value of m. To do that we'll need to derivate R with respect to m. (this is similar to finding the stationary point of a curve)

$\dfrac{d}{dm}\left(R(m)\right)=\dfrac{d}{dm}\left((m-1)^2+4+(2-m)^2\right)$