Ask question

Ask question

All categories

Dahasolnce [82]

2 years ago

8

The equation of the graph shown plotted in the coordinate plane is y = x + 4. What statement is not true about this graph? A. Ea

ch value of y in an ordered pair that lies on this graph, except 0, is paired with two values of x. B. The value of x in every ordered pair that lies in this graph is less than the corresponding value of y. C. The slope between any two points on this graph is constant. D. The graph of this equation is a function.

2 answers:

frez [133]2 years ago

7 0

Definately option C

As it's mentioned y=x+4 it must be function
So in function every domain has its unique range.
Hence here the slope is not constant

As

The graph shown is a parabola which is not y=x+4

So option C is wrong

GREYUIT [131]2 years ago

6 0

Answer:

C

Step-by-step explanation:

The graph shown plotted in the coordinate plane is $y=x^2+4$

<u>True Statements</u>

A. Each value of y in an ordered pair that lies on this graph, except 0, is paired with two values of x.

If you draw a horizontal for y > 0 you will see that there are two values of x.

B. The value of x in every ordered pair that lies in this graph is less than the corresponding value of y.

D. The graph of this equation is a function.

<u>False Statements</u>

C. The slope between any two points on this graph is constant.

This is true for a linear function. As this is a quadratic function, this is untrue.

You might be interested in

3 gallons of paint are used to paint 16 wooden signs how much paint did each sign use, between what two whole numbers does the a

ipn [44]

The wnswercshould be

4 0

3 years ago

Question 1 (Multiple Choice Worth 1 points)

Oksi-84 [34.3K]

Sorry I don't know but i bet more can exolain.

4 0

2 years ago

Work out the area of ABCD.

lapo4ka [179]

Answer:

efghijklmnopqrstuvwxyz, thank me later

7 0

2 years ago

CAN SOMEONE HELP ME PLEASE ASAP!?

Viefleur [7K]

Answer:

axc is 90 degrees. hope that helps

6 0

2 years ago

Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

3 0

3 years ago