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Roman55 [17]
3 years ago
11

Which graph has a rate of change of zero

Mathematics
2 answers:
Hatshy [7]3 years ago
5 0

Answer:

When the graph has a horizontal line.

Step-by-step explanation:

Zero rate of change. When the value of increases, the value of remains constant. That is, there is no change in value and the graph is a horizontal line.

SOURCE: GOOGLE



GaryK [48]3 years ago
5 0

Answer:

B.)

Step-by-step explanation:

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Dexter runs around a track at 10 laps in 20 minutes. How many laps per minute is that?
IRINA_888 [86]

Answer:

2

Step-by-step explanation:

20/10=2

5 0
3 years ago
Read 2 more answers
Simplify to create an equivalent expression of<br> 2(−n−3)−7(5+2n)
musickatia [10]
<h3>Answer:   -16n-41</h3>

Work Shown:

2(-n-3) -7(5+2n)

2(-n)+2(-3) - 7(5)-7(2n) ... distribute

-2n - 6 - 35 - 14n

(-2n-14n) + (-6-35)

-16n - 41

6 0
3 years ago
How do you find vertical, horizontal and oblique asymptotes for f(x) = (5x-15 )/ (2x 14)?
Andrei [34K]
For vertical asymptotes, find the values which make the function indetermine in this case x=-7,so this is the only vertical asymptote.
For horizontal asymptotes, find the limit when x tends to infinity:
=(5x/x-15/x)/(2x/x+14/x) = 5/2, this is the horizontal asymptote y=5/2
For obliques, you have to meet the degree of the numerator is exactly a greater degree than the denominator, in this case they are the same degree so no oblique asymptote.
6 0
3 years ago
What degree of rotation is represented on this matrix
Korvikt [17]

Answer:

Option B is correct

the degree of rotation is, -90^{\circ}

Step-by-step explanation:

A rotation matrix is a matrix that is used to perform a rotation in Euclidean space.

To find the degree of rotation using a standard rotation matrix i.e,

R = \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}

Given the matrix: \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}

Now, equate the given matrix with standard matrix we have;

\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} =  \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}

On comparing we get;

\cos \theta = 0       and -\sin \theta =1  

As,we know:

  • \cos \theta = \cos(-\theta)
  • \sin(-\theta) = -\sin \theta

\cos \theta = \cos(90^{\circ}) = \cos( -90^{\circ})

we get;

\theta = -90^{\circ}

and

\sin \theta =- \sin (90^{\circ}) = \sin ( -90^{\circ})

we get;

\theta = -90^{\circ}

Therefore, the degree of rotation is, -90^{\circ}

7 0
3 years ago
The perimeter of a rectangle garden is 54 feet. The width is 5 more feet than the length. What is the length of the garden
love history [14]

The Length of the rectangle is 11 ft and the width is 16 ft

3 0
3 years ago
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