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PtichkaEL [24]
2 years ago
14

What’s the answer to the question?

Mathematics
1 answer:
katen-ka-za [31]2 years ago
6 0

Answer: R=V

Step-by-step explanation: :p

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169/13 draw a quick picture with base ten blocks explain
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I wish i could show u how to work it out but just do it or work it out the same way as the fractional numbers.
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3 years ago
Use >, <, or = to solve the following problems below.
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(6^2)__>__(-6^2)

-5__=__ |-5|

0.372×1,000__<__37.2×100

3 0
2 years ago
Gloria talked on her cell phone for 320 minutes the first month,243 minutes the second month,and 489 minutes the third month.Her
ratelena [41]

Answer:

I believe she spent $150 for the first three months.

The hidden question i think is how many minutes did gloria spend in total OR how many packages did gloria need to buy. sorry i haven't done hidden questions in a long time.  

4 0
3 years ago
What is the vertex of the parabola with the equation y=1/5x^2 + 2x - 8
bixtya [17]

Answer:

(-5,-13)

Step-by-step explanation:

Find the vertex of the parabola y=1/5x^2+2x−8.  

In this equation a=15  and b = 2.  

x= -2/2(1/5) = -2/2/5 = -2/1 * 5/2 = -5

Substitute −5 into the equation y=1/5x^2+2x−8

y=1/5(−5)^2+2(−5)−8

y=5−10−8

y=−13

The vertex is (−5 , −13)

8 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
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