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sveta [45]
3 years ago
11

PLEEEEEEEEEEEEEEEEEEEEEEEEEAAAAAAAAAAAAAASEEEEEEEEEEE ANYONE HELP.DO A FAVOUR AND ANSWER THIS QUESTION. IT IS A SIMPLE QUESTION

SO PLEEEEASE. THE QUESTION WILL BE INCLUDED IN THE ATACHMENT.
PS. I WILL GIVE YOU A BRAINLEST MARK.

IT IS QUESTION C

Mathematics
1 answer:
adell [148]3 years ago
4 0
Here's how to figure out this problem: add up all the numbers in the mark category and divide by the # of numbers there are.

So....

6 + 7 + 8 + 9 + 10 ÷ 5 = 8

8 is your final answer.
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2 years ago
How old is Neil if 200 reduced by 2 times his<br> age is 144?
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200-144=56
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3 years ago
The diagram shows a square with perimeter 20 cm.
olga_2 [115]

The percentage of the area inside the square that is shaded is 55%.

<h3>What is the perimeter of the square?</h3>

Given:

The perimeter of the square is 20Cm

Each side of the square is = 5Cm

Area = ( Length x Width )

(5 x 5) = 25 =Area

Area = 25

Area of triangle 1 = (1/2) x 5 x (5 - 3) = 5 cm²

Area of triangle 2 = (1/2) x 5 x (5 - 3.5) = 6.25 cm²

Area of shaded portion = 25 - (5 + 6.25) = 13.75 cm²

Percentage of shade portion

= (13.75 cm² / 25 cm²) x 100%

= 55%

The percentage of the area inside the square that is shaded is 55%.

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8 0
1 year 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
What is a signal 9 on a police scanner? They just said she "dropped a signal 9"
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