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3 years ago

Solve and show work.

Mathematics

1 answer:

Mice21 [21]3 years ago

8 0

Hope this helps have a great day!!

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I got the answer D is this correct? will vote brainliest

Marina86 [1]

Answer:

yes its a correct answer hope it will help u.......

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2 years ago

Person A can paint the neighbor's house 6 times as fast as Person B. The year A and B worked together it

uranmaximum [27]

Person A takes 2.3 days to paint the house working alone

Person B takes 13.8 days to paint the house working alone

Solution:

Let "x" be the number of days it takes person A to paint the house

Person A can paint the neighbor's house 6 times as fast as Person B

Therefore.

Number of days it takes person B to paint the house = 6x

Their rates of painting are added to get the rate working together

$\frac{\text{1 house}}{\text{x days}} + \frac{\text{1 house}}{\text{6x days}} = \frac{\text{1 house}}{\text{2 days}}$

$\frac{1}{x} + \frac{1}{6x} = \frac{1}{2}\\\\\frac{1 \times 6}{6x} + \frac{1}{6x} = \frac{1}{2}\\\\\frac{7}{6x} = \frac{1}{2}\\\\\ x = \frac{7}{3}= 2.3$

Thus person A takes 2.3 days to paint the house working alone

Person B = 2x = 6(2.3) = 13.8

Thus person B takes 13.8 days to paint the house working alone

3 0

3 years ago

Dana says that all even numbers are composite. Is she correct?

Ksenya-84 [330]

Answer:

yes

Step-by-step explanation:

all even numbers greater than two are composite numbers

8 0

3 years ago

The percentile of 1.2 equals?

laila [671]

It’s 0.3 because that’s the answer

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