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3 years ago

How would you represent a rotation of 360 algebraically?

Mathematics

2 answers:

SSSSS [86.1K]3 years ago

5 0

Answer:

(x,y)(x,-y)

Step-by-step explanation:

ALGEBRAIC REPRESENTATIONS OF ROTATIONS

To rotate a figure in the coordinate plane, rotate each of its vertices. Then connect the vertices to form the image.

We can use the rules shown in the table for changing the signs of the coordinates after a reflection about the origin.

Example 1 :

The triangle XYZ has the following vertices X(0, 0), Y(2, 0) and Z(2, 4). Rotate the triangle XYZ 90° counterclockwise about the origin.

Solution :

Step 1 :

Trace triangle XYZ and the x- and y-axes onto a piece of paper.

Step 2 :

Let X', Y' and Z' be the vertices of the rotated figure.

Since the triangle is rotated 90° counterclockwise about the origin, the rule is

(x, y) ------> (-y, x)

Step 3 :

X(0, 0) ------> X'(0, 0)

Y(2, 0) ------> Y'(0, 2)

Z(2, 4) ------> Z'(-4, 2)

Step 4 :

Sketch the image X'Y'Z' using the points X'(0, 0), Y'(0, 2) and Z'(-4, 2).

Example 2 :

The triangle PQR has the following vertices P(0, 0), Q(-2, 3) and R(2,3). Rotate the triangle PQR 90° clockwise about the origin.

Solution :

Step 1 :

Trace triangle PQR and the x- and y-axes onto a piece of paper.

Step 2 :

Let P', Q' and R' be the vertices of the rotated figure.

Since the triangle is rotated 90° clockwise about the origin, the rule is

(x, y) ------> (y, -x)

Step 3 :

P(0, 0) ------> P'(0, 0)

Q(-2, 3) ------> Q'(3, 2)

R(2, 3) ------> R'(3, -2)

Step 4 :

Sketch the image P'Q'R' using the points P'(0, 0), Q'(3, 2) and Z'(3, -2).

Example 3 :

A quadrilateral has the following vertices A(0, 0), B(1, 2), C(4, 2) and D(3, 0). Rotate the quadrilateral 180° clockwise about the origin.

Solution :

Step 1 :

Trace the quadrilateral ABCD and the x- and y-axes onto a piece of paper.

Step 2 :

Let A', B', C' and D' be the vertices of the rotated figure.

Since the quadrilateral is rotated 180° clockwise about the origin, the rule is

(x, y) ------> (-x, -y)

Step 3 :

A(0, 0) ------> A'(0, 0)

B(1, 2) ------> B'(-1, -2)

C(4, 2) ------> C'(-4, -2)

D(3, 0) ------> D'(-3, 0)

Step 4 :

Sketch the image A'B'C'D' using the points A'(0, 0), B'(-1, -2), C(-4, -2) and D'(-3, 0).

Example 4 :

The triangle XYZ has the following vertices X(0, 0), Y(2, 0) and Z(2, 4). Rotate the triangle XYZ 270° counterclockwise about the origin.

Solution :

Step 1 :

Trace triangle XYZ and the x- and y-axes onto a piece of paper.

Step 2 :

Let X", Y" and Z" be the vertices of the rotated figure.

Since the triangle is rotated 270° counterclockwise about the origin, the rule is

(x, y) ------> (y, -x)

Step 3 :

X(0, 0) ------> X"(0, 0)

Y(2, 0) ------> Y"(0, -2)

Z(2, 4) ------> Z"(4, -2)

Step 4 :

Sketch the image X"Y"Z" using the points X"(0, 0), Y"(0, -2) and Z"(4, -2).

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9th

GenaCL600 [577]3 years ago

3 0

Answer:It means turning around until you point in the same direction again. Other ways of saying it: "Doing a 360" means spinning around completely once (spinning around twice is a "720").

...

A full rotation is 360 degrees.

Step-by-step explanation:

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From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for the mean would be:

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And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

$P( z>2.143) = 1-P(Z$

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