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DENIUS [597]
3 years ago
6

A 2X2 square is centered at the origin. It is dilated by a factor of 3. What are coordinates of the vertices of the square?

Mathematics
1 answer:
Vesna [10]3 years ago
7 0

Answer:

The vertices are:

A' = (-3, -3)

B' = (3, -3)

C' = (3, 3)

D' = (-3, 3)

Step-by-step explanation:

Given:

A 2 x 2 square is centered at the origin.

So, the center of the square is (0, 0)

Since it is 2 x 2 square, the side of the square is 2 units.

So, the vertices of the 2 x 2 square are A (-1, -1),  B(1, -1), C(1. 1), D(-1, 1)

The above square is dilated by a factor of 3.

Let's name the dilated square A'B'C'D'

To find the coordinates of the vertices of dilated square, we need to multiply each vertices of ABCD by 3.

A(-1, -1) = 3(-1, -1) = A'(-3, -3)

B(1, -1) = 3(1, -1) = B'(3, -3)

C(1, 1) = 3(1, 1) = C'(3, 3)

D(-1, 1) = 3(-1, 1) = D'(-3, 3)

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Determine the general solution of:<br><br> sin 3x - sin x = cos2x
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Sin³ x-sin x=cos ² x

we know that:
sin²x + cos²x=1  ⇒cos²x=1-sin²x
Therefore:

sin³x-sin x=1-sin²x
sin³x+sin²x-sin x-1=0

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z³+z²-z-1=0

we divide by Ruffini method:
              1     1     -1     -1
        1           1      2      1                z=1
-------------------------------------
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       -1         -1      -1                       z=-1
--------------------------------------
              1     1       0                       z=-1

Therefore; the solutions are z=-1 and z=1

The solutions are:
if z=-1, then
sin x=-1   ⇒x= arcsin -1=π+2kπ    (180º+360ºK)   K∈Z


if z=1, then
sin x=1   ⇒ x=arcsin 1=π/2 + 2kπ   (90º+360ºK)   k∈Z

π/2 + 2kπ    U   π+2Kπ=π/2+kπ     k∈Z    ≈(90º+180ºK)

Answer: π/2 + Kπ    or     90º+180ºK          K∈Z
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