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3 years ago

Identify the coordinates of the vertices of each figure after the similarity transformation.

Mathematics

1 answer:

bija089 [108]3 years ago

8 0

Answer:

After translation:

Translation: (x, y) → (x-6, y-8)

W'X'Y'Z' = W'(-10, 0), X'(-10, -8), Y'(-18, -4), Z'(-14, 8)

After dilation:

Dilation: (x, y) → (2x, 2y), centered at (0, 0)

W''X''Y''Z'' = W'' (-20, 0), X'' (-8, 0) , Y'' (-24, 8), Z'' (-16, 32)

Step-by-step explanation:

Given the vertices of the Quadrilateral WXYZ

W(-4, 8)

X(-4, 0)

Y(-12, 4)

Z(-8, 16)

As the rule of translation is given by

Translation: (x, y) → (x-6, y-8)

So, after the translation the vertices of Quadrilateral WXYZ will be translated as:

W(-4, 8) → (x-6, y-8) ⇒ (-4-6, 8-8) = W'(-10, 0)

X(-4, 0) → (x-6, y-8) ⇒ (-4-6, 0-8) = X'(-10, -8)

Y(-12, 4) → (x-6, y-8) ⇒ (-12-6, 4-8) = Y'(-18, -4)

Z(-8, 16) → (x-6, y-8) ⇒ (-8-6, 16-8) = Z'(-14, 8)

Therefore,

W'X'Y'Z' = W'(-10, 0), X'(-10, -8), Y'(-18, -4), Z'(-14, 8)

Next, the rule of dilation is given by

Dilation: (x, y) → (2x, 2y), centered at (0, 0)

W'(-10, 0) → (2x, 2y) ⇒ (2(-10), 2(0)) = W'' (-20, 0)

X'(-4, 0) → (2x, 2y) ⇒ (2(-4), 2(0)) = X'' (-8, 0)

Y'(-12, 4) → (2x, 2y) ⇒ (2(-12), 2(4)) = Y'' (-24, 8)

Z'(-8, 16) → (2x, 2y) ⇒ (2(-8), 2(16)) = Z'' (-16, 32)

Therefore,

W''X''Y''Z'' = W'' (-20, 0), X'' (-8, 0) , Y'' (-24, 8), Z'' (-16, 32)

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Convert,the complex number into polar form: 4+4i

kow [346]

Z = a + bi
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r² = a² + b²
r² = (4)² + (4)²
r² = 16 + 16
r² = 32
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r = 4(1.414)
r = 5.656

$cos\theta = \frac{a}{r}$
$cos\theta = \frac{4}{4\sqrt{2}}$
$cos\theta = \frac{4}{4\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}$
$cos\theta = \frac{4\sqrt{2}}{4\sqrt{4}}$
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$cos\theta = \frac{\sqrt{2}}{2}$
$2(cos\theta) = 2(\frac{\sqrt{2}}{2})$
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$sin\theta = \frac{b}{r}$
$sin\theta = \frac{4}{4\sqrt{2}}$
$sin\theta = \frac{4}{4\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}$
$sin\theta = \frac{4\sqrt{2}}{4\sqrt{4}}$
$sin\theta = \frac{4\sqrt{2}}{4(2)}$
$sin\theta = \frac{4\sqrt{2}}{8}$
$sin\theta = \frac{\sqrt{2}}{2}$
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$2sin\theta = 1.414$

z = a + bi
z = rcosθ + (rsinθ)i
z = r(cosθ + i sinθ)

z = 4 + 4i
z = 5.656cosθ + (5.656sinθ)i
z = 5.656(cosθ + i sinθ)
z = 5.656(cos45 + i sin45)

$\theta = tan^{-1}\frac{b}{a}$
$\theta = tan^{-1}\frac{4}{4}$
$\theta = tan^{-1}(1)$
$\theta = 45$

The polar form of 4 + 4i is approximately equal to 5.656(cos45 + i sin45).

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3 years ago

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netineya [11]

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To do this, multiply equation one by 2, and equation two by 5 (this will make the coefficient 10 in both).

14x + 10y = 38
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Eliminate the y variable by adding the equations from each other

(14x + 10y = 38)
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We now have -21x = -42. The y variable has been eliminated.

Solve for x

-21x = -42

Divide both sides by -21 to get x on its own.

x = 2

Substitute x into the equation to find y.

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14+5y=19

Subtract 14 from both sides.

5y=19-14

5y=5

Divide by 5 on both sides to get y on its own.

y=5/5

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8_murik_8 [283]

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