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2 years ago

quadrilateral ABCD is dilated by a scale factor of 2 centered around (2,2). Which statement is true about the dilation?

Mathematics

1 answer:

Setler [38]2 years ago

6 0

Answer:

The vertices of the image will have the coordinates

A(-5, 1) → A'(2x-2, 2y-2) → A'(-12, 0)

B(-4, 3) → B'(2x-2, 2y-2) → B'(-10, 4)

C(1, 2) → C'(2x-2, 2y-2) → C'(0, 2)

D(-3, 0) → D'(2x-2, 2y-2) → D'(-8, -2)

Step-by-step explanation:

Note: You missed to mention the vertices of a quadrilateral ABCD. So, I am assuming the following vertices. It would anyways clear your concept.

Let us suppose

A(-5, 1)

B(-4, 3)

C(1, 2)

D(-3, 0)

We know that the rule of the dilation with the center of dilation at (2,2) by a scale factor of 2 is:

(x, y) → (2x-2, 2y-2)

Therefore, the vertices of the image will have the coordinates

A(-5, 1) → A'(2x-2, 2y-2) → A'(-12, 0)

B(-4, 3) → B'(2x-2, 2y-2) → B'(-10, 4)

C(1, 2) → C'(2x-2, 2y-2) → C'(0, 2)

D(-3, 0) → D'(2x-2, 2y-2) → D'(-8, -2)

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3 years ago

Lim (n/3n-1)^(n-1) n → ∞

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Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

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then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

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So, the overall limit is indeed 0:

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3 years ago

X is a normally Distributed random variable with a standard deviation of 4.00. Find the mean of X when 64.8% of the area lies to

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Answer:

Step-by-step explanation:

σ = 4 ; μ =?

8.52 to the left of X

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P(X < 8.52) = 0.648

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(x - μ) / σ

P(Z < (8.52 - μ) / 4)) = 0.648

The Z value of 0.648 of the lower tail is equal to 0.38 (Z probability calculator)

Z = 8.52 - μ / 4

0.38 = 8.52 - μ / 4

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Step-by-step explanation:

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