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evablogger [386]
2 years ago
12

Quadrilateral EFGH was dilated by a scale factor of 2 from the center (1, 0) to create E'F'G'H'. Which characteristic of dilatio

ns compares segment E'F' to segment EF?​

Mathematics
1 answer:
Juliette [100K]2 years ago
3 0

The answer choice which is the characteristic of dilations comparing both segments is; A segment in the image is proportionally longer or shorter than its corresponding segment in the pre-image

<h3>Which answer choice compares segment E'F' to segment EF?</h3>

By consider the coordinates of the quadrilaterals EFGH and E'F'G'H' as given in the task content image, it follows that the coordinates are as follows;

  • E(0, 1), F(1, 1), G(2, 0), and H(0, 0)

  • E'(-1, 2), F'(1, 2), G'(3, 0), and H'(-1, 0)

Upon computation of the length of the segments, it follows that the two segments are in proportions. Hence, the answer choice which is correct is; A segment in the image is proportionally longer or shorter than its corresponding segment in the pre-image.

Remark:

  • A segment that passes through the center of dilation in the pre-image continues to pass through the center of dilation in the image.
  • A segment in the image has the same length as its corresponding segment in the pre-image.
  • A segment that passes through the center of dilation in the pre-image does not pass through the center of dilation in the image.
  • A segment in the image is proportionally longer or shorter than its corresponding segment in the pre-image.

Read more on length of segments;

brainly.com/question/24778489

#SPJ1

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Anvisha [2.4K]

Answer:

x = 15

Step-by-step explanation:

I am going to use the "butterfly" method for proportions

(there is another type of butterfly method that is different, I am not sure if there is another name for this method, sorry!)

\frac{7}{5} = \frac{21}{x}

Multiply the:

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and then:

- set them equal to each other

So we will <u>multiply 7 by x and 21 by 5</u> (if this doesn't make sense tell me and I will explain it even more!)

7x = 21 * 5

7x = 105

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Hope this helps, have a nice day! :D <3

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How to use trig to find missing side or angle of right triangles.
andrew11 [14]
Explanation

Right triangles are triangles with one right angle i.e. one angle with a measure of 90° so when we deal with this type of figures we know at least one of its internal angles. The opposite side to the right angle is known as the hypotenuse whereas the other two sides are called legs.

Trigonometric functions in right triangles give us the following identities:

\begin{gathered} \tan x=\frac{\text{opposite side}}{\text{adjacent side}} \\ \cos x=\frac{\text{adjacent side}}{\text{hypotenuse}} \\ \sin x=\frac{\text{opposite side}}{\text{hypotenuse}} \end{gathered}

We also have the Pythagorean theorem that states that the square of the hypotenuse is the sum of the squares of the legs.

Let's imagine that we have a right triangle like the following:

If we want to find the missing angles A and/or B we need to know at least two sides of the triangle.

For example, if we want to find B and we have b and a we can use the tangent of B and its inverse the arctangent (tan^(-1)):

\tan B=\frac{b}{a}\rightarrow B=\tan^{-1}(\frac{b}{a})

If we know a and c we can find B using its cosine:

\cos B=\frac{a}{c}\rightarrow B=\cos^{-1}(\frac{a}{c})

If we know b and c then we can use the sine:

\sin B=\frac{b}{c}\rightarrow B=\sin^{-1}(\frac{b}{c})

We can also do the same for angle A, we just need to use its corresponding opposite side and adjacent side. And that's how you find a missing angle using trigonometry in a right triangle.

If we are looking for a missing side we can also use trigonometry. In order to find a missing side we need two other sides or another side and an angle. For example, let's assume that we want to find side a.

If we have b and c then we just need to use the Pythagorean theorem:

c^2=a^2+b^2\rightarrow a=\sqrt{c^2-b^2}

If we have b and A then we can use the tangent of A to built an equation for a:

\tan A=\frac{a}{b}\rightarrow a=b\cdot\tan A

If we have b and B we can use the tangent of B:

\tan B=\frac{b}{a}\rightarrow a=\frac{b}{\tan B}

If we have A and c we can use the sine of A:

\sin A=\frac{a}{c}\rightarrow a=c\cdot\sin A

If we have B and c we can use the cosine of B:

\cos B=\frac{a}{c}\rightarrow a=c\cdot\cos B

And that's how you find a missing side using trigonometry.

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