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Annette [7]
3 years ago
12

BRAINLIEST!!

Mathematics
2 answers:
Natali5045456 [20]3 years ago
7 0
A dialation produces a shape that is the same shape but different size. so we draw a new triangle about point B where A'B=2AB
so B and B' are the same and
AB and BC will be parts of the sides. and A' is located on the ray
Oksanka [162]3 years ago
4 0

Answer:

1st, 2nd and 4th statements are correct choices.

Step-by-step explanation:

We have been given that Joseph is creating a dilation through point B with a scale factor of 2. We are asked to choose the correct statements about the dilation.

Since we know that in dilation the side of pre-image and corresponding sides of the image and always proportional. All the sides of image are stretched or compressed by same scale factor.

Let us see our given choices one by one.

1. A' will be located on ray BA.

Since our pre-image is dilated by a factor of 2 to get the new image, so the length of side B'A' will be 2 times the length of BA and we can get point A' by extending the ray BA.  

Therefore, 1st statement is true.

2. B and B' are the same point.  

Since dilation multiplies the distance of every point from the center of dilation by the scale factor, so dilation move every point except the center of dilation. As B is the point of dilation, therefore, point B and B' will coincide and 2nd statement is true as well.

3. Line segment B'C' will be 1/2 as long as line segment BC.

Since factor of dilation is 2, so all the side lengths of new image will be 2 times the corresponding side lengths of pre-image.

As line segment B'C' is side length of new image, therefore, it will be 2 times the length of BC and 3rd statement is false.

4. Segment AB and BC will be the part of the image's sides.

Since B is the point of dilation, so side lengths of new figure will be 2 times the side lengths of pre-image, therefore, 4th statement is true.

5. The image will be inside the pre-image.

Since scale factor of dilation is 2, so the image will be bigger than given pre-image and will be outside pre-image. Therefore, 5th statement is false.


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It's difficult to make out what the force and displacement vectors are supposed to be, so I'll generalize.

Let <em>θ</em> be the angle between the force vector <em>F</em> and the displacement vector <em>r</em>. The work <em>W</em> done by <em>F</em> in the direction of <em>r</em> is

<em>W</em> = <em>F</em> • <em>r</em> cos(<em>θ</em>)

The cosine of the angle between the vectors can be obtained from the dot product identity,

<em>a</em> • <em>b</em> = ||<em>a</em>|| ||<em>b</em>|| cos(<em>θ</em>)   ==>   cos(<em>θ</em>) = (<em>a</em> • <em>b</em>) / (||<em>a</em>|| ||<em>b</em>||)

so that

<em>W</em> = (<em>F</em> • <em>r</em>)² / (||<em>F</em>|| ||<em>r</em>||)

For instance, if <em>F</em> = 3<em>i</em> + <em>j</em> + <em>k</em> and <em>r</em> = 7<em>i</em> - 7<em>j</em> - <em>k</em> (which is my closest guess to the given vectors' components), then the work done by <em>F</em> along <em>r</em> is

<em>W</em> = ((3<em>i</em> + <em>j</em> + <em>k</em>) • (7<em>i</em> - 7<em>j</em> - <em>k</em>))² / (√(3² + 1² + 1²) √(7² + (-7)² + (-1)²))

==>   <em>W</em> ≈ 5.12 J

(assuming <em>F</em> and <em>r</em> are measured in Newtons (N) and meters (m), respectively).

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Answer:

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