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Anastasy [175]
3 years ago
9

Select all of the following that would be an acceptable set builder notation for the set {7, 8, 9, 10, 11, 12, 13, 14}.

Mathematics
1 answer:
ludmilkaskok [199]3 years ago
5 0
<h3>Answer: Choice B</h3>

{x | x is a natural number with 6 < x < 15}

===========================================================

Explanation:

The given set is in roster notation. Think of a baseball or football team (or any sports team) how their roster lists out all of the players. In this case, each number is like a player.

The idea here is to simplify the set so we don't have to write out every single item. Instead we have a rule to make things a bit simpler.

In this case, the set {7, 8, 9, 10, 11, 12, 13, 14} describes all natural numbers from 7 to 14

This means we can say {x | x is a natural number with 6 < x < 15}

Note how the endpoints x = 6 and x = 15 are not included. This is because there isn't a "or equal to" as part of the inequality sign. Sure enough 6 and 15 are not part of the original set.

An equivalent set would be \{ x | \text{ x is a natural number with } \ 7 \le x \le 14\} and here we have "or equal to" involved. For this example, the endpoints x = 7 and x = 14 are included.

--------------

Something like choice A and choice D are not complete answers because we don't know if x is a whole number, natural number, rational number, real number, etc. If x was say a real number, then x = 10.75 would be involved with choice A. But 10.75 is not part of the original set.

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Jessie draws triangle ABC on a coordinate grid. The slope of line segment AB is Jessie then transforms triangle
balu736 [363]

Answer:

1) Supports Jessie's Claim

2) Does Not Support Jessi's Claim

3) Supports Jessie's Claim

4) Does Not Support Jessi's Claim

Step-by-step explanation:

The given transformations are;

1) Rotation of 180° around the origin

For a rotation of 180° around the origin, either clockwise or anti clockwise, for a given coordinate of the preimage (x, y), the coordinate of the image is (-x, -y)

Therefore, whereby the slope of the preimage, given two points (0, 0) and (2, 2), = (2 - 0)/(2 - 0) = 1

For the image with the points (0, 0) and (-2, -2), we have;

(-2 - 0)/(-2 - 0) = 1

Therefore, the slope of the preimage and the image are equal

Therefore, supports Jessie's Claim

2) For a reflection across the line y = 2, we have

We note that the line y = 2 is parallel to the x-axis

For a reflection across the x-axis, for a preimage (x, y), we have the coordinates of the image (x, -y)

However for the reflection across the line y = 2, we have;

For a preimage, (x, y), the coordinate of the image is (x, -y+4)

Given two points, of the preimage (0, 0) and (2, 2), we have the image given as (0, 4) and (2, -2 + 4) = (2, 2);

The slope of the preimage is (2 - 0)/(2 - 0) = 1

The slope of the image is (2 - 4)/(2 - 0) = -1

The slope of the line of the preimage and the image are different

Therefore, does Not Support Jessi's Claim

3) For a translation up 1.25 units, we note that the difference in the y and x values of the coordinates of the preimage and the image will be equal when finding the slope, and therefore, the slope of the figure of the preimage and the slope of the figure of the image will be equal

Therefore, supports Jessie's Claim

4) For a reflection across the x-axis, a point on the preimage, with coordinates (x, y) will form a point on the image with coordinates (x, - y)

For a preimage with points (0, 0) and (2, 2), we have the image as (0, 0) and (2, -2)

The slope of the preimage is (2 - 0)/(2 - 0) = 1

The slope of the image is (-2 - 0)/(2 - 0) = -1

The slope of the line of the preimage and the image are different

Therefore, does Not Support Jessi's Claim

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