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frez [133]
3 years ago
13

Consider the table of values.

Mathematics
1 answer:
Anestetic [448]3 years ago
8 0

Answer:

B.  The relationship between the variables in the table is quadratic because the second differences are constant.

Step-by-step explanation:

We have to look for the differences in the y values of the given table.

The y values are:

0, 2, 6, 12, 20, 30, 42

The differences between the consecutive values are:

2, 4, 6, 8, 10, 12

These are the first differences and are not constant. This means the relationship is not linear. Now we again find the differences between the consecutive values:

2, 2, 2, 2, 2

These are the second differences. The second differences are constant. So this means the given relationship is quadratic.

Therefore, option B gives the correct answer. The relationship between the variables in the table is quadratic because the second differences are constant.

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Side AB has a length of 4, and side BC has a length of 11.7

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3 years ago
When solving an equation like x3 = 64, why does our solution not include the negative cube root?
MAVERICK [17]
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2 years ago
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What percent of 63.8 is 37.8?
lakkis [162]
To find percentage, you take 37.8 and divide it by 63.8 then multiply by 100

(37.8/63.8) × 100 = 59.25%
6 0
3 years ago
5 + 14 x y = 9 x y - 5
Anon25 [30]

Answer:

xy = -2

Step-by-step explanation:

To solve:

5 + 14xy = 9xy - 5

=> First subtract 9xy from each side of "="

5 + 5xy = -5

=> Now, subtract 5 from each side:

5xy = -10

=> Finally divide 5 on each side:

xy = -10/5

xy = -2

Hope this helps!

4 0
3 years ago
Read 2 more answers
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

6 0
3 years ago
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