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lina2011 [118]
3 years ago
11

Drag the ordered pair that are 5.5 units apart.

Mathematics
1 answer:
Alexus [3.1K]3 years ago
6 0

Answer:

  (-1.5, -3.5) and (-1.5, 2)

Step-by-step explanation:

The x-coordinates in increasing order are ...

  -3, -1.5, 5

The distances between pairs of these coordinates are ...

  -1.5 -(-3) = 1.5

  5 -(-3) = 8

  5 -(-1.5) = 6.5

Clearly, any pair with x = 5 is too far away from any pair with a different x-value.

<u>∆x = 1.5</u>

If the distance between pairs is 5.5, then the distance formula tells us the y-coordinate differences for pairs with x-coordinates of -3 and -1.5 must be ...

  d = √((x2 -x1)^2 +(y2 -y1)^2)

  5.5 = √(1.5^2 +(y2 -y1)^2) . . . substitute known values

  30.25 = 2.25 +(y2 -y1)^2 . . . . square both sides

  √28 = (y2 -y1) . . . . . . . . . . . . . . subtract 2.25 and take the square root

This value is irrational. Clearly none of the y-coordinates is irrational, so there are no point pairs with x-coordinates -3 and -1.5 that are 5.5 units apart.

__

<u>∆x = 0</u>

If the x-coordinates are the same, then the y-coordinates must differ by 5.5 in order for the points to be 5.5 units apart. The coordinates in order are ...

  for x = -1.5, y = -3.5, 2, 2.5 . . . . . differences of 5.5, 6, 0.25

  for x = 5, y = -3.5, 1.5 . . . . . . . difference of 5

The only pair we can find that is 5.5 units apart is ...

  (-1.5, -3.5) and (-1.5, 2).

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Which statement describes the translation of y = −1/5 (x + 5)2 + 2 from standard position? Standard position is when the vertex
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