Question

Her house is at (-4, 10). Her school is at (-4, 3). The community center is at (2, 3) The grocery store is at (-4, -6) Part A: Use absolute values to calculate the distance in units from Nina's house to her school. Show your work. (4 points) Part B: Is the total distance from Nina's house to the school to the grocery store greater than the total distance from Nina's house to the school to the community center? Justify your answer. (6 points)

Answer 1

For part A, we will see that the distance is 7 units. For part B, the distance from Nina's house to the school to the grocery store is larger.

How to find the distances?

Remember that the distance between two points (a, b) and (c, d) is:

$D = \sqrt{(a - c)^2 + (b - d)^2}$

A) Her house is at (-4, 10), and her school is at (-4, 3), so the distance is:

$D = \sqrt{(-4 - (-4))^2 + (10 - 3)^2} = \sqrt{7^2} = 7$

The distance is 7 units.

B) In the first case, the distance between the school and the grocery store is:

$D' = \sqrt{(-4 - (-4))^2 + (3 - (-6))^2} = 9$

So the total distance is 9 + 7 = 16

And the distance between the school and the community center is:

$D'' = \sqrt{(-4 - 2)^2 + (3 - 3)^2} = 6$

So the total distance is 6 + 7 = 13

So we can see that the total distance from Nina's house, to the school, to the grocery store is larger than the other.

If you want to learn more about distance:

brainly.com/question/23848540

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