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21 day ago

Part B

Mathematics

1 answer:

AnnZ [9.1K]21 day ago

7 0

Answer:

Part A) The straight-line distance between the school and the Fire Station is $4.6\ miles$

Part B) The straight-line distance between the school and the hospital is $2.1\ miles$

Step-by-step explanation:

The full question can be found in the attached image.

Let

The positive x-coordinates represent the East

The positive y-coordinates represent the North

The negative x-coordinates indicate the West

The negative y-coordinates indicate the South

Treat point A (0,0) as the Middle School (reference point).

Part A) The Town Hall is situated 4.3 miles directly east of the Middle School,

thus the coordinates for Town Hall are B(4.3,0).

The Fire Station is positioned 1.7 miles directly north of Town Hall, therefore

the coordinates for the Fire Station are C(4.3,1.7).

What is the straight-line distance between the school and the Fire Station?

Keep in mind that

the distance formula between two points is indicated as

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

We have

A(0,0) and C(4.3,1.7).

Now substitute

$d=\sqrt{(1.7-0)^{2}+(4.3-0)^{2}}$

$d=\sqrt{(1.7)^{2}+(4.3)^{2}}$

$d=4.6\ miles$

Part B) Now considering

The hospital is 3.1 miles west of the fire station.

The coordinates for the hospital are D(4.3-3.1,1.7).

D(1.2,1.7)

What is the straight-line distance between the school and the hospital?

Remember that

the distance formula applies here as well,

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

with points A(0,0) and D(1.2,1.7).

Substituting the values gives us

$d=\sqrt{(1.7-0)^{2}+(1.2-0)^{2}}$

$d=\sqrt{(1.7)^{2}+(1.2)^{2}}$

$d=2.1\ miles$

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