Question

The map shows the location of a mall, library, and school in a city:

Answer 1

Answer:

Option: A is the correct answer.

A) 8 miles.

Step-by-step explanation:

It is given that:

Britney traveled from the school to the mall and then from the mall to the library.

Let S denotes the school.

L denotes the library.

and M denotes the Mall.

The school is located at (10,-4)

and Mall is located at (10,6)

and library is located at (-14,6)

So, the distance traveled by Britney is equal to length of line segment SM+ Length of line segment ML.

length of line segment SM is calculated as:

$=\sqrt{(10-10)^2+(6-(-4)^2}\\\\=\sqrt{(10)^2}\\\\=10\ miles$

and length of line segment ML is calculated as:

$=\sqrt{(-14-10)^2+(6-6)^2}\\\\=\sqrt{(24)^2}\\\\=24\ miles$

Hence, the total distance traveled by Britney is:

10+24=34 miles.

Now, the distance traveled by Alice is equal to the length of the line segment SL.

i.e. it is calculated as:

$=\sqrt{(10-(-14))^2+(6-(-4))^2}\\\\=\sqrt{(24)^2+(10)^2}\\\\=\sqrt{576+100}\\\\=\sqrt{676}\\\\=26\ miles$

Hence, distance traveled by Alice is:

26 miles.

Hence, the distance that is more traveled by Britney is:

34-26=8 miles.

Hence, the miles that Britney travels more than Alice is:

8 miles.

Answer 2

Answer:

A) 8 miles.

Step-by-step explanation:

We  haven been given a map that shows the location of a mall, library, and school in a city.

First of all we will find the distance traveled by Britney from school to mall and mall to library using distance formula.

$\text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Upon substituting the coordinates of school and mall in distance formula we will get,

$\text{Distance between school and mall}=\sqrt{(10-10)^2+(-4-6)^2}$

$\text{Distance between school and mall}=\sqrt{(0)^2+(-10)^2}$

$\text{Distance between school and mall}=\sqrt{0+100}$

$\text{Distance between school and mall}=10$  

Now let us find distance between mall and library.

$\text{Distance between mall and library}=\sqrt{(10--14)^2+(6-6)^2}$

$\text{Distance between mall and library}=\sqrt{(10+14)^2+(0)^2}$

$\text{Distance between mall and library}=\sqrt{(24)^2+(0)^2}$

$\text{Distance between mall and library}=\sqrt{576+0}$

$\text{Distance between mall and library}=24$

$\text{Total distance covered by Britney}=\text{10 miles +24 miles}$

$\text{Total distance covered by Britney}=\text{34 miles}$

Therefore, Britney traveled 34 miles.

Now let us find the distance covered by Alice by substituting coordinates of school and library in distance formula.

$\text{Distance covered by Alice}=\sqrt{(10--14)^2+(-4-6)^2}$

$\text{Distance covered by Alice}=\sqrt{(10+14)^2+(-10)^2}$

$\text{Distance covered by Alice}=\sqrt{(24)^2+(-10)^2}$

$\text{Distance covered by Alice}=\sqrt{576+100}$

$\text{Distance covered by Alice}=\sqrt{676}$

$\text{Distance covered by Alice}=26$

Therefore, Alice traveled 26 miles.

$\text{The number of miles Britney traveled more than Alice}=\text{34 miles -26 miles}$

$\text{The number of miles Britney traveled more than Alice}=\text{8 miles}$

Therefore, Britney traveled 8 miles more than Alice and option A is the correct choice.