6 of 76 of 7 Items Question You are a city planner working on a new park design project. The city is considering adding a new playground and a fountain in a local park. You have mapped out two potential locations for these features: Playground (Location A) is situated at the coordinates (3,5) on the city park map. Fountain (Location B) is planned to be at (7,9). Calculate the distance between the Playground (Location A) and the Fountain (Location B). Show all your calculations using the steps provided. (3 points) Step 1. D= (x2−x1)2+(y2 − y1)2−−−−−−−−−−−−−−−−−−−√ = Step 2. D= (x difference)2+(y difference)2−−−−−−−−−−−−−−−−−−−−−−−−−−−√ = Step 3. D= (x difference squared)+(y difference squared)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√ = Step 4. D= final answer = The final answer could have a square root or round to the nearest tenth. Skip to navigation page 5 of 5 Formula keypad has been closed. Press Control + Backslash to open it again.

Answer 1

To calculate the distance (D) between the Playground (Location A) at coordinates (3,5) and the Fountain (Location B) at coordinates (7,9), we will follow the steps outlined.

Step 1:

The distance formula is given by:

\[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

In this case:

• \( (x_1, y_1) = (3, 5) \) (coordinates of Location A)
• \( (x_2, y_2) = (7, 9) \) (coordinates of Location B)

Step 2:

First, we calculate the differences for x and y:

• \( x \text{ difference} = x_2 - x_1 = 7 - 3 = 4 \)
• \( y \text{ difference} = y_2 - y_1 = 9 - 5 = 4 \)

Substituting into the distance formula:

\[ D = \sqrt{(4)^2 + (4)^2} \]

Step 3:

Now we calculate the squares of the differences:

\[ D = \sqrt{(4^2) + (4^2)} = \sqrt{16 + 16} \]

Now, simplify it further:

\[ D = \sqrt{32} \]

Step 4:

Finally, we calculate the square root:

\[ D = \sqrt{32} \approx 5.7 \]

So, to the nearest tenth, the final answer is:

\[ D \approx 5.7 \]

Hence, the distance between the Playground (Location A) and the Fountain (Location B) is approximately 5.7 units.