Question

what is the approximate distance between seaside and crawford? based on the coordinate grid... approximately blank miles. (coordinate grid with x-axis from -8 to 8, y-axis from -8 to 8, points labeled seaside, landview, crawford)

Explanation:

Step1: Identify coordinates

From the graph, let's assume each grid square is 1 unit (mile).

• Coordinates of Seaside: Let's say from the grid, Seaside is at \((-2, 4)\) (since it's 2 units left of y - axis and 4 units up).
• Coordinates of Crawford: Crawford is at \((3, -1)\) (3 units right of y - axis and 1 unit down).

Step2: Apply distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)

Substitute \(x_1=-2,y_1 = 4,x_2=3,y_2=-1\) into the formula:

First, calculate \((x_2 - x_1)\) and \((y_2 - y_1)\)
\(x_2 - x_1=3-(-2)=3 + 2=5\)
\(y_2 - y_1=-1 - 4=-5\)

Then, calculate \((x_2 - x_1)^2+(y_2 - y_1)^2\)
\((5)^2+(-5)^2=25 + 25=50\)

Now, find the square root of 50: \(d=\sqrt{50}\approx7.07\approx7\) (or more accurately, \(\sqrt{50}\approx7.1\), but if we consider grid - based approximation, we can also use the Pythagorean theorem by counting the horizontal and vertical distances. The horizontal distance between - 2 and 3 is \(3-(-2)=5\) units, the vertical distance between 4 and - 1 is \(4-(-1)=5\) units (taking absolute value for distance). Then the distance is \(\sqrt{5^2 + 5^2}=\sqrt{25 + 25}=\sqrt{50}\approx7.1\), which is approximately 7 or 7.1 miles. If we assume each grid is 1 mile, the approximate distance is about 7 miles (or 7.1 miles, but 7 is a common approximation).

Answer:

Approximately \(\boldsymbol{7}\) (or \(\boldsymbol{7.1}\)) miles. (If we take the more precise value of \(\sqrt{50}\approx7.07\), we can say approximately 7 miles or 7.1 miles. The answer may vary slightly depending on the exact coordinates, but the key is to use the distance formula between the two points' coordinates.)