Question

From his home, Rudy would have to walk due north to get to his friend Javier's house and

due east to get to his friend Steve's house. It is 1 mile from Rudy's house to Steve's house
and a straight-line distance of 3 miles from Javier's house to Steve's house. How far is Rudy's
house from Javier's house? If necessary, round to the nearest tenth.
miles

Answer 1

We can use the Pythagorean theorem to solve this problem. Let x be the distance from Rudy's house to Javier's house. Then, the distance from Steve's house to Javier's house is √(3^2 - 1^2) = √8 miles.

Now we can apply the Pythagorean theorem to find the distance from Rudy's house to Javier's house:

x^2 + (√8)^2 = (3 miles)^2

x^2 + 8 = 9

x^2 = 1

x = 1 mile (rounded to the nearest tenth).

Therefore, Rudy's house is 1 mile away from Javier's house.

Answer 2

Rudy's house is approximately 2.8 miles away from Javier's house.

Explanation:

We can use the Pythagorean theorem to solve this problem. Let x be the distance from Rudy's house to Javier's house. Then, we can set up the following equations based on the given information:

x^2 + 1^2 = 3^2 (using the distance between Javier's and Steve's houses)

x^2 = 9 - 1

x^2 = 8

x ≈ 2.8

Therefore, Rudy's house is approximately 2.8 miles away from Javier's house.