To solve for the distance from Andy's house to Silvergrove, we can visualize the positions of the locations on a coordinate plane.
- Let Andy's house be at the origin \((0, 0)\).
- Since Springdale is due west of Andy’s house, it is located at \((-5.9, 0)\).
- Since Andy's house is due south of Silvergrove, we can place Silvergrove at \((0, y)\), where \(y\) is the distance north from Andy's house to Silvergrove.
We are given that the distance between Springdale and Silvergrove is 10 miles. This distance can be found using the distance formula:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of Springdale \((-5.9, 0)\) and Silvergrove \((0, y)\):
\[ 10 = \sqrt{(0 - (-5.9))^2 + (y - 0)^2} \] \[ 10 = \sqrt{(5.9)^2 + y^2} \]
Now, squaring both sides:
\[ 100 = (5.9)^2 + y^2 \] \[ 100 = 34.81 + y^2 \] \[ y^2 = 100 - 34.81 \] \[ y^2 = 65.19 \] \[ y = \sqrt{65.19} \] \[ y \approx 8.07 \]
Now that we have \(y\), which is the distance from Andy's house to Silvergrove, we can calculate the straight-line distance from Andy's house to Silvergrove:
Using the distance formula between Andy's house \((0, 0)\) and Silvergrove \((0, 8.07)\):
\[ d = \sqrt{(0 - 0)^2 + (y - 0)^2} = |y| = 8.07 \]
Finally, rounding this value to the nearest tenth gives:
\[ \text{Distance} \approx 8.1 \text{ miles} \]
Thus, the distance from Andy's house to Silvergrove is approximately \( \boxed{8.1} \) miles.