## Algebra 2 (1st Edition)

$18.33$ miles
Let it enter the layer at $(x_2,-4)$ and leave at $(x_1,-4)$. Then $x_1^2+(-4)^2=5^2\\x_1^2=9\\x_1=\pm3$ As we can see $x_1$ is negative, thus $x_1=-3$. $x_2^2+(-4)^2=10^2\\x_1^2=84\\x_1\approx\pm9.165$ As we can see $x_2$ is negative, thus $x_2=-9.165$. Then the distance is: $|x_1-x_2|=|-3-(-9.165)|=6.165$ Since the figure is symmetric it will cover the same amount of space in the fourth quadrant as well, thus the answer is: $2\cdot6.165=12.33$ But we must also add the result obtained in part c) because that is over this layer as well, thus we get: $18.33$ miles.